# If$A=\begin{bmatrix} 5 & 2 \\7 & 3 \end{bmatrix}$ and$B=\begin{bmatrix} 2 & -1 \\-1 & 1 \end{bmatrix}$ verify that $$(AB)^{-1}=B^{-1}A^{-1}$$

Toolbox:
• Let $A = [ a_{ij} ]$ be a square matrix x. Let $A_{ij}$ be the cofactor of $a_{ij}$. Then $[ A_{ij}]$ is the matrix of cofactors and $adj\: A$ ( or adjoint of the matrix A) is given by $adj\: A=[A_{ij}]^T$
• Let A be a square matrix of order 2. $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Then $adj\: A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ ( shortcut for finding the adjoint in multiple choice questions / objective questions )
• A determinant can be expanded by using the elements of any row or column.
• The inverse of a non-singular square matrix A is given by $A^{-1} = \frac{1}{|A|} adj\: A.$ A non-singular matrix is one whose determinant value is nonzero.
Step 1
AB = $\begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$
$= \begin{bmatrix} (5)(2)+(2)(-1) & (5)(-1)+(2)(1) \\ (7)(2)+(3)(-1) & (7)(-1)+(3)(1) \end{bmatrix}$
$= \begin{bmatrix} 10-2 & 5+2 \\ 14-3 & -7+3 \end{bmatrix} = \begin{bmatrix} 8 & -3 \\ 11 & -4 \end{bmatrix}$
$(AB)^T = \begin{bmatrix} 8 & 11 \\ -3 & -4 \end{bmatrix}$
Step 2
To find the inverses
$|A| = \begin{vmatrix} 5 & 2 \\ 7 & 3 \end{vmatrix} = 15-14=1$
$adj\: A = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$
$A^{-1} = \frac{1}{|A|} adj\: A = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$
$|B| = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} = 2-1 = 1$
$adj\: B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$
$B^{-1} = \frac{1}{|B|} adj\: B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$
$|AB| = \begin{vmatrix} 8 & -3 \\ 11 & -4 \end{vmatrix} = -32+33=1$
$adj\: AB = \begin{bmatrix} -4 & 3 \\ -11 & 8 \end{bmatrix}$
$(AB)^{-1} = \begin{bmatrix} -4 & 3 \\ -11 & 8 \end{bmatrix}$
Step 3
$B^{-1}\: A^{-1} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$
$= \begin{bmatrix} (1)(3)+(1)(-7) & (1)(-2)+(1)(5) \\ (1)(3)+(2)(-7) & (1)(-2)+(2)(5) \end{bmatrix}$
$= \begin{bmatrix} 3-7 & -2+5 \\ 3-14 & -2+10 \end{bmatrix} = \begin{bmatrix} -4 & 3 \\ -11 & 8 \end{bmatrix}$
From step 2 $(AB)^{-1} = B^{-1}A^{-1}$

edited May 23, 2013