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Inverse of $\begin{bmatrix}2& -1 \\-5 & 3 \end{bmatrix}$ is

\[\begin{array}{1 1}(1)\begin{bmatrix} 2& -1 & \\ -5& 3\end{bmatrix}&(2)\begin{bmatrix} -2 & 5 \\ 1& -3 \end{bmatrix}\\(3)\begin{bmatrix} 3 & -1 \\-5&-3 \end{bmatrix}&(4)\begin{bmatrix} -3 & 5 \\-2&1 \end{bmatrix}\end{array}\]

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Let $A=\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$
$|A|=6-5=1 \neq 0$
$\therefore A^{-1}$ exists
$A^{-1} =\large\frac{1}{|A|}$$ \;adj \;A$
Cofactor matrix of A
$\qquad =\begin{bmatrix} 2 & -5 \\ -1 & +3 \end{bmatrix}$
$adj A=\begin{bmatrix} 2 & -1 \\ -5 & +3 \end{bmatrix}$
$ A^{-1}=\large\frac{1}{1}$$\begin{bmatrix} 2 & -1 \\ -5 & +3 \end{bmatrix}$
$\qquad=\begin{bmatrix} 2 & -1 \\ -5 & +3 \end{bmatrix}$
Hence 1 is the correct answer.
answered May 5, 2014 by meena.p
 

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