Browse Questions

# Find the inverse of the matrix (if it exists): $\begin{bmatrix} -1&5 \\ -3&2 \\ \end{bmatrix}$

$\begin{array}{1 1} \frac{-1}{13}\begin{bmatrix}2& -5\\3 & -1 \end{bmatrix} \\ \frac{1}{13}\begin{bmatrix}2& -5\\3 & -1 \end{bmatrix} \\ 13\begin{bmatrix}2& -5\\3 & -1 \end{bmatrix} \\ -13\begin{bmatrix}2& -5\\3 & -1 \end{bmatrix} \end{array}$

Toolbox:
• (i)A square matrix is said to be singular if $|A|$ =0.
• (ii)A square matrix is invertible only if it is a non-singular matrix.
• (iii) $A^{-1}=\frac{1}{|A|}adj \;A$
• (iv)the adjoint of a square matrix A=[$a_{ij}]_{n\times n}$
• where $A_{ij}$ is the cofactor of the element $a_{ij}.$
Let A=$\begin{bmatrix}-1 & 5\\-3 & 2\end{bmatrix}$
The value of the determinant ,
|A|=$-1\times 2-5\times -3$
$\;\;\;=-2-(-15)=13.$
Since |A|$\neq 0$ inverse exists,
Now let us find the adjoint of the matrix.
$M_{11}=2$
$M_{12}=-3$
$M_{21}=5$
$M_{22}=-1$
$A_{11}=(-1)^{1+1}.2=2.$
$A_{12}=(-1)^{1+2}.(-3)=3.$
$A_{21}=(-1)^{2+1}.(5)=-5.$
$A_{22}=(-1)^{2+2}.(-1)=-1.$
Hence (adj A)=$\begin{bmatrix}A_{11} & A_{21}\\A_{12} &A_{22}\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}2& -5\\3 & -1 \end{bmatrix}$
The inverse of the matrix is $A^{-1}=\frac{1}{|A|}adj \;A$
Therefore inverse of A is $\frac{1}{13}\begin{bmatrix}2& -5\\3 & -1 \end{bmatrix}$