14.04 Inverse of a matrix

In mathematics, the inverse of a number is the number needed so that when you multiply them together the result is $1$1. In matrices this would be equivalent to the identity matrix, $I$I.

For instance, $6\times\frac{1}{6}=1$6×16​=1, where $\frac{1}{6}$16​ and $6$6 are inverses of one another. The concept of an inverse is often used when solving equations, like solving $5x=20$5x=20.

$5x$5x	$=$=	$20$20	(Writing down the equation)
$5x\times\frac{1}{5}$5x×15​	$=$=	$20\times\frac{1}{5}$20×15​	(Multiplying by the inverse of $5$5)
$x$x	$=$=	$4$4	(Simplifying both sides)

 

Notation

The inverse of $6$6 can also be written as $6^{-1}$6−1 . As with this example or when using $\sin^{-1}$sin−1 to represent inverse sine, we have that:

$A^{-1}$A−1 represents the inverse of matrix $A$A.

 

The determinant of a $2\times2$2×2 matrix

The determinant is a number that is used to find the inverse of a matrix (if it exists). If the determinant is zero then the inverse is said to be undefined, this can be useful when solving systems of linear equations later.

Generally, the determinant of a matrix $A$A is written as $det(A)$det(A) and is calculated as follows:

For a matrix $A$A$=$=

$a$a

$b$b

, the determinant is $det(A)$det(A)$=$=

$a$a

$b$b

$=ad-cb$=ad−cb

$c$c

$d$d

$c$c

$d$d

Practice question

Question 1

The inverse of a 2 x 2 matrix

To find the inverse of a $2\times2$2×2 matrix, say $A$A, we swap the entries along the main-diagonal, and multiply the entries in the off-diagonal by $-1$−1. Then we multiply the result by $\frac{1}{det(A)}$1det(A)​.

For a matrix $A$A$=$=

$a$a

$b$b

, the inverse is $A^{-1}=$A−1=

$\frac{1}{det(A)}$1det(A)​

$d$d

$-b$−b

$c$c

$d$d

$-c$−c

$a$a

 

The inverse matrix $A^{-1}$A−1 has the property that when we multiply it by $A$A, we get the identity matrix.

$AA^{-1}=A^{-1}A=I$AA−1=A−1A=I

Practice questions

Question 2

Question 3

Question 4

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.

$=$=

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:

	$=$=			(Writing down the equation)
	$=$=			(Simplifying the scalar multiplication)
	$=$=			(Moving the matrix to the other side)
	$=$=			(Simplifying the addition)

Example 2

Consider the following matrix equation.

$=$=

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:

	$=$=			(Writing down the equation)
	$=$=			(Collecting the terms with $X$X on one side)
	$=$=			(Simplifying the addition and subtraction on both sides)
	$=$=			(Multiplying both sides by $\frac{1}{2}$12​)
	$=$=			(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 5

Question 6

Question 7