This section covers:

**Introduction to the Matrix****Adding and Subtracting Matrices****Multiplying Matrices****Matrices in the Graphing Calculator****Determinants, the Matrix Inverse, and the Identity Matrix****Solving Systems with Matrices****Solving Word Problems with Matrices****Cramer’s Rule****Number of Solutions when Solving Systems with Matrices****More Practice**

# 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 2^{nd} row and it’s the 1^{st }entry.

# Adding and Subtracting Matrices

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:

# 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** (

**will be a matrix):**

*X***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

**. Watch the order when we multiply by the inverse (matrix multiplication is not commutative), and thank goodness for the calculator!**

*X*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,

**= the second number, and**

*y***= the third number. So here are the three equations:**

*z*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,

**= the number of tulips, and**

*t***= the number of lilies. So let’s put the money terms together, and also the counting terms together:**

*l*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,

**for manufacturing), 40 percent of it, or .40**

*y**x*pays to produce internal energy, and 25 percent of it, or .25

*x*pays for internal manufacturing. Of the value of the manufacturing produced, .25

*y*pays for its internal energy and .10

*y*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**,

**, and**

*y***, we need to get the determinants of**

*z***four**matrices, the first one being the 3 by 3 matrix that holds the coefficients of

**,**

*x***, and**

*y***. Let’s call this first determinant**

*z***;**

*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

**. So Similarly,**

*x*OK, now for the fun and easy part! To get the ** x**,

**, and**

*y***answers to the system, you simply divide the determinants So Now we know that**

*z***= 5,**

*x***= 1, and**

*y***= –2.**

*z*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 solution**s (called **dependen**t 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

**and**

*x***in terms of**

*y***: This would look like so the solution set for {**

*z***,**

*x***,**

*y***} is {5 – .375**

*z***, 3 + .875**

*z***,**

*z***}. (This may be a little advanced for high school 🙂 )**

*z*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!

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

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! (=^_^=)

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

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

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

great

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!

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

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

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

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.

Here is how I’d set this up: a + b + c = 50, 20a + 50b + 30c = 1500, a = c. Then you can put this in a matrix and solve using

Cramer’s Rule. Thanks, Lisaplz 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

Thanks for writing! Unfortunately, these questions are beyond the scope of this website. You might look at this site http://www.mathsisfun.com/algebra/matrix-inverse-minors-cofactors-adjugate.html, and see if that helps? So sorry – I will try to do more research to see what I can figure out thought! Lisa

any way thanku lisa for ur help … 🙂

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

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

5% rent from Metropolis Airport and return to Downtown

80% rent from Downtown and return to downtown

10% rent from Downtown and return to Metropolis Airport

10% rent from Suburban Airport and return to Metropolis Airport

5% rent from Suburban Airport and return to downtown

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!

Thanks for writing. I found this: https://answers.yahoo.com/question/index?qid=20140216223142AAddVGd and then it shouldn’t be too difficult solving the system with matrices. Let me know if you want it solved. Thanks, Lisa

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

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

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

Thanks for writing! I think I found this problem on Illustration 7 in the following link:

https://www.scribd.com/doc/19613606/Applications-of-Matrices-to-Business-and-Economics

hi everyone please help me with these..

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

Thanks for writing! I wasn’t sure how to do this exactly, but found this explanation: https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Matrices-and-determiminant.faq.question.1035222.html

Hope that helps!

Lisa

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

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

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

yes I did figure it out eventually, thanks for trying though

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

(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

Did you get ever get an answer to this? Not sure how to do it – sorry 😉 Lisa

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

Thanks for writing! I found a similar problem here (see link – Example 4) – does it make sense? Lisa

http://www.mhhe.com/math/precalc/barnettpc2/student/olc/graphics/barnett01paga_s/ch08/downloads/pc/ch08section6.pdf

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: = <1 + t1 1 + t2 1 + t3> and multiply it by C. You should then have the total cost.

[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> +

Maybe there’s any easier way? Lisa

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?

Sorry -I don’t do problems like these. Lisa

solve the matrix using example of intrest per year problem

Thanks for writing! Can you be more specific on what type of problem you want? Lisa

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

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

PLEASE HELP ME!!!!

Find the encoding matrix from this information:

– position 1,1 in the encoding matrix is an even number

– the decoding matrix only contains integers

– position 1,1 in the encoding matrix is the negative of the number in position 1,1 in the decoding matrix

– modulo arithmetic hasn’t been used

The encoded message received is – 79, -32, 85, -37, 173, -73.

PLEASE HELP!!!

Sorry – I’m not sure how to solve this problem ;( Lisa