Browse Questions

# If $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ write $A^{-1}$ in terms of A.

Toolbox:
• A matrix is said to be singular if $|A|$ =0.
• A matrix is said to be invertible only if $|A|\neq 0$.
• $A^{-1}=\frac{1}{|A|}adj \;A$
• 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 $[a_{ij}].$
Step 1:
$A=\begin{bmatrix}2 &3\\5&-2\end{bmatrix}$
$\mid A\mid=-4-15=-19$
$\neq 0$
Hence inverse exists
Step 2:
Adj A=$\begin{bmatrix}-2 &5\\3&2\end{bmatrix}$
$A^{-1}=\large\frac{1}{\mid A\mid}$$(adj A) \qquad=\large\frac{1}{-19}$$\begin{bmatrix}-2&5\\3 & 2\end{bmatrix}$