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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Determinants
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If $A=\begin{bmatrix}3&4\\2&4\end{bmatrix},B=\begin{bmatrix}-2&-2\\0&-2\end{bmatrix}$ then $(A+B)^{-1}$ is equal to

$\begin{array}{1 1}(a)\;is\;a\;skew-symmetric\;matrix\\(b)\;A^{-1}+B^{-1}\\(c)\;does\;not\;exist\\(d)\;none\;of\;these\end{array}$

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1 Answer

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we have $A+B=\begin{bmatrix}3&4\\2&4\end{bmatrix}+\begin{bmatrix}-2&-2\\0&-2\end{bmatrix}$
$\qquad\qquad\quad\;\;=\begin{bmatrix}1&2\\2&2\end{bmatrix}$
$\mid A+B\mid=\begin{vmatrix}1&2\\2&2\end{vmatrix}$
$\Rightarrow 2-4$
$\Rightarrow -2$
Also $adj (A+B)=\begin{bmatrix}2&-2\\-2&1\end{bmatrix}$
$(A+B)^{-1}=-\large\frac{1}{2}$$\begin{bmatrix}2&-2\\-2&1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}-1&1\\1&-\large\frac{1}{2}\end{bmatrix}$
$A^{-1}=\large\frac{1}{4}$$\begin{bmatrix}4&-4\\-2&3\end{bmatrix}$
$B^{-1}=\large\frac{1}{4}$$\begin{bmatrix}-2&2\\0&-2\end{bmatrix}$
$A^{-1}+B^{-1}=\begin{bmatrix}1&-1\\-\large\frac{1}{2}&\large\frac{3}{4}\end{bmatrix}+\begin{bmatrix}-\large\frac{1}{2}&\large\frac{1}{2}\\0&-\large\frac{1}{2}\end{bmatrix}$
$\Rightarrow \begin{bmatrix}\large\frac{1}{2}&-\large\frac{1}{2}\\-\large\frac{1}{2}&\large\frac{1}{4}\end{bmatrix}$
$\therefore (A+B)\neq A^{-1}+B^{-1}$
Hence (d) is the correct answer.
answered Nov 25, 2013 by sreemathi.v
 

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