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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Matrices
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If $A=\begin{bmatrix}0&1&0\\0&0&1\\1&-1&0\end{bmatrix}$ then $A^3+A$ equals

$\begin{array}{1 1}(A)\;\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\\(B)\;\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}\\(C)\;\begin{bmatrix}0&0&1\\0&0&1\\1&-1&0\end{bmatrix}\\(D)\;\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\end{array}$

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1 Answer

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$A^2=A.A=\begin{bmatrix}0&1&0\\0&0&1\\1&-1&0\end{bmatrix}\begin{bmatrix}0&1&0\\0&0&1\\1&-1&0\end{bmatrix}$
$\Rightarrow \begin{vmatrix}0&0&1\\1&-1&0\\0&1&-1\end{vmatrix}$
$A^3=A^2.A=\begin{bmatrix}0&0&1\\0&-1&0\\0&1&-1\end{bmatrix}\begin{bmatrix}0&1&0\\0&0&1\\1&-1&0\end{bmatrix}$
$\Rightarrow \begin{bmatrix}1&-1&0\\0&1&-1\\-1&1&1\end{bmatrix}$
$A^3+A=\begin{bmatrix}1&-1&0\\0&1&-1\\-1&1&1\end{bmatrix}+\begin{bmatrix}0&1&0\\0&0&1\\1&-1&0\end{bmatrix}$
$\Rightarrow \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
Hence (D) is the correct answer.
answered Apr 23, 2014 by sreemathi.v
 

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