# The Matrix and Solving Systems with Matrices

This section covers:

# Introduction to the Matrix

A matrix (plural matrices) is sort of like a “box” of information where you are keeping track of things both right and left (columns), and up and down (rows).  Usually a matrix contains numbers or algebraic expressions.  You may have heard matrices called arrays, especially in computer science.

As an example, if you had three sisters, and you wanted an easy way to store their age and number of pairs of shoes, you could store this information in a matrix.  The actual matrix is inside and includes the brackets:

Matrices are called multi-dimensional since we have data being stored in different directions in a grid.  The dimensions of this matrix are  “2 x 3” or “2  by 3”, since we have 2 rows and 3 columns.  (You always go down first, and then over to get the dimensions of the matrix).

Again, matrices are great for storing numbers and variables – and also great for solving systems of equations, which we’ll see later.  Each number or variable inside the matrix is called an entry or element, and can be identified by subscripts.  For example, for the matrix above, “Brett” would be identified as , since it’s on the 2nd row and it’s the 1st entry.

Let’s look at a matrix that contains numbers and see how we can add and subtract matrices.

Let’s say you’re in avid reader, and in June, July, and August you read fiction and non-fiction books, and magazines, both in paper copies and online.  You want to keep track of how many different types of books and magazines you read, and store that information in matrices.  Here is that information, and how it would look in matrix form:

We can add matrices if the dimensions are the same; since the three matrices are all “3  by  2”, we can add them.  For example, if we wanted to know the total number of each type of book/magazine we read, we could add each of the elements to get the sum:

Thus we could see that we read 6 paper fiction, 9 online fiction, 6 paper non-fiction, 5 online non-fiction books, and 13 paper and 14 online magazines.

We could also subtract matrices this same way.

If we wanted to see how many book and magazines we would have read in August if we had doubled what we actually read, we could multiply the August matrix by the number 2.  This is called matrix scalar multiplication; a scalar is just a single number that we multiply with every entry.   Note that this is not the same as multiplying 2 matrices together (which we’ll get to next):

# Multiplying Matrices

Multiplying matrices is a little trickier.  First of all, you can only multiply matrices if the dimensions “match”; the second dimension (columns) of the first matrix has to match the first dimension (rows) of the second matrix, or you can’t multiply them.  Think of it like the inner dimensions have to match, and the resulting dimensions of the new matrix are the outer dimensions.

Here’s an example of matrices with dimensions that would work:

Notice how the “middle” or “inner” dimensions of the first matrices have to be the same (in this case, “2”), and the new matrix has the “outside” or “outer” dimensions of the first two matrices (“3 by 5”).

Now, let’s do a real-life example to see how the multiplication works.   Let’s say we want to find the final grades for 3 girls, and we know what their averages are for tests, projects, homework, and quizzes.  We also know that tests are 40% of the grade, projects 15%, homework 25%, and quizzes 20%.

Here’s the data we have:

Let’s organize the following data into two matrices, and perform matrix multiplication to find the final grades for Alexandra, Megan, and Brittney.  To do this, you have to multiply in the following way:

Just remember when you put matrices together with matrix multiplication, the columns (what you see across) on the first matrix have to correspond to the rows down on the second matrix. You should end up with entries that correspond with the entries of each row in the first matrix.

For example, with the problem above, the columns of the first matrix each had something to do with Tests, Projects, Homework, and Quizzes (grades).  The row down on the second matrix each had something to do with the same four items (weights of grades).  But then we ended up with information on the three girls (rows down on the first matrix).

So Alexandra has a 90, Megan has a 77, and Brittney has an 87.  See how cool this is?   Matrices are really useful for a lot of applications in “real life”!

Now let’s do another example; let’s multiply the following matrices:

Don’t worry; probably most of the time you’ll be doing matrix multiplication will be in the calculator!

Oh, one more thing!  Remember that multiplying matrices is not commutative (order makes a difference), but is associative (you can change grouping of matrices when you multiply them).  Multiplying matrices is also distributive (you can “push through” a matrix through parentheses), as long as the matrices have the correct dimensions to be multiplied.

# Matrices in the Graphing Calculator

The TI graphing calculator is great for matrix operations!  Here are some basic steps for storing, multiplying, adding, and subtracting matrices:

(Note that you can also enter matrices using ALPHA ZOOM and the arrow keys in the newer graphing calculators.)

We’ll learn other ways to use the calculator with matrices a little later.

# Determinants,  the Matrix Inverse, and the Identity Matrix

Soon we will be solving Systems of Equations using matrices, but we need to learn a few mechanics first!

Most square matrices (same dimension down and across) have what we call a determinant, which we’ll need to get the multiplicative inverse of that matrix.  The inverse of a matrix is what we multiply that square matrix by to get the identity matrix.  We’ll use the inverses of matrices to solve Systems of Equations; the inverses will allow us to get variables by themselves on one side (like “regular” algebra).  You’ll solve these mainly by using your calculator, but you’ll also have to learn how to get them “by hand”.

Note that the determinant of a matrix can be designated by  $$\det \left[ \text{A} \right]$$  or  $$\left| \text{A} \right|$$,  and the inverse of a matrix by  $${{\text{A}}^{{-1}}}$$.

Let’s do some examples and first get the determinant of matrices (which we can get easily on a calculator!).  The determinant is always just a scalar (number), and you’ll see it with two lines around the matrix:

Now let’s use the determinant to get the inverse of a matrix.  We’ll only work with 2 by 2 matrices, since you’ll probably be able to use the calculator for larger matrices.  Note again that only square matrices have inverses, but there are square matrices that don’t have one (when the determinant is 0):

Note that a matrix, multiplied by its inverse, if it’s defined, will always result in what we call an Identity Matrix:  .  An identity matrix has 1’s along the diagonal starting with the upper left, and 0’s everywhere else.

When you multiply a square matrix with an identity matrix, you just get that matrix back:  .  Think of an identity matrix like “1” in regular multiplication (the multiplicative identity), and the inverse matrix like a reciprocal (the multiplicative inverse).

# Solving Systems with Matrices

Why are we doing all this crazy math?  Because we can solve systems with the inverse of a matrix, since the inverse is sort of like dividing to get the variables all by themselves on one side.

To solve systems with matrices, we use .  Here is why, if you’re interested in the “theory” (the column on the right provides an example with “regular” multiplication).  (I  is the identity matrix.)

Let’s take the system of equations that we worked with earlier and show that it can be solved using matrices:

Oh, and there’s another way to solve these in your calculator, but your teacher may not tell you.  I’m not going to go into the details, but it’s using a method called reduced row echelon form, where we can put everything in one matrix (called an augmented matrix).  Let’s try this for the following matrix:

A little easier, right?

# Solving Word Problems With Matrices

Now that we know how to solve systems using matrices, we can solve them so much faster!  Let’s do a couple of pure matrices problems, and then more systems problems :).

## Matrix Multiplication Problem

Solutions:

(a)  When we multiply a matrix by a scalar (number), we just multiply all elements in the matrix by that number.  So 2P =

(b)  When we square P, we just multiply it by itself.  Let’s do this “by hand”:

(c)  Since , we have  .  Let’s use our calculator to put P in [A] and   in [B].  Then  .

## Matrix Equation Problem:

This one’s a little trickier, since it doesn’t really look like a systems problem, but you solve it the same way:

Solve the matrix equation for X (X will be a matrix):

Solution:

Let’s add the second matrix to both sides, to get X and it’s coefficient matrix alone by themselves.  Then we’ll “divide” by the matrix in front of X.  Watch the order when we multiply by the inverse (matrix multiplication is not commutative), and thank goodness for the calculator!

We can check it back:  .  It works!

## Matrix Multiplication Word Problem:

The following matrix consists of a shoe store’s inventory of flip flops, clogs, and Mary Janes in sizes small, medium, and large:

The store wants to know how much their inventory is worth for all the shoes.  How should we set up the matrix multiplication to determine this the best way?

## Solution:

The trick for these types of problems is to line up what matches (flip flops, clogs, and Mary Janes), and that will be “in the middle” when we multiply.  This way our dimension will line up.   Another way to look at it is we need to line up what goes across the first matrix with what goes down the second matrix, and we’ll end up with what goes down the first matrix for these types of problems.

So our matrix multiplication will look like this, even though our tables look a little different (I did this on a calculator):

So we’ll have $1050 worth of small shoes,$2315 worth of medium shoes, and $1255 worth of large shoes for a total of$4620.

## Another Matrix Multiplication Word Problem:

A nut distributor wants to know the nutritional content of various mixtures of almonds, cashews, and pecans.  Her supplier has provided the following nutrition information:

Her first mixture, a protein blend, consists of 6 cups of almonds, 3 cups of cashews, and 1 cup of pecans. Her second mixture, a low fat mix, consists of 3 cups of almonds, 6 cups of cashews, and 1 cup of pecans. Her third mixture, a low carb mix consists of 3 cups of almonds, 1 cup of cashews, and 6 cups of pecans. Determine the amount of protein, carbs, and fats in a 1 cup serving of each of the mixtures.

## Solution:

Sometimes we can just put the information we have into matrices to sort of see what we are going to do from there.  It makes sense to put the first group of data into a matrix with Almonds, Cashews, and Pecans as columns, and then put the second group of data into a matrix with information about Almonds, Cashews, and Pecans as rows.  This way the columns of the first matrix lines up with the rows of the second matrix, and we can perform matrix multiplication.  This way we get rid of the number of cups of Almonds, Cashews, and Pecans, which we don’t need.  So here is the information we have in table/matrix form:

Then we can multiply the matrices (we can use a graphing calculator) since we want to end up with the amount of Protein, Carbs, and Fat in each of the mixtures.  The product of the matrices consists of rows of Protein, Carbs, and Fat, and columns of the Protein, Low Fat, and Low Carb mixtures:

But we have to be careful, since these amounts are for 10 cups (add down to see we’ll get 10 cups for each mixture in the second matrix above).  Also, notice how the cups unit “canceled out” when we did the matrix multiplication (grams/cup time cups = grams).

So to get the answers, we have to divide each answer by 10 to get grams per cup.  So the numbers in bold are our answers:

## Matrix Word Problem when Tables are not Given:

Sometimes you’ll get a matrix word problem where just numbers are given; these are pretty tricky.  Here is one:

An outbreak of Chicken Pox hit the local public schools.  Approximately 15% of the male and female juniors and 25% of the male and female seniors are currently healthy, 35% of the male and female juniors and 30% of the male and female seniors are currently sick, and 50% of the male and female juniors and 45% of the male and female seniors are carriers of Chicken Pox.

There are 100 male juniors, 80 male seniors, 120 female juniors, and 100 female seniors.

Using two matrices and one matrix equation, find out how many males and how many females (don’t need to divide by class) are healthy, sick, and carriers.

Solution:

The best way to approach these types of problems is to set up a few manual calculations and see what we’re doing.  For example, to find out how many healthy males we would have, we’d set up the following equation and do the calculation:   .15(100) + .25(80) = 35.   Likewise, to find out how many females are carriers, we can calculate:  .50(120) + .45(100) = 105.

We can tell that this looks like matrix multiplication.  And since we want to end up with a matrix that has males and females by healthy, sick and carriers, we know it will be either a 2 x 3 or a 3 x 2.  But since we know that we have both juniors and seniors with males and females, the first matrix will probably be a 2 x 2.  That means, in order to do matrix multiplication, the second matrix that holds the %’s of students will have to be a 2 x 3, since there are 3 types of students, healthy (H), sick (S), and carriers (C).  Notice how the percentages in the rows in the second matrix add up to 100%.  Also notice that if we add up the number of students in the first matrix and the last matrix, we come up with 400.

So we can come up with the following matrix multiplication:

So there will be 35 healthy males, 59 sick males, and 86 carrier males, 43 healthy females, 72 sick females, and 95 carrier females.  Pretty clever!

## Matrix Multiplication when Diagonals are Answers:

The first table below show the points awarded by judges at a state fair for a crafts contest for Brielle, Brynn, and Briana.   The second table shows the multiplier used for the degree of difficulty for each of the pieces the girls created.  Find the total score for each of the girls in this contest.

Solution:

This one’s a little trickier since it looks like we have two 3 x 2 matrices (tables), but we only want to end up with three answers: the total score for each of the girls.

If we were to do the matrix multiplication using the two tables above, we would get:

Hmm….this is interesting; we end up with a matrix with the girls’s names as both rows and columns.  It turns out that we have extraneous information in this matrix; we only need the information where the girls’ names line up.  So we only look at the diagonal of the matrix to get our answers:  Brielle had 86.8 points, Brynn 79.2 points, and Briana 110 points.

What we really should have done with this problem is to use matrix multiplication separately for each girl; for example, for Brielle, we should have multiplied  and so on.  Oh well, no harm done; and now you’ll know what to do if you see these types of matrices problems.

# Using Matrices to Solve Systems

Solve these word problems with a system of equations.  Write the system, the matrix equations, and solve:

## Finding the Numbers Word Problem:

The sum of three numbers is 26.  The third number is twice the second, and is also 1 less than 3 times the first.  What are the three numbers?

Solution:

Let’s translate word-for-word from English to Math that we learned in the Algebra Word Problem Section here.

Let x = the first number, y = the second number, and z = the third number.  So here are the three equations:

Note that, in the last equation, “one less than” means put the –1 at the end (do this with real numbers to see why).

We need to turn these equations into a matrix form that looks like this:

So we need to move things around so that all the variables (with coefficients in front of them) are on the left, and the numbers are on the right.  (It doesn’t matter which side; just watch for negatives).  If we just have the variable in the equation, we put a 1 in the matrix; if we don’t have a variable or a constant (number), we put a 0 in the matrix.  So we get:   and in matrix form:

Putting the matrices in the calculator, and using the methods from above, we get:

So the numbers are 5, 7, and 14.  Much easier than figuring it out by hand!

## A Florist Must Make 5 Identical Bridesmaid Bouquets Systems Problem

Here’s a problem from the Systems of Linear Equations and Word Problems Section; we can see how much easier it is to solve with a matrix.

A florist is making 5 identical bridesmaid bouquets for a wedding.  She has $610 to spend (including tax) and wants 24 flowers for each bouquet. Roses cost$6 each, tulips cost $4 each, and lilies cost$3 each.  She wants to have twice as many roses as the other 2 flowers combined in each bouquet.  How many roses, tulips, and lilies are in each bouquet?

Solution:

Let’s look at the question that is being asked and define our variables:  Let r  = the number of roses, t  = the number of tulips, and l  = the number of lilies.  So let’s put the money terms together, and also the counting terms together:

Now let’s put the system in matrices (let’s just use one matrix!) and on the calculator:

So for all the bouquets, we’ll have 80 roses, 10 tulips, and 30 lilies.

So for one bouquet, we’ll have  of the flowers, so we’ll have 16 roses, 2 tulips, and 6 lilies.

## An Input Output Problem

Input-output problems are seen in Economics, where we might have industries that produce for consumers, but also consume for themselves.  An application of matrices is used in this input-output analysis, which was first proposed by Wassily Leontief; in fact he won the Nobel Prize in economics in 1973 for this work.

We can express the amounts (proportions) the industries consume in matrices, such as in the following problem:

The following coefficient matrix, or input-output matrix, shows the values of energy and manufacturing consumed internally needed to produce $1 of energy and manufacturing, respectively. In other words, of the value of energy produced (x for energy, y for manufacturing), 40 percent of it, or .40x pays to produce internal energy, and 25 percent of it, or .25x pays for internal manufacturing. Of the value of the manufacturing produced, .25y pays for its internal energy and .10y pays for manufacturing consumed internally. The inputs are the amount used in production, and the outputs are the amounts produced. (a) If the capacity of energy production is$15 million and the capacity of manufacturing production is $20 million, how much of each is consumed internally for capacity production? (b) How much energy and manufacturing must be produced to have$8 million worth of energy and $5 million worth of manufacturing available for consumer use? Solution: (a) If production capacities are$15 million for energy and $20 million for manufacturing, the amount consumed internally is . So$11 million of energy is consumed internally and $5.75 million of manufacturing is consumed internally. This makes sense, for example, since we’re multiplying the proportion of energy consumed internally (.4) by the production capacity for energy ($15 million) and adding that to the proportion of energy needed for internal manufacturing (.25) by the production capacity of manufacturing ($20 million) to get the total dollar amount of energy needed or consumed internally ($11 million).  Then we do the same for manufacturing.

(b) The amount of energy and manufacturing to be produced to have $8 million worth of energy and$5 million worth of manufacturing available for consumer (non-internal) use is solved using the following equation (we want what’s “left over” after the internal consumption, so it makes sense): .   To get , we can use the formula .  So the two industries must produce $17.7 million worth of energy and$10.5 million worth of manufacturing, respectively.

# Cramer’s Rule

Sometimes you’ll have to learn Cramer’s Rule, which is another way to solve systems with matrices.  Cramer’s Rule was named after the Swiss mathematician Gabriel Cramer, who also did a lot of other neat stuff with math.

Cramer’s rule is all about getting determinants of the square matrices that are used to solve systems.  It’s really not too difficult; it can just be a lot of work, so again, I’ll take the liberty of using the calculator to do most of the work 🙂

Let’s just show an example; let’s solve the following system using Cramer’s rule:

To solve for x, y, and z, we need to get the determinants of four matrices, the first one being the 3 by 3 matrix that holds the coefficients of x, y, and z.  Let’s call this first determinant D;

Now we’ll get a matrix called , which is obtained by “throwing away” the first (x) column, and replacing the numbers with the “answer” or constant column.  So

You can probably guess what the next determinant we need is: , which we get by “throwing away” the second column (y) of the original matrix and replacing the numbers with the constant column like we did earlier for the x.  So   Similarly,

OK, now for the fun and easy part!   To get the x, y, and z answers to the system, you simply divide the determinants   So  Now we know that x = 5, y = 1, and z = –2.

Note that, like the other systems, we can do this for any system where we have the same numbers of equations as unknowns.

# Number of Solutions when Solving Systems with Matrices

Most systems problems that you’ll deal with will just have one solution.  (These equations are called independent or consistent).  But, like we learned in the Systems of Linear Equations and Word Problems Section here, sometimes we have systems where we either have no solutions or an infinite number of solutions.

Without going too much into Geometry, let’s look at what it looks like when three systems (each system looks like a “plane” or a piece of paper) have an infinite number of solutions, no solutions, and one solution, respectively:

Systems that have an infinite number of solutions (called dependent or coincident) will have two equations that are basically the same.  One row of the coefficient matrix (and the corresponding constant matrix) is a multiple of another row.  Then it’s like you’re trying to solve a system with only two equations, but three unknowns.

A system that has an infinite number of solutions may look like this:

Systems with no solutions (called inconsistent) will have one row of the coefficient matrix a multiple of another, but the coefficient matrix will not have this.  So a system that has no solutions may look like this:

When you try to these types of systems in your calculator (using matrices), you’ll get an error since the determinant of the coefficient matrix will be 0.  This is called a singular matrix and the calculator will tell you so:

Also, if you put these systems in a 3 by 4 matrix and use RREF, you’ll be able to see what is happening.  For the systems with infinite solutions, you can see you won’t get an identity matrix, and that 0 always equals 0.   You can actually define the set of solutions by just allowing z to be anything, and then, from the other rows, solve for x and y in terms of z:   This would look like   so the solution set for {x, y, z} is {5 – .375z, 3 + .875z, z}.  (This may be a little advanced for high school 🙂 )

For the system with no solutions, you’ll get this, where you can see that you still don’t have an identity matrix, and 0 can never equal 1 from the last row:

Learn these rules, and practice, practice, practice!

Use the MathType keyboard to enter a Limit problem, and then click on Submit (the arrow to the right of the problem) to solve the problem. You can also click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software.  You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to Introduction to Linear Programming  – you are ready!

## 184 thoughts on “The Matrix and Solving Systems with Matrices”

• Tony,
Thank you so much for pointing out the error in my Matrices section; I’ve since corrected it.

Do you want to do more proofreading? 🙂
Lisa

1. I have been having so much trouble setting this problem up
Determine the solution by setting up and solving the matrix equation.
A nut distributor wants to determine the nutritional content of various mixtures of pecans, cashews, and almonds. Her supplier has provided the following nutrition information:
Almonds. Cashews. Pecans
Protein. 26.2g/cup. 21.0g/cup. 10.1g/cup
Carbs. 40.2g/cup. 44.8g/cup. 14.3g/cup
Fat. 71.9g/cup. 63.5g/cup. 82.8g/cup

Her first mixture, protein blend, contains 6 cups of almonds, 3 cups of cashews, and 1 cup of pecans. Her second mixture, low fat mix, contains 3 cups almonds, 6 cups cashews, and 1 cup of pecans. Her third mixture, low carb mix contains 3 cups almonds, 1 cup cashews, and 6 cups pecans. Determine the amount of protein, carbs, and fats in a 1 cup serving of each of the mixtures.

I have solved this by multiplying both of the matrices then dividing each element by 10 but that’s not the way I am supposed to solve this as there are no equations being set up.

• As the school psychologist, you are consulting with the staff nutritionist about one of your student’s low energy and propensity for minor illness. the nutritionist has suggested a daily routine of 350 units of vitamin C, 4200 units of Vitamin A, and 400 units of Vitamin D. She also recommends the following sources of these vitamins: Megadose Supra contains 50 units of C, 1000 units of A, and 100 units of D per capsule. Crazy Caps contains 100 units of C, 200 units of A, and 100 units of D. Formula I contains 50 units of C, 500 units of A and no Vitamin D per capsule. How many of each capsule should the student take to get the recommended dosages?

Can you solve this using determinants? or just write it in matrix form. Thank you!

2. This is great, and also can you show when matrices are used in repetition, like occurrence over a few days? I love this website, I will have to bookmark it \(-^^-)/

• Thanks so much for writing!! Good point – I just added this in the Matrix Multiplication section:
Remember that multiplying matrices is not commutative (order makes a difference), but is associative (you can change grouping of matrices when you multiply them). Multiplying matrices is also distributive (you can “push through” a matrix through parentheses), as long as the matrices have the correct dimension to be multiplied.
So if you have matrices in repetition, you can multiply the first 2, then the 3rd, or the second 2 first. You cannot change the order of multiplying though. Does this make sense? Lisa

3. Can u help me solve this problem??

Mr. Tan want to support his wife on her dietary plan. He decided to investigate the carbohydrate content of the vegetable they plant. Together , i cup of raw cabbage, 5 raw cauliflowers, and 1 cup of raw tomatoes contain 12 gram of carbohydrates. One cup of raw cabbage and 6 cauliflowers have one-half the carbohydrates of 1 cup of raw tomatoes. One cup each of raw cabbage and raw tomatoes has 3 times the carbohydrate content of 6 raw cauliflowers.Find the number of grams of carbohydrates in the given portion size of each vegetable

• Thanks for the problem! I set it up with the following equations: r + 6c + t = 12 (I think for it to work, you meant 6 raw cauliflowers in the 3rd sentence?)
r + 6c = .5t
r + t = 3(6)c

Then I put the following matrix in my calculator and used RREF: [1 6 1 12
1 6 -.5 0 to get 1 cup of raw cabbage, 1/2 of a cauliflower, and 8 cups of tomatoes.
1 -18 1 0]
Does this make sense?
Lisa

• thank you so much for helping me out..i’ve really lost it try to do this problem..thank you again..:)!!..hehe

4. B=2by2 matriz a=1 b=1 c=2 d=1
The question asks me to find A in the equation B(I+A^-1)=I
A^-1 is the inverse of A and the I is the identity matrix.
Can you help me out

• Thanks for writing! I just “divided” by B (using inverse matrix) and then subtracted the identity matrix and got A^-1 = -2 1 2 -2. Then took the inverse of that and got A = -1, -1/2, -1, -1. (a b d). Does that make sense? Lisa

5. A chemical manufacturer wants to lease a fleet of 24 railroad tank cars with a combined carrying capacity of 420.000 gallons.. Tank cars with three different carrying capacities are available : 7.000 gallons, 14.000 gallons, and 28.000 gallons. How many each type of tank car should be leased ?.
use gaussian elimination method to solve it.
please any help me out to solve me this problem

Electricity consumed internally
Steel consumed internally

Matrix A= .25 .20
.50 .20
If the production capacity of electricity is $15 million and the production capacity for steel is$20 million how much of each is consumed internally for capacity production?

How much electricity and steel must be produced to have $5 million worth of electricity and$8 million worth of steel available for consumer use.

Thank you

• Hi and thanks for writing! I had to look up how to do this, but here’s what I got:
You may have also found it in the Finite Mathematics book, by Howard Rolf. Hope that helps!
Lisa

A feeding station in a field study provides for three types of deer: young deer, adult females and adult males. There are 12 deer in all. In total, the deer consume 18.5 units of food and 14.5 units water each day. Suppose each young deer consumes 1 unit of food and 0.5 unit of water each day. The corresponding values for an adult female are 1.5 and 1, and for an adult male are 2 and 2. The ecologist needs to figure out how many young deer, adult females and adult males are in the herd.

Thank you!

• Good problem! Let y = young deer, f = adult females, and m = adult males. Then you have:
Total Deer: y + f + m = 12
Food: 1y + 1.5f + 2m = 18.5
Water:.5y + 1f + 2m = 14.5

Put in Matrix form and solve to get 3 young deer, 5 female deer, and 4 male deer.

8. Could you use the inverse of the matrix to find that answer?
I do not know whats the purpose of using the inverse?
Thank you

• Great Question! If you have the same number of variables as equations, you can get the answer to the system of equations by taking the inverse of the coefficient matrix [A] (the matrix with the numbers before the x, y, and z, for example) and multiplying by the constant matrix [B] (the matrix with the numbers on the right hand side of the equal signs). So the answer would be [A]^-1 * [B], where [A]^-1 is the inverse of [A]. I’m not sure we can do that with the candy problem below since we have more variables than equations. You can look here to see how I explain it (and up a little further to see about inverses).

9. matrices used to give me worst nightmares, it was like a ghost always giving me hard times. The day I discovered this website wow, I felt the heavy burden on my mindset disappearing, recently, I’m in Love with matrices and I’m good.

10. one more question ask u frd ………………………i have one question
Q1. let i m taking three vehicle like car,jeep , motorcycle they are all move to 10 km .how much they will take time at each different path. i m mention below
i takes three different s path like ……. rough . smooth , zing-zag

name of vehicle smooth rough Zing-zag
motorcycle 5 mint 10mt 7mt
Car 6mt 7mt 9mt
jeep 7mt 7mt 8mt

type of weight

Smooth 20%
rough 50%
Zing- Zag 30%

which vehicle best ride for 10km …….i just think and type .if u want any thing change pzl do give me ans ……………if u have more example like cricket player , cloth , pen, room pict, home fan , etc any thing we see in normal . can we apply matrix .if we apply give me few example ……………….my email is krrish_vishist@yahoo.co.in
i m waiting ur reply i hop u will back me reply soon ………..

• Thanks for writing. I’m not sure I totally understand the problem, but I used matrices to multiply the type of vehicle by time on each type of road matrix (3×3) by a matrix of the weights of each type of road (3×1) and got a 3×1 matrix of the times for each type of vehicle that is [8.1 7.4 7.3]. So the jeep would have the shortest time? Does this make sense?

11. I am making presentation on real life problems; Which can be solved by matrices.
I want 15 examples.
Hoping a response as early as possible.

• Thanks for writing! There are many examples: Cost analyses, Surveys, Airline Schedules – anything where you have two different types of data. You may want to google “real life examples of matrices”. Lisa

12. Hi, lisa

I am working on Linear algebra project
& I want 10 real life example of matrices
Can you pls help me…

Pls make it fast

13. Pls can you help me solve this problem,
The sum of 1200usd is invested in three projects x,y,z at the ratio of 4:5:6 respectively,the expected cash inflow per annum is 700usd,the combined income for projects x and y is 140usd more than the income from project z.
calculate the amount invested in project z?
what is the cash inflow for project z?
what is the combined cash inflow for projects x and y?
calculate the amount invested in project y?

Pls don’t write the solution in the calculator form because i don’t have it,thanks.

• Thanks for writing; I will try. Since the income is at a ratio of 4:5:6, we have 4x + 5x + 6x = 1200. Solve for x and get 80. So the income for x is 4×80 = $320, the income for y is 5×80=$400, and the income for z is 6×80=$480. It looks like the cash inflow for the properties are all in the same ratio (700/1200), so we have the cash inflow for z = (700/1200)x480 =$280. Since the combined cash inflow for x and y is $140 more than z, their combined inflow is$420. I think that answers all the questions. Lisa

14. thanks for quick response,help me with these also
1)A salesman has below record for products sold during peak season of october to december.The products are in three variants A B C.How many units of product C only is he expected to sell to earn 900 commission?
2)The buying price of a basket of oranges is $1000 and selling price is$5 per orange.What is the profit per basket if 300 oranges are found in the basket?
What is the break even point(quantity),if the buying price of a basket remains $1000 and the selling price is$5?
What is the profit per basket if 250 oranges are found in the basket?

• For number 1), there is not enough information – are you missing a table? For 2), we have Profit = Revenues – Cost, so P = 5(300) – 1000 = $500. For the break even quantity, 0 = 5x – 1000; x =$200. For the profit if there are 250 oranges, 5(250) – 1000 = $250. Hope that makes sense. Lisa 15. Hello Lisa, do you think you could help me with this question without a calculator? A dietitian wishes to plan a meal around three foods. The meal is to include 12120 units of vitamin B, 26440 units of vitamin C, and 17260 units of Vitamin E. The number of units of the vitamins in each unit of the foods is summarised: Food I ￼ Food II Food III Vitamin B 120 300 420 Vitamin C 240 700 940 Vitamin E 160 450 610 Determine the amount of each food the dietitian should include in the meal in order to exactly meet the vitamin requirements. I tried using your section on solving systems with matrices. However, some of the values are very big and of negative values. I’d really appreciate it if you could help me! Thank you! 16. Hi Lisa, Question about solving for an unknown matrix… When multiplying both sides of an equation by the matrix inverse, like in your problem “Find [Q] when [P]*[Q] = solution” If ‘solution’ had also been a (2×2) matrix, what would be the order of the factors? [Q] = [inverse P]*[2×2 solution] or [Q] = [2×2 solution]*[inverse P] I cannot use a graphing calculator for my tests, so I really appreciate seeing all of the methods on your fabulous site! Thanks • Thanks for writing and thanks so much for using She Loves Math! The first way you described it is correct: [Q] = [inverse P]*[2×2 solution] . Try that and see if you get the correct answers! (You can check by multiplying the matrices back). Hope that helps, Lisa 17. hii need help with this question A hotel rents double rooms at$320 per day and single rooms at $260 per day. If 46 rooms were rented one day for a total of$13,760.

1. Display the information given in a system format. [1 mark]
2. Write the system in matrix from Ax = b. [1 mark]
3. Find the determinant of matrix A in (2) above. [1 mark]
4. Find the inverse of matrix A in (2) above.

• Thanks for writing! Here’s what I got – hope it helps 🙂 Lisa

1) d + s = 46
320d + 260s = 13760
2) | 1 1 | x |d| = |46|
|320 260| |s| |13760|
3) Determinant is 1*260 – 1*320 = -60
4) Inverse matrix is (1/-60) |260 -1| = |-13/3 1/60|
|-320 1| |16/3 -1/60|

• x+y=46 640x+260y=13760 2)/a/=-380

• For the first systems of equations, I got x = 4.737 and y = 41.263. For 2, I get no solution, since an absolute value can’t be negative. Hope that makes sense! Lisa

18. Hi this might not have anything to do with the topic but can you help?

Equations
a. If the price for a ticket was $200 TT per ticket, 6000 persons were willing and able to attend the Mother’s Day Ball but only 1000 persons were if the price increased to$300 TT. Derive the demand curve, for tickets of the Mother’s Day Ball. [1 mark]
b. If the price for a ticket was $200 TT per ticket, the promoter was willing and able to offer 4000 Mother’s Day Ball tickets for sale and 8000 ticket if the price was$600 TT. Derive the supply curve, for tickets to the Mother’s Day Ball. [1 mark]
c. Hence, determine whether there will be a surplus or shortage or neither for Mother’s Day Ball tickets if the price of a ticket was $240. [2 marks] Functions In Fantasy Island an individual pays 10% tax on gross annual income not exceeding$200,000. For every dollar on the next $300,000 the tax is 15%. Any income in excess of this is taxed at 20 cents in the dollar. Using T to represent annual tax liability and x gross annual income: a. Determine the annual tax liability function T(x) for an income earner in Fantasy Island. [2 marks] b. What is the tax liability of a person who earns a gross annual income of$350,000? [1 mark]
c. What is the tax liability of a person who earns a gross annual income of $500,000? [1 mark] Thanks in advance • Hello – Thanks for writing! I’m not sure how to work these demand curve/supply/tax liability problems, but I’m investigating and will try to answer soon! Lisa 19. I’m having a really hard time answering this problem and it would be great if you could help me with it: Lawn Co. produces three grades of commercial fertilizers. A 100-lb bag of grade A fertilizer contains 18 lb of nitrogen, 4 lb of phosphate, and 5 lb of potassium. A 100-lb bag grade B fertilizer contains 20 lb of nitrogen and 4 lb each of phosphate and potassium. A 100-lb bag of grade C fertilizer contains 24 lb of nitrogen, 3 lb of phosphate, and 6 lb of potassium. How many 100-lb bags of each of the three grades of fertilizers should Lawn Co. produce if 24,000 lb of nitrogen, 4900 lb of phosphate, and 6200 lb of potassium are available and all the nutrients are used? P.S. Your site really helped me a lot in my mathematics subjects. But I just gave up on these one. Thank you in advance!! ^_^ • Thanks for writing! It’s usually a good idea to look at what they asking to figure out the variables, so I let x = number of 100 lb bags of Grade A fertilizer, y = number of 100 lb bags of Grade B fertilizer, and z = number of 100 lb bags of Grade C fertilizer. So I got 18x + 20y + 24z = 240, 4x + 4y + 3z = 49, and 5x + 4y + 6z = 62. (Divide each of total amts of fertilizer by 100 since we’re dealing with 100 lb bags). Solving by matrices, I get x = 8.8, y = 2.4, and z = 1.4. But usually in these problems, you don’t get decimal answers for bags of things. So I think maybe the problem should have stated 264000 of nitrogen? Then the answers turn out better: x = 4, y = 6, and z = 3. Hope that helps! Lisa 20. System of Linear Equation Real Estate Cantwell Associates, a real estate developer, is planning to build a new apartment complex consisting of one bedroom units and two- and three-bedroom townhouse. A total of 192 units is planned, and the number of family units (two- and three-bedroom townhouse) will equal the number of one-bedroom units. If the number of one-bedroom units will be 3 times the number of three-bedroom units. Find how many units of each type will be in the complex. • Let’s first look at what they are asking: how many units of each type. So let x = one bedroom units, y = two bedroom, and z – three bedroom. Then we have x + y + z = 192, x = y + z, and x = 3z. We can either use substitution or Matrix Multiplication to solve to get x = 96, y = 64, and z = 32. Hope that helps! 21. please help me out with this matrix question! Protein, carbohydrates, and fats can be obtained from three foods. Each ounce of Food I contains 5 grams of protein, 10 grams of carbohydrates, and 40 grams of fat. Each ounce of Food II contains 10 grams of protein, 5 grams of carbohydrates, and 30 grams of fat Each ounce of Food III contains 15 grams of protein, 15 grams of carbohydrates, and 80 grams of fat. a. How many ounces of each of the three foods are required to yield 300 grams of protein, 300 grams of carbohydrates, and 1500 grams of fat? b. Find the per unit cost of production of each product if the per unit cost of protein, carbohydrates, and fats are 100, 100, and 1000 respectively? • Here’s how I set it up: 5x + 10y + 15z = 300, 10x + 5y + 15z = 300, 40x + 30y + 80z = 1500. I then used matrices to get x = 10, y = 10, and z = 10. For b) I created a matrix of the matrix of the the total amount in each product for each of the protein, carbs, and fats that looks like [50 100 400 (next row) 100 50 300 (next row) 150 150 800] and multiplied this by a matrix (that goes down) [100 100 1000] to get [415000 315000 830000]. Not sure if this is right, though! Hope it helps, Lisa 22. Please help me out with this one as well: Ms. Rose a business woman invested different amounts at 8%,8.75%, and 9%, all at simple interest. Altogether, Ms. Rose invested$40,000 and earns $3,455 per year. In addition, she has$4,000 more invested at 9% and 8%.
a. Using the information above, set out the system of linear equations, which should be solved to obtain the amount that will be invested at respective interest rates.
b. By matrix notation, find the amount that Ms. Rose needs to invest at each interest rate.

• For this one, I got equations .08x + .0875y + .09z = 3455, x + y + z = 40000, and x + 4000 = z. Using matrices, I got x = $11000, y =$14000, and z = $15000. Lisa 23. Please help me out with these questions…….please attach some images to help 1. A manufacturer produces two types of products X and Y. Each product is first processed in a machine M1 and sent to another machine M2 for finishing. Each unit of X requires 20 minutes time in M1 and 10 minutes time in M2 while the corresponding time for Y are 10 minutes in M1 and 20 minutes in M2. The total time available on each machine is 600 minutes. Calculate the number of units of X and Y produced by constructing a matrix equation of the form AX=B and then solve by matrix inversion method. 2. An amount of$5,000 is put into three investments at the rates of 6%, 7%, and 8% per year respectively. The total annual income is $358. If the combined income from the first two investments is$70 more than the income from the third, find the amount of each investment by using matrix methods.

• Here’s how I set up the problems, and the answers I got. Hope this helps 🙂 lisa
x + y + z = 5000
.06x + .07y + .08z = 368
.06x + .07y – .08z = 70

Solving the matrix:
1 1 1 | 5000
0.06 0.07 0.08 | 358
0.06 0.07 -0.08 | 70

you get: (in $) x = 1000$
y = 2200 $z = 1800$
to the total investment of 5000

• er, Lisa…
problem part 2 above, states:
“….the combined _income_ (=$) from the first two investments is$70 more than the _income_ (=$) from the third… find the _amount_ (=$), of each investment x, y, z .”

Shouldn’t your 2nd eq. (above), be:
.06x +.07y = .08z + 70 ($value eq) ? so that, .06x + .07y – .08z = 70 Solving the matrix: 1 1 1 | 5000 0.06 0.07 0.08 | 358 0.06 0.07 -0.08 | 70 you get: (in$)
x = 1000 $y = 2200$
z = 1800 $which all adds up to the total investment of 5000$ (=check OK!).

After all. you are the Teacher
and the Ghana gentleman and myself

1. In a two-industry economy,it is known that industry I uses 10 units of its own product and 60 units of another commodity to produce a dollar worth of commodity A; industry II uses 50 units of its own product but uses none of any other commodity in producing a dollar worth of commodity B. The market demands $1000 of commodity A and$ 2000 of commodity B.
a. State the input matrix equations for the economy
b. Find the input levels by using cramer’s rule

2. The output levels of machinery, electricity, and oil of a small country are 3000, 5000, and 2000 respectively. Each unit of machinery requires input of 0.3 units of electricity, and 0.3 units of oil. Each unit of electricity requires input of 0.1 unit of machinery and 0.2 units of oil. Each unit of oil requires input of 0.2 units of machinery and o.1 units of electricity. Find the total amount exported by each industry?

• Thanks for writing. This is a little beyond what I’m used to, but here is what I got. I’m not sure if they are correct though, so please check them. Thanks, Lisa

25. I was working on a expression involving trigonometry so i discovered that writing it in matrix shows the result of multiplying the former expression by -1. what must have happened.

• Thanks for writing! I’m not sure exactly what you mean; could you explain your problem more? Thanks, Lisa

26. Please help me in placing this question in a system of equations so that i can write it in matrice form.

A security firm borrowed $900,000 for business expansion. Some of the money was borrowed at 5%, some at 6% and some at 7% simple annual interest. The total annual interest is$52,000 and the amount borrowed at 6% is one and a half of that borrowed at 7%.

• Thanks for writing; here’s what I got: x + y + z = 900000, .05x + 06y + .07z = 52000, y = 1.5z. Putting these in a matrix (3 by 4) [1 1 1 900000 .05 .06 .07 52000 0 1 -1.5 0] and solving, I get x = 400000, y = 300000, and z = 200000. Hope that helps! Lisa

• Incredible! I worked it out and got the same answer! U da best lisa! i find it hard to make the equations. Here is another one please; INEQUALITIES

A company makes two types of space savers, English and Italian. Each English type requires 3 hours, 2 hours and 2 hours, while each Italian requires 2 hours, 1 hour and 4 hours for cutting, assembling and finishing respectively.
The company has 300 hours available for cutting, 240 hours for assembling and 120 hours finishing. Each English type earns $85 in profit while the Italian type B earns$75.
a. If you are required to find the combinations of both types of space savers that will maximize profit, list the inequalities that would be necessary.

• Here’s what I got for your second problem: x = English, y = Italian. Maximize 85x + 75y such that: 3x + 2y < = 300, 2x + y <= 240, and 2x + 4y <= 120. Also, x >= 0, and y>= 0. Did you need to solve this too? Lisa

• yes i do.
b. From a graph of the inequalities in part a. list the vertices of the solution sets.

c. Which of the combinations of the two types of space savers will maximize profits?

d. How much will be the maximum profit?

• This is what I got: When I graphed the inequalities, I got the last graph under all the other ones, so the vertices of the solution sets I got are (0, 0), (0, 30), and (60, 0). When plugging in these points into 85x + 75y, I got the (60, 0) would maximize the profits at a profit of $5100. Does this make sense? Lisa • Hi, do you have the working for the questions that David posted below? I didn’t seem to quite understand how you got the answers for it. regards and thanks a million! • Hello i was working out the question David asked and I got out that y = 3000000 but I cant get out x and z. Can you please help. 27. PLEASE HELP ME WITH THIS Fion invested$42000 in three different accounts; savings, time deposit and bond. The interests were 5%, 7% and 9% respectively. His total annual interest was $2600 and the interest from savings account was$200 less than the total interest from the other two investments. How much did he invest at each rate? Use matrix to solve this.
THANK YOU

• I got s + t + b = 42000, .05s + .07t + .09b = 2600, and s = (t+b) – 200. Putting them in matrix form, I got [1 1 1, .05 .07 .09, 1 -1 -1] (this is a 3 by 3) times [s t b] (this is a 3 by 1) = [42000 2600 -200] (this is a 3 by 1). The answers I got are s = 20900, t = 17200, and b = 3900. Does this make sense? Lisa

• lisa I realy appreciate u for supporting many others and also my self . pls if u can make it clarify ur solution method is not clear how do u get z answer

• Thanks for writing! Could you please tell me which problem you are referring to? Thanks, Lisa

28. I need help?A security firm borrowed $900 000 for a business expansion. Some of the money was borrowed at 5% some at 6% and some at 7% simple annual interest. The total annual interest is$52 000 and the amount borrowed at 6% is one and a half of that borrowed at 7%.
A Create a system of equations which captures the essence of the information given
b. write the system in matrix form
c. Using the inverse method find the amount borrowed at each rate
I’m confused i’m used to getting the matrix straight forward so now that its worded its proven difficult. can u help with at least the first question

• Thanks for writing! Here’s what I got: x = amt at .05, y = amt at .06, and z = amt at .07. .05x + .06y + .07z = 52000 (this is the interest). Then we also have x + y + z = 900000, and y = 1.5z. You can put these in a matrix by putting all the variables on one side, and the numbers on the other side. Does that make sense? Lisa

29. Hi Lisa the question that David asked about to do with the security firm can you work it out completely. I don’t understand matrix to good.

Using the Cramer Rule how would you find the amount borrowed at each rate?

Thanks

30. hi..please, can u help me to solve this question? tqvm

Let A = {a, b, c, d, e} and S, T, U, and V relations on A where
S = {(a, a), (a, b), (b, c), (b, d), (c, e), (e, d), (c, a)}
T = {(a, a), (b, a), (b, c), (b, d), (e, e), (d, e), (c, b)}
U = {(a, b), (a, a), (b, c), (b, b), (e, e), (b, a), (c, b), (c, c), (d, d), (a, c), (c, a)}
V = {(a, b), (b, c), (b, b), (e, e), (b, a), (c, b), (d, d), (a, c), (c, a)}

Find the represenation matrices for S, T, U, and V. Then uses these matrices to determine
which of the relations are symmetric, reflexive, transitive, or/and antisymmetric.

31. hi frd…….. i have one problum ………………i need to solve matrix like order (3*3*3) and to how to find transpose and inverse of that type of order of matrix…………..plz give me reply fast its so urgent every one know about this type of matrix plz give me reply if any body have martial about this topic .plz send me at my email id( krrish_vishist@yahoo.co.in ) . i am waiting ……………

Mr. Tan want to support his wife on her dietary plan. He decided to investigate the carbohydrate content of the vegetable they plant. Together , i cup of raw cabbage, 5 raw cauliflowers, and 1 cup of raw tomatoes contain 12 gram of carbohydrates. One cup of raw cabbage and 6 cauliflowers have one-half the carbohydrates of 1 cup of raw tomatoes. One cup each of raw cabbage and raw tomatoes has 3 times the carbohydrate content of 6 raw cauliflowers.Find the number of grams of carbohydrates in the given portion size of each vegetable

• I found this same problem on my blog; here is what I got: Thanks for the problem! I set it up with the following equations: r + 6c + t = 12 (I think for it to work, you meant 6 raw cauliflowers in the 3rd sentence?)
r + 6c = .5t
r + t = 3(6)c

Then I put the following matrix in my calculator and used RREF: [1 6 1 12
1 6 -.5 0 to get 1 cup of raw cabbage, 1/2 of a cauliflower, and 8 cups of tomatoes.
1 -18 1 0]
Does this make sense?
Lisa

33. Which related equation can be used to solve the equation?
Column A Column B
1.
m + 5 = –7.3
2.
m + 5 = 7.3
3.
m + 7.3 = –5
4.
m + 7.3 = 5
A.
m = –7.3 – 5
B.
m = 5 – 7.3
C.
m = 7.3 – 5

It confuses me on how to do this can someone help me

• Thanks for writing, but I think I’m missing part of the problem? I’m not sure what the problem is asking. Lisa

34. Can you help me to solve this problem ? i asked many people but they just gave different answer. i do not know with one is the solution

Mrs. Tan have an obesity problem and her personal doctor recommends her to taking care of her dietary habits at home by first controlling the lunch menus. She must plan a lunch menu that provides 350 calories, 50 g carbohydrates, and 40 mg of calcium. A 3-oz serving of roasted meat contains 225 calories, 0 g oh carbohydrates, and 12 mg of calcium. One bowl of salad contains 150 calories, 35 g of carbohydrates and 9 mg of calcium. A one cup serving of orange contains 55 calories, 12 g of carbohydrates and 28 mg of calcium. How many servings of each required to provide the desired of nutrition values that suitable for Mrs. Tan?

• Here’s how I set this up: Let x = serving of 3 oz meat, y = salad bowl, and z = cup orange. The equations I got are: 225x + 150y + 55z = 350, 0x + 35y + 12z = 50, 12x + 9y + 28z = 40. When I solved though, I got weird numbers: x = .59, y = 1.15, and z = .8. Hope this help! Lisa

35. Hi Lisa, thanks for such a helpful page. I’ve been doing practice problems and I stumbled upon this one. Do you have any advice on where to start?

An outbreak of chicken pox hits Fairfax County High Schools. Approximately 15% of the male and female juniors and 25% of the male and female seniors are healthy, 35% of the male and female juniors and 30% of the male and female seniors are currently sick, and 50% of the male and female juniors and 45% of the male and female seniors are carriers of the disease. There are 103 males juniors, 79 male seniors, 108 female juniors, and 105 female seniors. Using two matrices of values and ONE matrix equation, determine how many males and how many females (no need to divide by class) are healthy, sick, and carriers.

Thanks for the help!
-Jessie

36. Hi,

Here it goes….

Cash is 13% of Marketable Securities
Total Current Assets (Cash + Marketable Securities +Inventory + Receivables) = $1,750,000 LT Assets = 0.40xMarketable Securities + 2.24xInventory + 2.15xReceivables Inventory+1.5xReceivables = 0.54xLong Term Assets Cash +1.2xInventory+0.56xReceivables=$1,100,000

Cash = c
Total current assets = t
Marketable securities = m
Inventory = i
Receivables = r

This is how I set up the problem with the respective variables.
c = 0.13m
t (c + m + i + r) = 1,750,000
l = 0.40m + 2.24i + 2.15r
i + 1.5r = 0.54l
c + 1.2i + 0.56r = 1,100,000

Can anyone tell me if I’m on the right track?

Thank you!

Lily

• Lily,
Thanks for writing. You are on the right track, but the second equation is just c + m + i + r = 1750000. Since you have 5 equations with 5 unknowns use lt assets (l) instead of total current assets as a variable), you should be able to solve. Here’s what I get when I put it all in a matrix and solve (although I had to round the numbers:) c = 61889.73, l = 2846842.57, m = 476074.85, i = 561516.30, and r = 650519.12. Does that make sense? Lisa

37. Hi Lisa,
I try to find an efficient solution to this problem:
A sum of 100000$has to be invested into 4 stocks (A,B,C,D), based on historic performance of the stocks. I have to test and find out what mix of those stocks (% of sum) would give the best portfolio. Investment in each stock can vary from 0 to 100% of the sum, with increments of 10%. I need help how to set an algorithm that will draw all possible portfolios in term of stocks weight in the portfolio. Ex. A-10%, B-20%, C-30%, D-40% –> 100% I hope that you could help as it seems to me a matrix problem. Many thanks, Srul • Thanks for writing. I’m not sure I understand this problem exactly, since isn’t this more a permutation problem? I’m sure there could be a way to set it up as a matrix, but I’m just not sure how to do that. Did you get an answer? Sorry I can’t help more, Lisa 38. Hi Lisa, Many thanks for your try. The problem is: to find all unique sets of 4 numbers A,B,C,D that each number can get the value from 0.0 to 1.0 incremented by 0.1 and A+B+C+D=1.0 . We have a total of 11 possible values and 4 numbers and that gives a total of 11 power 4 = 14,641 possible unique sets, while the number of the sets that sums up to 1 is 258 only. The challenge is to find a solution (template?) that will enable us to form only the sets that sum up to 1.0. This would have saved 14,383 wasted iterations (try on error). Best regards, Srul • Hello Srul, I actually just drew a tree diagram and am getting that there are 286 different permutations of A, B, C, and D (each can be 0%, 10%, 20%, 30%, 40%, 50%, 60%, 70%, 80%, 90%, or 100%) so they will add up to 100%. I’m not sure how to put this into a matrix, but I can send you the tree diagrams in a picture if you want. Lisa 39. Find the inverse of the matrix below; [2 1 1] [5 1−2] [−4 3 4] Hence or otherwise solve the following system of three simultaneous equations 2×1+x2+x3=2 5×1−x2−2×3=9 −4×1+3×2+4×3=4 • Thanks for writing! To find the inverse matrix, you can either do it by hand, or in the calculator. I did it in the calculator and found it to be: [10/27 -1/27 -1/9] [-4/9 4/9 1/3] [19/27 -10/27 -1/9] For the second problem, I also put it in the calculator to get x1 = -3, x2 = 40, x3 = -32. Hope that’s what you got 🙂 Lisa • thanks Lisa, could you also assist me with this.. detailed calculation please Two companies A and B are in competition and are currently holding 30% and 40% of the total market share respectively. Company A retains 80% of its customers and loses the rest to B. Company B retains 70% of its customers and loses the rest to A. i) Formulate the transition probability matrix ii) What will each company market share be in one year time and two years’ time? iii) Determine the market share in the long-run • Thanks for writing again! I believe this is an example of a Markov Chain model (but I haven’t done this in a long time :). I got the transition probability matrix is: [.8 .7] [.2 .3] Initial market share of A: [.30 .70] and initial market share of B: [.40 .60]. So A’s market share in 1 year will be [.30 .70] [.8 .7] [.2 .3] = [.38 .42], or 38% after one year. Similarly B’s will be .44 or 44% after one year. In two years, we’ll do the same, but replace the first matrix by the new market share to get A’s at 38.8% and B’s at 44.4%. I’m not sure how exactly to get the market share in the long run, but by doing this a large number of times, it looks like A’s is 38.88888% and B’s is 44.4444%. Hope that helps! Lisa 40. Please help me in solving this: A salesman has below record for products sold during peak seson of October to December. The products are in three variants A, B & C. How many units of product C only is he expected to sell to earn 900 commission? 41. Here is what I am working on in my class I truly do not understand how to set up the system of linear equations. I am trying to figure this out by looking at and reworking the following problem. Figure 1 shows the intersections of five one-way streets and the number of cars that enter each intersection from both directions. For example, I1 shows that 400 cars per hour enter from the top and that 450 cars per hour enter from the left. The letters a, b, c, d, e, f, and g represent the number of cars moving between the intersections. To keep the traffic moving smoothly, the number of cars entering the intersection per hour must equal the number of cars leaving per hour. 1. Describe the situation. 2. Create a system of linear equations using a, b, c, d, e, f, and g that models continually flowing traffic. 3. Solve the system of equations. Variables f and g should turn out to be independent. 4. Answer the following questions: a. List acceptable traffic flows for two different values of the independent variables. b. The traffic flow on Maple Street between I5 and I6 must be greater than what value to keep traffic moving? c. If g = 100, what is the maximum value for f? d. If g = 100, the flows represented by b, c, and d must be greater than what values? In this situation, what are the minimum values for a and e? e. This model has five one-way streets. What would happen if the model had five two-way streets? 42. Rank of matrix – A matrix is said to be of rank r; when 1. there is at least one minor of A of order r which does not vanish. 2. Every minor of A of order ( r+1) or higher vanishes. 43. i really don’t know how to solve this. Can you help? 2. A local roadside café serves beef burgers, eggs, chips and beans in four combination meals: Slimmer – 150 g chips 100 g beans 1 burger Normal 1 egg 250 g chips 150 g beans 1 burger Jumbo 2 eggs 350 g chips 200 g beans 2 burgers Veggie 1 egg 200 g chips 150 g beans – A party orders 1 slimmer, 4 normals, 2 jumbos and 2 veggies meals. What is the total amount of materials that the kitchen staffs need to cook? One of the customers sees the size of a jumbo meal and changes his order to a normal meal. How much less materials will the kitchen staff need? • Thanks for writing! I’d set up a 4 by 4 matrix with Eggs, Chips, Beans, and Burger going down one side, and Slimmer Normal, Jumbo and Veggie going left to right. Then multiply this 4 by 4 matrix by a 4 by 1 matrix with the number of Slimmers, Normals, Jumbos, and Veggies meals going down. Then you will get the amount for the eggs, chips, beans and burgers. For the second part, just change the 4 normals to 5, and the 2 jumbos to 1. Does this make sense? Let me know if you want to see an answer. Lisa 44. Hi Lisa! I don’t know how to solve the following question; i know it’s simple, but don’t know how to go about solving it. The question is to find the rank of a matrix: (1 0 -1) A = (3 4 5) (0 -6 -7)? It would be very kind of you if you could show the solution step by step. Thanks in advance for your help. Have a great day!!! • Thanks for writing! I’m not sure exactly what your matrix is, but to find the rank, you can use RREF in the calculator, and the rank of the matrix is equal to the non-zero rows in RREF. If I read your matrix correctly, it’s a 3 by 3, and I get the rank of it as 3. Does that make sense? Lisa • Hi! When I solve this, I get x = 2 and y = 1 (I used substitution). Let me know if you want any more of the work. Lisa 45. Great Work Lisa…..I am a Software Guy but didn’t had the basic understanding of the Matrices,your explanation was from the basics and of great help. Now I can solve the matrices coding problems quite easily. BWT, which books would you recommend for kids between 10-15 years(for nephews)…when i was small all books started off with “matrix(or any other concept) is used in large applications and should be studied seriously”…..but never-ever did any book explain how to use it in daily basis.(the drawback of having too much knowledge by writers…i suppose). For comment readers: Computer science for kids. http://www.laurenipsum.org/ Once again i would congratulate you for the great work!!! Rohit • Thanks so much for taking a look at the site and writing! Honestly, if you’re just teaching the kids math topics, I always like the “… for Dummies” books or the “Idiot’s Guide to …” books. They explain things really in a more direct way, and give lots of examples. Keep using my site, and please let me know if you see any ways it can be better 🙂 Lisa 46. I would like to really thank you so much for this, it really helped me revise for my maths test that is coming up this week!! Keep doing this it is truly amazig! (=^_^=) 47. A salesman has below record for products sold during peak seson of October to December. The products are in three variants A, B & C. How many units of product C only is he expected to sell to earn 900 commission?Months Sales of Units A B C Total Commission drawn (N) January 90 100 20 800 200 February 130 50 40 900 300 March 60 100 30 850 400 48. A salesman has below record for products sold during peak seson of October to December. The products are in three variants A, B & C. How many units of product C only is he expected to sell to earn 900 commission?Months Sales of Units A =January 90 100 20,commission=800 200 B=February130 50 40,commission 900 300 C=March 60 100 30,commission 850 400 • Thanks for writing, but I’m not sure I understand this problem. Are the commission numbers for October to December? It looks like they are from January to March? Lisa 49. Please, I need help with solving the following problems: Problem 1: (Performance Test) A teacher estimates that of the students who pass a test 80% will pass the next test, while of the students who fail a test, 50% will pass the next test. Let x an y denote the number of students who pass and fail a given test, and let u and v be the corresponding numbers for the following test. (a) write a matrix equation relating x & y to u & v. (b) suppose that 25 of the teacher’s students pass the third test and 8 fail the third test. How many students will pass the fourth test. Approximately how many passed the second test? Problem 2: an out of shape athlete runs 6 miles per hour, swims 1 mile per hour, and bikes 10 miles per hour. He entered a triathlon, which requires all 3 events and finished it in 5 hours and 40 minutes. A friend, who runs 8 miles per hour, swims 2 miles per hour, and bikes 15 miles per hour finished the course in 3 hours and 35 minutes. The total course was 32 miles. How many miles was each segment (running, swimming and biking)? Problem 3: New parents Jim and Lucy want to start saving for their son”s college education. They have$5000 to invest in three different types of plans. A traditional savings account pays 3% annual interest, a certificate of deposit (CD) pays 6% annual interest, and a prepaid college plan pays 7½ % annual interest. If they want to invest the same amount in the prepaid college fund as in the other two plans together, how much should they invest in each plan to realize an interest income of $300 for the first year? • Thanks for writing. For now, I will do problem 3; let me know if you still need help on the other problems. I’d set up like this: .03x + .06y + .075z = 300. x + y + z = 5000, and z = x + y. If I put this into a matrix and solve, I get x =$1250, y = $1250, and z =$2500. Does that make sense? Lisa

• Thank you so much for your help, those were the answers I got, but needed to verify. Can you please help witb question 2 which I’m still having trouble fuguring it out. Thanks!

• an out of shape athlete runs 6 miles per hour, swims 1 mile per hour, and bikes 10 miles per hour. He entered a triathlon, which requires all 3 events and finished it in 5 hours and 40 minutes. A friend, who runs 8 miles per hour, swims 2 miles per hour, and bikes 15 miles per hour finished the course in 3 hours and 35 minutes. The total course was 32 miles. How many miles was each segment (running, swimming and biking)?
Here’s how I did this – let t1, t2, t3 be times for first guy, t4, t5, t6 be times for second. t1 + t2 + t3 = 5 40/60. t4 + t5 + t6 = 3 +35/60. Also, 6t1 + t2 + 10t3 = 32, and 6t1 = 8t4, t2 = 2t5, and 10t3 = 15t6. Solving by matrices, I get t1 = 1 2/3, t2 = 2, and t3 = 2. So then I get the distances of 10, 2, and 20 for running, swimming and biking. Does that make sense? Lisa

50. Hello there! I need some help with this word problem matrix. I’ve been working on it for hours and have gotten absolutely nowhere!
A health shop owner made trail mix containing dried fruit, nuts, & carob chips. The dried fruit sells for $5.50/lb, the nuts for$7.50/lb, & $8.50/lb for carob chips. The shop owner mixed 50lbs of trail mix & sells it for$6.70/lb. If the amount of nuts is 5lbs more than the carob chips, how much of each item was used in the trail mix?

• Thanks for writing! Here’s how I’d do this problem: 5.5f + 7.5n + 8.5c = 50(6.7), f + n + c = 50, n = 5 + c. Then I used the following matrix: [5.5 7.5 8.5 335 1 1 1 50 0 1 -1 5] (This is a 4 by 3 matrix, and I use RREF in the graphing calculator to get the answers). I get 25 lbs of fruit, 15 of nuts, and 10 of carob chips. Does this make sense? Lisa

• OH MY GOSH, thank you so so much! I was using the $6.70 for the second equation and 50lbs just for the first instead of multiplying 50 by$6.70. I appreciate it so much! This website is amazing thank you for using so many examples for even upper level math courses. My Trig & Calculus final are in a few weeks & this will save my life for sure!

51. Aisha has RM10000 to invest. As her financial consultant, you recommend that she invest in Treasury bills that yield 6%. Treasury bonds that yield 7% and corporate bonds that yield 8%. Aisha wants to have an annual income of RM680, and the amount invested in corporate bond must be half that invested in Treasury bills. Find the amount of each investment.

miss what the way to make this in matrix?i not able to answer it and if you know to make it in c coding can u share it.

• Here’s how I’d do this: x + y +z = 10000. .06x + .07y + .08z = 680. z = .5x. The matrix would look like [ 1 1 1 10000
.06 .07 .08 680
-.5 0 1 0]
I get 4000 in treasury bills, 4000 in treasury bonds, and 2000 in corporate bonds. Does that make sense? Lisa

52. Hey Lisa, I really need your help!

A roadside fruit stand sells mangoes at Php 75 a kilo, pomelos at Php 90 a kilo, and star apples at Php 60 a kilo. Karla buys 18 kilos of fruits at a total cost of Php 1380. Her pomelos and star apple together cost Php 180 more than her mangoes. How many kilos of each kind of fruit did she buy? Write a system of equations using Cramer’s Rule.

• Here’s how I’d set up this problem: 75m + 90p + 60a = 1380. m + p + a = 18. 90p + 60a = 75m + 180. We can then set up in a matrix: [ 75 90 60 1380
1 1 1 18
-75 90 60 180]
I then get 8 mangos, 6 pomelos, and 4 star apples. You can do this with Cramer’s Rule using the information found here. Does that make sense? Lisa

53. How much copper and how much iron should be added to 100 lb of an alloy containing 25% copper and 40% iron in order to obtain an alloy containing 30% copper and 50% iron? Write a system of equations using Cramer’s Rule.

• Thanks for writing! I found this problem here and you can solve it using Cramer’s Rule the way I explained here. Does that help? Lisa

54. Jane is asked to buy three sizes of bottled water: A, B and C. The total number of bottles she needs to buy is 50. She has a budget of Php 1500. Size A is Php 20 each, size B is Php 50 each, and size C is Php 30 each. Additionally, the number of bottles of size A should be equal to that size of C. How many of each size should she buy? Write a system of equations using Cramer’s Rule.

55. plz solve this question : show by considering minors that the matrice A inverse , A transpose inverse , have the same rank as A ….. it will be easy for me if u attach image of the work out

56. 1.The buying price of a basket of oranges is #1000 and the selling price is #5 per orange. what is the profit per basket if 300 oranges are found in the basket.?
a. #150
b.#200
c.#300
d. #500.
2. what is the break even point(quantity)if the buying price of a basket remains #1000 and the selling price is #5

• Thanks for writing! Here’s how I’d do this problem: Let x = the number of oranges. Then we have Profit = Revenue – Cost. So Profit = 5x – 1000. If 300 oranges are in the basket, we have Profit = 5(300) – 1000 = #500 (d). The break even point would be when profit = 0, so 0 = 5x – 1000, or x = 200 oranges. Does that make sense? Lisa

57. Hello, I see that you have been very helpful with assisting with systems of equations and matrices. Could you please help me set this up?
You have been hired by a consultant for Crazy Al’s Car Rentals in the city of Metropolis. Crazy Al’s car rentals has a total of 2200 cars that it rents from three locations within the city: Metropolis Airport, downtown and Suburban Airport.
Following is the weekly rental and return patterns:
90% of costumers who rent from Metropolis Airport return their cars to Metropolis Airport
How many of his cars should be at each of his three locations at the start of each week so that the same number of cars will be there at the end of the week (and hence at the start of the next week).
Use systems of equations and matrices to set up a systems of equations representing this situation and solve this problem.
Include math steps.

Any help would be appreciated. I don’t know where to even start
Thanks!

58. 1) The 7th term of an A.P is 15 and the fourth term is 9. Find the common difference of the sequence
2) The 7th term of an A.P is 15 and the fourth term is 9. find the sequence fifth term.
3) The 7th term of an A.P is 15 and the fourth term is 9. Find the sequence tenth term.
4) The 7th term of an A.P is 15 and the fourth term is 9. Find the sequence first term
5) Find the 7th term of an A.P whose first term is 102 and common difference is -3,
I await your respose liza. Thanks

• So let’s use the equation for an arithmetic sequence: an = a1 + d(n – 1). Since 15 – 9 is 6, and 7 – 4 is 3, we can see that the common difference is 6/3 = 2. (See how this would mean every term goes up by 2?) So we have an = a1 + 2(n – 1). Let’s plug in a “point” to get what a1 is: 15 = a1 + 2(7 – 1), so a1 = 3. So the sequence is an = 3 + 2(n – 1). So the 5th term or a5 is 3 + 2(5 – 1), or 11. See if you can do the other problems? Thanks 😉 Lisa

59. i need help..
question is : An automobile company uses three types of steel S1, S2 and S3 .For producing three types of cars c1, c2 , c3. Steel Requirements (in tons) for each type of car are given below:

type of cars
C1 C2 C3
S1 2 3 4
S2 1 1 2 TYPE OF STEEL
S3 3 2 1

Determine the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.

• Thanks for writing! I put this in a matrix and got C1=2, C2=3 and C3=4. Here is the matrix I solved: [2 3 4 29
1 1 2 13
3 2 1 16]
Does that make sense?
Lisa

60. An automobile company uses three types of steel S1, S2 and S3 .For producing three types of cars c1, c2 , c3. Steel Requirements (in tons) for each type of car are given below:
Type of car
C1 C2 C3
S1 2 3 4
S2 1 1 2
S3 3 2 1

Type of steel

Q1. Determine the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively

suppose three companies: A, B, C dominates the market for a certain product and are competing against each other for a large share of the market. currently comapany A has 2/9 of the market, comapany B has 4/9 of the market and company C has 1/3 of the market. the market survey indicates that every 6 months company A retains 3/4 of its customer and loss 1/6 to company B and 1/12 to company C. company B retains 1/2 of its customer and loss 1/3 to company A and 1/6 to company C. company C retains 3/8 of its customer and loss 1/4 to company A and 3/8 to company B. find the share of the market that each company will have.
1. one year later
2. in the long run

62. Hi Lisa,kindly help me solve this. Guabodia furniture company sells executive chairs,dinning chairs and dinning tables. The monthly sales figures are executive chairs 700, dinning chairs 1000, and dinning tables 250. The price of each executive chair is#500 and #200 for dinning chairs. His costs of production are as follows; #200 for executive chairs, #125 for dinning chairs and #600 for dinning tables.
I. Express his TC in vector notation
Ii. Express his TR in vector notation
Iii express his profit in vector notation
If.find his TC,TR and profit.

• Here’s how I’d set this up: Total revenue – 500(700) + 200(1000) + ?(250), Total cost – 200(700) + 125(1000) + 600(250). Then to get the profit, subtract the cost from the revenue. Does that make sense? Lisa

63. a shopkeeper sells three products vise x,y,z during a particular month 20units of x, 30 units of y and 45 units of z were sold at rs 120 ,rs 100 and rs 225 respectively . the cost of x ,y,z to the shopkeeper is rs95,rs 125 and rs 185respectively .find the profit made by him using matrix algebra

• Thanks for writing. Here’s how I’d do this: profit = revenue – cost profit = [120 – 95 100 – 125? 225 – 185] [20
30
45]
I’m not sure if you have a typo since the cost of product y is greater than the profit?
Anyway, if you do this matrix multiplication, you should get a 1 x 1 matrix that shows the profit. Does this make sense? Lisa

64. Hi Lisa, need some help with this question, Thanks in advance

An cleaning company performs vacuum cleaning, dusting and general clean-up work. For each task they charge by the hour – vacuum cleaning $30, dusting$50 and general clean-up work $60. They have four employees: Allan, Bryan, Casey and Danny. Each one is capable of handling all three types of work. Allan is paid$25 an hour while Bryan, Casey and Danny are paid $14,$11.50 and $8 respectively. Table A shows the number of hours each employee spends, on average, on each task in a normal week. Table B shows the hours spent on each of the three types of work for each of four clients Eddy, Freddy, Gretel and Hannah in a particular week. Table A Employee Vacuum cleaning (hours) Dusting(hours) General clean-up (hours) Allan 0 2 4 Bryan 1 5 2 Casey 3 4 1 Danny 3 3 0 Table 2 Client Vacuum cleaning (hours) Dusting (hours) General clean-up (hours) Eddy 0 6 1 Freddy 1 2 3 Gretel 2 1 0 Hannah 2 2 1 Using matrix representation and methods, determine: a) the income generated by each of the four employees b) the cost, in terms of salary, of each of the three types of work performed c) the charge made to each client for a week’s work • Thanks for writing! I haven’t had time to do this problem – did you get it worked out? Lisa 65. Hi Lisa. Pls. do solve this mathematical problem for me. 1) Consider the simple macro model described by the following equations Y= C + A0 (1) C = a + b(Y – T) (2) T = d + tY (3) Where Y is income, T is tax revenue, C is consumption, A0 is the constant autonomous expenditure, and a, b, d, and t are all positive parameters. Find the equilibrium values of the endogenous variables Y, C, and T by writing the equations in matrix form and applying Cramer’s rule. • Thanks for writing! I’m not sure how to do this problem, although I do talk about Cramer’s Rule here. Lisa 66. (a) A simple society economy consists of Wool Production, Butchery and Hides Tanning. 60% of each unit wool output goes towards wool production, 10% towards running of the butchery and the rest towards hides tanning. Each of output from the butchery is shared among the three sectors Wool, Butchery and Hide Tanning in the ration of 3:5:2. 20% of each of the hides tanning output goes towards Wool production, 10% towards running the Butchery and the rest towards hides tanning. Given that the external demand for the output from Wool Processing, Butchery and Hides Tanning are respectively 600Units, 1500 Units and 900 Units, Find:- Kindly help me derive the technical coefficient matrix from this 67. Hi,kindly help me solve this An investment advisor has two types of investments available for clients Investment Type A that pays 4% P.A and B of a higher risk that pays 8% P.A clients may divide the investment btn the two alternatives to achieve any total returns desired between the two. However the higher the desired returns the higher the risk. clients clients 1 2 3 TOTAL INVEST MENT 20000 50000 10000 ANNUAL RETURNS DESIRED 1200 37500 500 R1 6% 7.50% 5% R2 Using Inverse method how should each client invest to achieve the indicated returns 68. Hi, I can’t wrap my head around this problem… Can you help? Let x = (x1 , . . . , xn )′ be a vector containing the number of units purchased of each of a variety of grocery items. Let y = (y1, . . . , yn)′ be a vector of unit prices, such that yi = the price/unit of item i. For example, x = (4, 3, 2)′ and y = (.95, .25, 6.50)′ might represent 4 dozen eggs at$0.95 per dozen, 3 lbs. of apples at $0.25/lb, and 2 cans of pate de fois gras at$6.50 per can (cheap, if it’s entier).
(a) Formulate a matrix expresstion for the total (net) cost of the commodities in x.

The answer to this is easy: Total cost = x’y

(b) Suppose each commoditiy is subject to a particular rate of tax, these being given by a vector, t = t1, . . . , tn so that if commodity i is taxed at 5%, ti = 0.05. Formulate an expression in terms of matrices and vectors for the total cost of x including taxes. [Remember, cost after tax = net cost × (1 + t).]

This is what is so confusing!

• Here’s how I’d do this one – although, there may be an easier way! For b), create 3 by 3 matrix that contains the numbers of units, so it would look like:
[4 0 0
0 3 0
0 0 2]. Call this matrix A. Then create 3 by 1 matrix with prices, so it would look like:
[.95
.25
6.5]. Call this matrix B.
Then multiply A by B and transpose it, so you’ll have (A * B)^T, to get 1 by 3 matrix: [3.8 .75 13]. Call this matrix or vector C.
Then create vector <1 1 1> + = <1 + t1 1 + t2 1 + t3> and multiply it by C. You should then have the total cost.
Maybe there’s any easier way? Lisa

69. Lisa,
I have an input-output problem with a twist. Three services (1, 2, & 3) are used to produce 2 products (4 & 5). The outputs for each service and product are 1 = 10, 2 = 100, 3 = 500, 4 = 24, and 5 = 36. The input-output relationships are shown in a square matrix as follows:
[10 -20 -150 0 0
-6 100 -100 0 0
-4 -30 500 0 0
0 -40 -50 24 0
0 -10 -200 0 36]
Thus, each service and product’s output is shown as a positive number on the diagonal and its consumption by the other services & products are shown as negative numbers in the columns.

Each service and product also incur direct costs of 1 – $500, 2 –$750, 3 – $900, 4-$1,100 and 5 – $400. I can calculate the cost of the two products (I think) by inverting the matrix and multiplying it by the direct cost vector. Product 4 then costs$90.38 and Product 5 = $41.14. Here is the twist: If my actual sales result in a different product mix (let’s assume 36 for 4 and 24 for 5) how can I calculate what my output levels for services 1, 2 and 3 should have been based on the actual product mix realized? I guess I am trying to determine the diagonal values for the 3 services based on the realized diagonal values of products 4 and 5 given the known consumption relationships of services? 70. solve the matrix using example of intrest per year problem 71. Hi Lisa! I just had a question about matrices- this problem has stumped me and I just don’t how to solve it. I would be greatly appreciative if you’d help me out 🙂 “An investment company recommends that a client invest in AAA. AA. and A rated bonds. The average annual yield on AAA bonds is 6%, on AA bonds is 7%, and on A bonds is 10%. The client tells the company she wants to invest twice as much in AAA bonds as in A bonds. How much should be invested in each type of bond if the client has a total of$50,000 to invest, and wants an annual income (that is, earned interest) of \$3,620 yearly?”

How do I put this into matrix form?

• Thanks for writing! Here’s how I’d do this: .06x+.07y+.1z=3620, x + y + z = 50000, x = 2z. The expanded matrix will be
[.06 .07 .1 3620]
[ 1 1 1 50000]
[ 1 0 -2 0]
Solving, I get x = 24000, y = 14000, z = 12000. (x = AAA, y = AA, z = A bonds)
Does that make sense? Lisa

72. After going through i really appreciate.. that I have been able to get something..