# $\text{Let A = } \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \text{and B = } \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}. \text{Verify that } (AB^{-1}) = B^{-1}A^{-1}$

Toolbox:
• (i)A matrix is said to be singular if $|A|$ =0.
• (ii)A matrix is said to be invertible only if $|A|\neq 0$.
• (iii) $A^{-1}=\frac{1}{|A|}adj \;A$
• (iv)The adjoint of a square matrix A=[$a_{ij}]_{n\times n}$ is defined as the transpose of the matrix$[A_{ij}]n\times n$
• where $A_{ij}$ is the cofactor of the element $[a_{ij}].$
Given$A=\begin{bmatrix}3 & 7\\2 & 5\end{bmatrix}$

The value of determinant of A is

$|A|=3\times 5-7\times 2$

$\;\;\;=15-14=1.$

The adj A of a square matrix of order 2 can also be obtained by interchanging $a_{11}$ and $a_{11}$ and by changing signs of $a_{12}$ and $a_{21}$

(i.e)$adj \;A=\begin{bmatrix}5 & -7\\2 & 3\end{bmatrix}$

Now $A^{-1}=\frac{1}{|A|}(adj A)$,we know |A|=1.

$A^{-1}=1\begin{bmatrix}5 & -7\\2 & 3\end{bmatrix}$

Similarly $B=\begin{bmatrix}6 & 8\\7 & 9\end{bmatrix}$

The value of the determinant B is

$|B|=6\times 9-8\times 7=54-56=-2.$

We can obtain the adjoint of B as we did before,

$(adj B)=\begin{bmatrix}9 & -8\\-7 & 6\end{bmatrix}$

Hence $B^{-1}=\frac{1}{|B|}(adj B),$we know |B|=-2.

$B^{-1}=\frac{1}{-2}\begin{bmatrix}9 & -8\\-7 & 6\end{bmatrix}$

$\;\;\;=\begin{bmatrix}\frac{-9}{2} & 4\\\frac{7}{2} & 3\end{bmatrix}$

Now let us obtain $B^{-1}A^{-1}$

Matrix multiplication can be done by multiplying the rows and corresponding columns.

$B^{-1}A^{-1}=\begin{bmatrix}\frac{-9}{2} & -4\\\frac{7}{2} & 3\end{bmatrix}\begin{bmatrix}5 & -7\\-2 & 3\end{bmatrix}$

$\;\;\;=\begin{bmatrix}\frac{-45}{2}-8 & \frac{63}{2}\\\frac{35}{2}+6 & \frac{-49}{2}-9\end{bmatrix}=\begin{bmatrix}\frac{-61}{2} & 87/2\\\frac{47}{2} & \frac{-67}{2}\end{bmatrix}$

Hence $B^{-1}A^{-1}=\begin{bmatrix}-61/2 & 87/2\\47/2 & -67/2\end{bmatrix}$-------(1)

Let us obtain $(AB)=\begin{bmatrix}3 & 7\\2 & 5\end{bmatrix}\begin{bmatrix}6 & 8\\7 & 9\end{bmatrix}$

We can multiply these two matrices as we did before

$\;\;\;=\begin{bmatrix}18+49 & 24+63\\12+35 & 16+45\end{bmatrix}$

$\;\;\;=\begin{bmatrix}67 & 87\\47 & 61\end{bmatrix}$

Now |AB| can be obtained as follows :

$|AB|=67\times 61-87\times 47$

$\;\;\;=4087-4089=-2.$

Hence |AB|=-2.

We can obtain adj(AB) as before:

$adj(AB)=\begin{bmatrix}61 & -87\\-47 & 67\end{bmatrix}$

Therefore $(AB)^{-1}=\frac{1}{|AB|}adj \;AB$

$\;\;\;=\frac{1}{-2}\begin{bmatrix}61 & -87\\-47 & 67\end{bmatrix}$

$\;\;\;=\begin{bmatrix}-61/2 & 87/2\\47/2 & -67/2\end{bmatrix}$----(2)

Hence from equ(1) and equ(2) we have

$B^{-1}A^{-1}=(AB)^{-1}$

Hence the result is verified.

edited Feb 28, 2013