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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Matrices
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If $A=\begin{bmatrix}1&-2\\4&5\end{bmatrix}$ and $f(1)=t^2-3t+7$ then $f(A)+\begin{bmatrix}3&6\\-12&-9\end{bmatrix}$ is equal to

$\begin{array}{1 1}(a)\;\begin{bmatrix}1&0\\0&1\end{bmatrix}&(b)\;\begin{bmatrix}0&0\\0&0\end{bmatrix}\\(c)\;\begin{bmatrix}0 &1\\1&0\end{bmatrix}&(d)\;\begin{bmatrix}1&1\\0&0\end{bmatrix}\end{array}$

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1 Answer

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Given :
$A=\begin{bmatrix}1&-2\\4&5\end{bmatrix}$
$f(t)=t^2-3t+7$
Now $A^2=\begin{bmatrix}1&-2\\4&5\end{bmatrix}\begin{bmatrix}1&-2\\4&5\end{bmatrix}$
$\qquad\;\;\;=\begin{bmatrix}-7&-12\\24&17\end{bmatrix}$
Now $f(A)=A^2-3A+7$
$\Rightarrow \begin{bmatrix}-7&-12\\24&17\end{bmatrix}-3\begin{bmatrix}1&-2\\4&5\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}-3&-6\\12&9\end{bmatrix}$
$f(A)+\begin{bmatrix}3&6\\-12&-9\end{bmatrix}=\begin{bmatrix}-3&-6\\12&9\end{bmatrix}+\begin{bmatrix}3&6\\-12&-9\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
Hence (b) is the correct answer.
answered Nov 22, 2013 by sreemathi.v
 

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