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 know that $AA^{-1}=A^{-1}A=I$AA−1=A−1A=I where $A^{-1}$A−1 is the inverse of $A$A and $I$I is the identity matrix.
Division of matrices
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 multiplication by using the inverse, provided the inverse of a matrix exists.
Because $AA^{-1}=A^{-1}A$AA−1=A−1A we can use this property to simplify by carry out pre-multiplication and post-multiplication in equations instead of division. This is because we are aiming to multiply a matrix and its inverse to form the identity matrix.
For instance consider the matrix equation $AX=B$AX=B. To solve for $X$X we need to pre-multiply. This means placing $A^{-1}$A−1 out the front of both sides of the equation.
| $AX$AX | $=$= | $B$B | (Writing down the equation) |
| $A^{-1}AX$A−1AX | $=$= | $A^{-1}B$A−1B | (Pre-multiplying both sides by $A^{-1}$A−1) |
| $IX$IX | $=$= | $A^{-1}B$A−1B | (Using the fact that $A^{-1}A=I$A−1A=I) |
| $X$X | $=$= | $A^{-1}B$A−1B | (Using the fact that $IX=X$IX=X) |
For the matrix equation $XA=B$XA=B, to solve for $X$X we need to post-multiply. This means placing $A^{-1}$A−1 at the end of both sides of the equation.
$XAA^{-1}=BA^{-1}$XAA−1=BA−1 . Here $A^{-1}$A−1 is post-multiplied on both sides.
| $XA$XA | $=$= | $B$B | (Writing down the equation) |
| $XAA^{-1}$XAA−1 | $=$= | $BA^{-1}$BA−1 | (Post-multiplying both sides by $A^{-1}$A−1) |
| $XI$XI | $=$= | $BA^{-1}$BA−1 | (Using the fact that $AA^{-1}=I$AA−1=I) |
| $X$X | $=$= | $BA^{-1}$BA−1 | (Using the fact that $XI=X$XI=X) |
Worked examples
Example 1
Consider the following matrix equation.
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$=$= |  |
Solve the equation for $X$X.
Think: Solving a matrix equation is similar to solving an ordinary equation. We want to first simplify the scalar multiplication on the left, and then move the resulting matrix to the other side of the equation.
Do:
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$=$= | |
(Writing down the equation) | |
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$=$= | |
(Simplifying the scalar multiplication) | |
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$=$= |  |
(Moving the matrix to the other side) | |
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$=$= |  |
(Simplifying the addition) |
Example 2
Consider the following matrix equation.
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$=$= |  |
Solve the equation for $X$X.
Think: Solving a matrix equation is similar to solving an ordinary equation. We want to collect the terms containing $X$X on one side and the matrices on the other side.
Do:
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$=$= | |
(Writing down the equation) | |
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$=$= |  |
(Collecting the terms with $X$X on one side) | |
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$=$= |  |
(Simplifying the addition and subtraction on both sides) | |
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$=$= |  |
(Multiplying both sides by $\frac{1}{2}$12) | |
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$=$= | |
(Simplifying the scalar multiplication on both sides) |
Example 3
Make $X$X the subject of the matrix equation $AXC=B$AXC=B.
Think: We want to make $X$X the subject of the matrix equation by using pre-multiplication and post-multiplication.
Do:
| $AXC$AXC | $=$= | $B$B | (Writing down the equation) |
| $A^{-1}AXC$A−1AXC | $=$= | $A^{-1}B$A−1B | (Pre-multiplying by $A^{-1}$A−1 on both sides) |
| $IXC$IXC | $=$= | $A^{-1}B$A−1B | (Using the fact that $A^{-1}A=I$A−1A=I) |
| $XCC^{-1}$XCC−1 | $=$= | $A^{-1}BC^{-1}$A−1BC−1 | (Post-multiplying by $C^{-1}$C−1 on both sides) |
| $XI$XI | $=$= | $A^{-1}BC^{-1}$A−1BC−1 | (Using the fact that $CC^{-1}=I$CC−1=I) |
| $X$X | $=$= | $A^{-1}BC^{-1}$A−1BC−1 | (Using the fact that $XI=X$XI=X) |
Practice questions
Question 1
$A$A, $B$B and $C$C are matrices such that $AB=C$AB=C. Using matrix algebra, fill in the gaps to solve for matrix $B$B.
Multiply both sides of the equation by the inverse of $\editable{}$: $\left(\editable{}\right)^{-1}\editable{}B=\left(\editable{}\right)^{-1}C$()−1B=()−1C The product of any matrix and its inverse results in the identity matrix: $\editable{}B=\left(\editable{}\right)^{-1}C$B=()−1C The product of any matrix and the identity matrix is the matrix itself: $\editable{}=\left(\editable{}\right)^{-1}C$=()−1C
Question 2
$A$A, $B$B and $C$C are matrices such that $BA-C=0$BA−C=0. Using matrix algebra, fill in the gaps to solve for matrix $B$B.
Perform a reverse operation to eliminate matrix $\editable{}$. $BA-C+\editable{}=0+\editable{}$BA−C+=0+ Simplify both sides of the equation. $BA=\editable{}$BA= Multiply both sides of the equation by the inverse of $\editable{}$. $BA\left(\editable{}\right)^{-1}=C\left(\editable{}\right)^{-1}$BA()−1=C()−1 The product of any matrix and its inverse results in the identity matrix. $B\editable{}=CA^{-1}$B=CA−1 The product of any matrix and the identity matrix is the matrix itself. $\editable{}=CA^{-1}$=CA−1
Question 3
| Let $M$M$=$= |
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, and $N$N$=$= |
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. |
Find $X$X, if $XM=N$XM=N, in its most simplified form.
$X$X$=$= $\editable{}$ $\editable{}$ $\times$× $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $=$= $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
Solving simultaneous equations
It is possible to solve simultaneous equations using matrices. Inverse matrices are the main tool that we use to do this.
Remember that simultaneous equations are two equations that when solved simultaneously (at the same time) provide a solution that is true for each of the equations. You will have already solved these using graphical and algebraic methods. Now we will use matrices.
First, represent the information using matrices.
Take this pair of equations.
To write these using matrices split up the system into $3$3 parts.
- a coefficient matrix (numbers in front of the pronumerals)
- a variable matrix (also referred to as pronumerals)
- an answer matrix (values on the other side of the equals sign)
To verify that this matrix representation is indeed equivalent, let's expand the multiplication.
By equating equivalent positions, the elements $2x+3y$2x+3y must equal $16$16, and $-3x+y=-13$−3x+y=−13.
Let's now look at how to solve it.
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
Recall that when multiplying matrices, the order is important.
This means the inverse of 
is required.
We already know how to find the inverses:
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$=$= |  |
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$=$= |  |
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$=$= |  |
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$=$= |  |
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$=$= |  |
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$=$= |  |
So this means that the solution to
is $x=5$x=5, and $y=2$y=2. If graphing the two equations the intersection point would be $\left(5,2\right)$(5,2) .
Try this yourself before you look at the solution.
What happens when you try and solve, using matrices, a set of equations that have no solution.
Part A - Explain what kind of lines, and hence their system of equations, that would not have a point of intersection (also known as no solution).
Part B - Explore and investigate what happens if you try to solve a set of these equations using matrices.
Practice questions
Question 4
Given the following linear equations:
$x+3y=23$x+3y=23
$4x+2y=32$4x+2y=32
Express the system of equations in matrix form.
$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $×$× $x$x $y$y $=$= $\editable{}$ $\editable{}$ Find the determinant of the coefficient matrix.
Solve the system of equations using matrices.
$x$x $y$y $=$= $\frac{1}{\editable{}}$1 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\times$× $\editable{}$ $\editable{}$ $=$= $\frac{1}{\editable{}}$1 $\editable{}$ $\editable{}$ $=$= $\editable{}$ $\editable{}$
Question 5
Given the following linear equations:
$4x+6y=23$4x+6y=23
$8x+12y=61$8x+12y=61
Express the system of equations in matrix form.
$\editable{}$ $6$6 $\editable{}$ $\editable{}$ $×$× $x$x $y$y $=$= $\editable{}$ $61$61 Calculate the determinant of the coefficient matrix.
The determinant of the coefficient matrix is equal to zero. What can you conclude?
The coefficient matrix does not have an inverse and there are multiple solutions for the system of equations.
AThe coefficient matrix has an inverse and there are multiple solutions for the system of equations.
BThe coefficient matrix has an inverse and there are no solutions for the system of equations.
CThe coefficient matrix does not have an inverse and there is no solution for the system of equations.
D
Question 6
We have two numbers $x$x and $y$y, where $x>y$x>y.
Their sum is $42$42, while their difference is $16$16.
Write two equations that describe the relationship between $x$x and $y$y in the form $ax+by=c$ax+by=c.
Write each equation on the same line, separated by a comma.
Express the system of equations in matrix form.
$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $×$× $x$x $y$y $=$= $\editable{}$ $16$16 Calculate the determinant of the coefficient matrix.
Solve the system of equations using matrices.
$x$x $y$y $=$= $\frac{1}{\editable{}}$1 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\times$× $\editable{}$ $\editable{}$ $=$= $\frac{1}{\editable{}}$1 $\editable{}$ $\editable{}$ $=$= $\editable{}$ $\editable{}$