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# Find the inverse of the matrix (if it exists): $\begin{bmatrix} 2&-2 \\ 4&3 \\ \end{bmatrix}$

$\begin{array}{1 1} -14\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix} \\ 14\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix} \\ \frac{1}{14}\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix} \\ \frac{-1}{14}\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix} \end{array}$

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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 $a_{ij}.$
• Adjoint is denoted by adj A.
Let A=$\begin{bmatrix}2 & -2\\4 & 3\end{bmatrix}$
The value of the determinant ,
|A|=$(2\times 3)-(4\times -2)$
$\;\;\;=6+8=14.$
Since |A|$\neq 0$ inverse exists,
Now let us find the adjoint of the matrix.
$M_{11}=3$
$M_{12}=4$
$M_{21}=-2$
$M_{22}=2$
$A_{11}=(-1)^{1+1}.3=3.$
$A_{12}=(-1)^{1+2}.4=-4.$
$A_{21}=(-1)^{2+1}.(-2)=2.$
$A_{22}=(-1)^{2+2}.2=2.$
Hence (adj A)=$\begin{bmatrix}a_{11} & a_{21}\\a_{12} &a_{22}\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix}$
The inverse of the matrix is $A^{-1}=\frac{1}{|A|}adj \;A$
We have already found |A|=14.
Therefore inverse =$\frac{1}{14}\begin{bmatrix}3& 2\\-4 & 2 \end{bmatrix}$
answered Mar 6, 2013