10.03 Matrix equations and solving systems with matrices

Inverses and determinants

In mathematics an inverse of a number is the number needed so that when you multiply them together the result is $1$1.

For example, $6\times\frac{1}{6}=1$6×16​=1, where $\frac{1}{6}$16​ is the inverse of $6$6 (and vice-versa).

In our real number system we have used inverses a lot, specifically when we were solving equations.  

For example:

$5x$5x	$=$=	$20$20	we would normally divide by $5$5, but consider what happens if we multiply by $\frac{1}{5}$15​ instead
$5x\times\frac{1}{5}$5x×15​	$=$=	$20\times\frac{1}{5}$20×15​	it has the same effect because $5\times$5×$\frac{1}{5}=1$15​=1.   (hence $\frac{1}{5}$15​ is the inverse of $5$5.
$x$x	$=$=	$4$4

Matrix inverses are a little more complicated, to derive the inverse algebraically work through the next activity.

As noted above, $\frac{1}{6}$16​ is the inverse of $6$6. Just as we can write $6^-1$6−1 to represent $\frac{1}{6}$16​, we use the notation $A^{-1}$A−1 to denote the inverse of matrix $A$A. 

Now, let's look at the determinant and inverse of a 2 x 2 Matrix.

 

The Determinant of a 2x2 Matrix

Before we learn the formula for the inverse of a matrix, we need to learn about a concept called the determinant.

The determinant is the difference between the products of the diagonals.

		$a$a	$b$b		$\text{Determinant =}$Determinant =$a\times d-c\times b$a×d−c×b
		$c$c	$d$d

 

Often, the determinant of a matrix $A$A is written as $\text{det(A) }$det(A)

Notice, that is uses vertical bars | | around either side, rather than square parentheses [ ]. This is how we know that we are referring to the determinant, rather than the matrix itself.

The Inverse of a 2x2 Matrix

For a matrix:	$A$A	$=$=	$a_{11}$a11​ $a_{12}$a12​         $a_{21}$a21​ $a_{22}$a22​
The inverse of $A$A is given by:	$A^{-1}$A−1	$=$=	$1$1     $a_{22}$a22​ $-a_{12}$−a12​     $det(A)$det(A)     $-a_{21}$−a21​ $a_{11}$a11​

 

So, we just derived that for any matrix

that 

such at $AA^{-1}=A^{-1}A=I$AA−1=A−1A=I

The component out the front, , is called the determinant.  It determines if a matrix inverse exists.

If $ad-bc=0$ad−bc=0, then the  is undefined and the matrix will not have an inverse. We call matrices that do not have an inverse singular. 

Practice questions

Question 1

Question 2

Question 3

Question 4

Solving matrix equations

Just as we have equations that we can solve using algebraic manipulation, we also have matrix equations we can solve using manipulation according to the laws of matrices.  

We already know that 

$AA^{-1}=A^{-1}A=I$AA−1=A−1A=I

Where $A^{-1}$A−1 is the multiplicative inverse of $A$A and $I$I is the identity matrix. 

 

Division of matrices

We have looked at adding and subtracting matrices, scalar multiplication and multiplying matrices, but we haven't yet discussed division.

Well, division for matrices is actually not defined, but this doesn't mean we can't manipulate equations.  Anything that we would normally need to do using division, we instead do using multiplication by inverses, provided the inverse of a matrix exists.

Because $AA^{-1}=A^{-1}A$AA−1=A−1A we can use this property to carry out pre-multiplication and post-multiplication in equations.  This is instead of division. 

 

Pre-multiplication

For $AX=B$AX=B, to solve for $X$X we need to pre-multiply, (which means put the $A^{-1}$A−1 at the front)

$A^{-1}AX=A^{-1}B$A−1AX=A−1B - its important to note here that the $A^{-1}$A−1 is pre-multiplied on both sides. 

$IX=A^{-1}B$IX=A−1B

$X=A^{-1}B$X=A−1B

 

Post-multiplication

For $XA=B$XA=B, to solve for $X$X we need to post-multiply, (which means put the $A^{-1}$A−1 at the rear)

$XAA^{-1}=BA^{-1}$XAA−1=BA−1 - its important to note here that the $A^{-1}$A−1 is post-multiplied on both sides. 

$XI=BA^{-1}$XI=BA−1

$X=BA^{-1}$X=BA−1

 

Worked examples

Question 5

Solve for Matrix $X$X if

 

	$=$=
	$=$=
$X$X	$=$=
$X$X	$=$=

In solving this equation we used the properties of addition and subtraction, and scalar multiplication.  

Question 6

Solve for matrix X if

	$=$=
	$=$=	$3X-X$3X−X
	$=$=	$2X$2X
	$=$=	$\frac{1}{2}\times2X$12​×2X
	$=$=	$X$X

In solving this matrix equation, we had to use matrix addition/subtraction, scalar multiplication and collecting like terms. 

Question 7

Make X the subject of the matrix equation

$AXC$AXC	$=$=	$B$B
$A^{-1}AXC$A−1AXC	$=$=	$A^{-1}B$A−1B	pre-multiply by $A^{-1}$A−1 on both sides
$IXC$IXC	$=$=	$A^{-1}B$A−1B	$A^{-1}A=I$A−1A=I
$XCC^{-1}$XCC−1	$=$=	$A^{-1}BC^{-1}$A−1BC−1	post-multiply by $C^{-1}$C−1 on both sides
$XI$XI	$=$=	$A^{-1}BC^{-1}$A−1BC−1	$CC^{-1}=I$CC−1=I
$X$X	$=$=	$A^{-1}BC^{-1}$A−1BC−1

 

Practice questions

Question 8

Question 9

Augmented matrices and systems of equations

An augmented matrix uses the coefficients of the variables and the constant terms and writes them in one augmented matrix. 

Let's look at the system of equations  

Using our first matrix equation set up we convert this the following matrix equation. 

and as an augmented matrix we would write it like this.

Can you see how we have basically just shorthanded the matrix equation form?

We are just going to practice now converting systems into augmented matrix form, then we will move on to performing operations on the augmented matrix. 

Worked Examples

QUESTION 10

QUESTION 11

QUESTION 12

Matrices and systems of equations

Now we know how to both represent information using matrices, and how to multiply using matrices we can look at how to solve simultaneous equations using matrices. 

Remember that systems of equations, are two equations that when we solve simultaneously (at the same time), we find their point of intersection. You will have already solved these using graphical and algebraic methods, now we will use matrices. 

First, we need to know how to represent the information using matrices.

Take this pair of equations. 

If we write these using matrices we split up the system into $3$3 parts.

• a coefficient matrix (numbers in front of the variables)
• a variable matrix (only variables with coefficients of $1$1)
• a solution matrix (values on the other side of the equals sign)

To verify that this matrix representation is indeed equivalent, let's just quickly distribute the multiplication.

And then by equating equivalent positions, the elements $2x+3y$2x+3y must equal $16$16, and $-3x+y=-13$−3x+y=−13.  

Now that we have confirmed that this is the correct matrix representation for our system, let's look at how to solve it. 

Remember how for matrix equations, if $MX=C$MX=C, we can isolate matrix $X$X, by pre-multiplying both sides by the inverse of $M$M.  Hence  $X=M^{-1}C$X=M−1C

This means we now need the inverse of 

We already know how to find the inverses, recall that 

	$=$=
	$=$=
	$=$=
	$=$=
	$=$=
	$=$=

So this means that the intersection of the two equations:

is at $x=5$x=5, and $y=2$y=2.  And thus $\left(5,2\right)$(5,2) is the solution.  

Practice questions

Question 13

Question 14

Graphing calculators and matrices

Matrices lend themselves to automation by electronic devices. Systems of linear equations, expressed in matrix form, are solved very quickly using hand-held graphing calculators or other computer applications running on various devices .

There are typically a great many individual arithmetic operations involved if the calculation is done manually but a graphing calculator or other device hides these from the user. A relatively simple matrix notation expresses certain complicated problems in a concise way that can be unraveled by the algorithms built into the machine being used.

How this is done in practice depends on the particular device. Users should consult the instruction manual for their device for specific information about inputting matrices. However, an overall understanding of the processes involved is needed as a first step.

Systems of linear equations

The following is an example of a system of three linear equations in four variables. On the left of each equation is a sum of terms in which a numerical coefficient multiplies a variable that is not raised to any power other than $1$1. 

$w+3x-2y-5z$w+3x−2y−5z	$=$=	$-19$−19
$2w+x+y-z$2w+x+y−z	$=$=	$3$3
$-w-x+4y+3z$−w−x+4y+3z	$=$=	$21$21

By a solution to this system of equations, we mean a set of particular values of $w$w, $x$x, $y$y and $z$z that satisfy all three equations. 

In this case, we would expect to find that there are many such solutions. There is certainly no unique solution and it is possible that there is no solution at all. To have a unique solution, the system would need to have four equations each expressing a different relation among the variables.

The following system of four equations has a unique solution.

$w+3x-2y-5z$w+3x−2y−5z	$=$=	$-19$−19
$2w+x+y-z$2w+x+y−z	$=$=	$3$3
$-w-x+4y+3z$−w−x+4y+3z	$=$=	$21$21
$w+x+y+z$w+x+y+z	$=$=	$10$10

In matrix form, we need a $4\times4$4×4 matrix of coefficients. In your calculator, this will need a label $A,B,C,...$A,B,C,... . The convention is to identify matrices with capital letters. Similarly, the numbers on the right of the equations will form a column matrix which also needs to be entered into the calculator.

Check that the matrices given below are equivalent to the original system of equations when put into the matrix equation $AX=Y$AX=Y.

To solve the matrix equation, we will make use of the inverse $A^{-1}$A−1 of matrix $A$A. After entering the matrix $A$A into your calculator, you will need to input the command for calculating the inverse.

The following steps show how the equation will be solved.

$AX$AX	$=$=	$Y$Y
$A^{-1}AX$A−1AX	$=$=	$A^{-1}Y$A−1Y
$IX$IX	$=$=	$A^{-1}Y$A−1Y
$X$X	$=$=	$A^{-1}Y$A−1Y

This means that if we start with a matrix equation $AX=Y$AX=Y, then the solution, if it exists, is given by $X=A^{-1}Y$X=A−1Y. That is, we must multiply the column $Y$Y from the left by the inverse of matrix $A$A.

If there is no solution, your calculator will give an error message when you attempt to find the inverse of matrix $A$A. It may inform you that $A$A is singular or not invertible.

The inverse of matrix $A$A is shown below, together with the required matrix multiplication and the solution to the original system of equations.

We see that the solution is $w=1,x=2,y=3,z=4$w=1,x=2,y=3,z=4.

 

In the case of the system of three equations in four unknowns that we began with, we need to perform a sequence of row operations in order to obtain information about the solution set. (There are infinitely many solutions but they are restricted to a particular $1$1-dimensional subspace  of the $4$4-dimensional space defined by the variables.)

The matrix equation we wish to investigate is

We write it in the form of a table and carry out the indicated operations. There are procedures for this specific to your particular model of calculator. This example was created using an traditional spreadsheet.

The final rows of this process show that

$w$w	$=$=	$1$1
$x-1.4z$x−1.4z	$=$=	$-3.6$−3.6
$y+0.4z$y+0.4z	$=$=	$4.6$4.6

In matrix form, this general solution can be written

We only need to choose a value for the parameter $t$t to obtain a particular solution. Observe what happens when $t=4$t=4.

Solving systems of equations using matrices and a graphing calculator

With the power of a graphics calculator, we can quickly use the above method to find a solution to a system of equations. This is achieved by entering the matrices directly into the calculator, then getting the calculator to performing the matrix inversion and multiplication.

$w+3x-2y-5z$w+3x−2y−5z	$=$=	$-19$−19
$2w+x+y-z$2w+x+y−z	$=$=	$3$3
$-w-x+4y+3z$−w−x+4y+3z	$=$=	$21$21
$w+x+y+z$w+x+y+z	$=$=	$10$10

Above is the same system of linear equations we solved earlier in this lesson. Now we are going to solve them again, with the help of the Texas Instrument TI-Nspire CAS Calculator. However, this method will be very similar for any graphics calculator model you are currently using.

In the Calculator section of your graphics calculator, you need to enter a matrix. On this calculator, there is a button that allows you to enter a matrix with a chosen no. of columns and rows.

Remember we previously defined a $4\times4$4×4 co-efficient matrix, which we labeled as $A$A. We now want to enter this matrix into the calculator, so we will first create a matrix with 4 rows and 4 columns.

Once we have entered all the elements of the matrix into the calculator, we now want to store this matrix into the calculator, by pressing CTRL then STO->. You can then choose a letter to represent your matrix. In the picture above, we have chosen the letter $A$A. Press "enter" and this matrix will now be stored in the calculator.

We also need to add matrix $B$B, which is the $4\times1$4×1 column matrix from before, that represented what each of the four equations are equal to. You do this using the exact same method.

Once both matrices are entered into the calculator, we can use the formula $X=A^{-1}B$X=A−1B from before, to find the $4\times1$4×1 solution matrix $X$X.

Enter the formula as above into your calculator and press ENTER. The calculator will then give you the solution to this system of equations.

The column matrix shown above provides the unique solution to this system of equations, as found before, which is $w=1$w=1, $x=2$x=2, $y=3$y=3 and $z=4$z=4.

Practice questions

Question 15

Question 16

Matrix applications

The rules of matrix algebra enable us to deal with some large and complicated sets of relations in a concise form of notation.

In particular, systems of linear equations involving several variables can be written in matrix form. Manipulation of the resulting matrix equation is then an efficient way to obtain information about the solution set of the original equations.

Problems that arise in practical situations are usually expressed in words or possibly in the form of a diagram of some kind. The first step for a mathematical treatment is that of writing a set of (linear) equations that represent the given information using symbols for any unknown or variable quantities. From here, a matrix representation can be constructed.

Worked examples

Question 19

Last week Willard sold $9$9 washing machines, $3$3 refrigerators and $4$4 dishwashers for a total sales revenue of $\$10650$$10650. In the same period, Seeley sold $7$7 washing machines, $4$4 refrigerators and $5$5 dishwashers for a total of $\$10500$$10500. 

What information about the average selling price of each item can be deduced from this data?

The number of items sold was $9+3+4+7+4+5=32$9+3+4+7+4+5=32 and the total revenue was $\$21150$$21150. So, the average price per item was $\frac{21150}{32}\approx$2115032​≈ $\$661$$661. However, we would like to know more about the individual prices of the items, not just the overall average price.

The revenue from the washing machine sales must be the number of machines times the average price of a washing machine, and similarly for the other items. So, we can form two equations letting $w$w be the price of a washing machine, $r$r be the price of a refrigerator and $d$d be the price of a dishwasher.

$9w+3r+4d=10650$9w+3r+4d=10650
$7w+4r+5d=10500$7w+4r+5d=10500

In matrix form this is

In preparation for using the standard row-reduction algorithm, we write the numbers as an array.

We perform the same operations on this array as we would in attempting to solve the original equations simultaneously.

In the following tableau, we multiplied the first equation by $7$7 and the second equation by $9$9. After that, we took the first equation away from the second. These moves were followed by a sequence selected from the following list of allowable row operations:

• two rows can exchange places
• a row can be multiplied by a number
• two rows can be added (subtracted)

The information in the last pair of rows is equivalent to the original equations. It says that if Willard had sold $15$15 washing machines and $1$1 dishwasher his sales revenue would have been $\$11100$$11100 and if Seeley had sold $15$15 refrigerators and $17$17 dishwashers, his revenue would have been $\$19950$$19950.

We appear to be not much closer to finding the solution. We can say, however, that the price of a washing machine is given by $w=\frac{1}{15}\left(11100-d\right)$w=115​(11100−d) and we can also say that the price of a refrigerator is given by $r=\frac{1}{15}\left(19950-17d\right)$r=115​(19950−17d). If we had a value for $d$d, the price of a dishwasher, we would be able to calculate the other two prices.

In fact, there are many solutions to the original pair of equations, one for each possible value of $d$d.

Question 20

In Example $1$1 we found that there was not enough information available to find a unique solution. Suppose we are told that in addition to the sales made by Willard and Seeley, there were sales figures generated by Eleanor. She sold $3$3 washing machines, $3$3 refrigerators and $11$11 dishwashers for a total revenue of  $\$10650$$10650.

Going through the same steps as in Example $1$1, we construct an initial array and perform a series of row operations as shown below. We used a spreadsheet to partly automate the calculations. (It is easy to make errors when doing a large number of steps by hand.)

The row reduction steps that were used are shown to the right of the tableau.

The final three rows of the tableau correspond to the equations, equivalent to the original equations, 

$1w+0r+0d=700$1w+0r+0d=700
$0w+1r+0d=650$0w+1r+0d=650
$0w+0r+1d=600$0w+0r+1d=600

In other words, the average selling prices for the three items were:
washing machine    $\$700$$700
refrigerator               $\$650$$650
dishwasher               $\$600$$600

Practice questions

Question 17

Question 18