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$A=\begin{bmatrix}1&0&0\\0&1&1\\0&-2&4\end{bmatrix}$ and $I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$ and $A^{-1}=[\large\frac{1}{6}$$(A^2+cA+dI)]$ then the value of $c$ and $d$ are

$(a)\;(-6,-11)\qquad(b)\;(6,11)\qquad(c)\;(-6,11)\qquad(d)\;(6,-11)$

1 Answer

We have $A=\begin{bmatrix}1 &0&0\\0&1&1\\0&-2&4\end{bmatrix}$
$I=\begin{bmatrix}1&0&0\\0 &1&0\\0&0&1\end{bmatrix}$
$A^{-1}=\large\frac{1}{6}$$[A^2+cA+dI]$
Multiply A on both the side of the equation
$\Rightarrow 6AA^{-1}=A^3+cA^2+dAI$
$\Rightarrow A^3+cA^2+dA-6I=0$
We find that $A^2=\begin{bmatrix}1 &0&0\\0&-1&5\\0&-10&14\end{bmatrix}$
We find that $A^3=\begin{bmatrix}1 &0&0\\0&-11&19\\0&-38&46\end{bmatrix}$
$\therefore A^3+cA^2+dA-6I=0$
$\Rightarrow \begin{bmatrix}1 &0&0\\0 &-11&19\\0 &-38&46\end{bmatrix}+c\begin{bmatrix}1&0&0\\0&-1&5\\0&-10&14\end{bmatrix}+d\begin{bmatrix}1&0&0\\0&1&1\\0&-2&4\end{bmatrix}-6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=0$
$\Rightarrow \begin{bmatrix}1+c+d-6&0&0\\0&-11-c+d-6&19+5c+d\\0&-38-10c-2d&46+14c+4d-6\end{bmatrix}$
$\Rightarrow \begin{bmatrix}0 &0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\Rightarrow 1+c+d-6=0$
$-11-c+d-6=0$
$\Rightarrow c+d=5$
$-c+d=17$
On solving we get $c=-6,d=11$
Hence (c) is the correct answer.
answered Nov 22, 2013 by sreemathi.v
edited Mar 19, 2014 by sharmaaparna1
 

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