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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Determinants
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If $A=\begin{bmatrix}1&2\\3&-5\end{bmatrix}$ then $A^{-1}$=

$\begin{array}{1 1}(A)\;\begin{bmatrix}-5&-2\\-3&1\end{bmatrix}\\(B)\;\begin{bmatrix}5/11&2/11\\3/11&-1/11\end{bmatrix}\\(C)\;\begin{bmatrix}-5/11&-2/11\\-3/11&1/11\end{bmatrix}\\(D)\;\begin{bmatrix}5&2\\3&-1\end{bmatrix}\end{array}$

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1 Answer

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Given $A=\begin{bmatrix}1&2\\3&-5\end{bmatrix}$
$\Rightarrow det A=-5-6$
$\Rightarrow -11$
$adj A=\begin{bmatrix}-5&-3\\-2&1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}-5&-2\\-3&1\end{bmatrix}$
Here $A^{-1}=\large\frac{1}{det A}$$adj\;A$
$\Rightarrow -\large\frac{1}{11}$$\begin{bmatrix}-5&-2\\-3&1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}5/11&2/11\\3/11&-1/11\end{bmatrix}$
Hence (B) is the correct answer.
answered Apr 23, 2014 by sreemathi.v
 

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