# If $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ show that $A^{-1} = \frac{1}{19}A.$

Toolbox:
• Inverse of a $2 \times 2$ matrix is
• $A^{-1}=\frac{1}{ |A| } \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end {bmatrix}$
• The determinant of a square matrix of order $2 \times 2$ is
• $|A|=a_{11} \times a_{22}- a_{12} \times a_{21}$
Given $A=\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$
Inverse of a $2 \times 2$ matrix can be obtained by interchanging the elements of $a_{11}$ and $a_{22}$ and the changing the symbols of $a_{12}\; and\; a_{21}$ and then divide it by |A|
Let us find |A|
$|A|=2 \times -2-3 \times 5$
$=-4-15$
$=-19$
Adj of $A =\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$
Hence $A^{-1}$ is
Therefore $A^{-1}=\frac{1}{-19} \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$
answered Apr 4, 2013 by