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Q17. If

$A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ and

$I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$ ,

find k so that $A^{2} = kA - 2I$ .

Answers (1)

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$A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

$A \times A= \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ $\times \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$A^{2} = \begin{bmatrix}9-8 &-6+4\\12-8&-8+4 \end{bmatrix}$

$A^{2} = \begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}$

$A^{2} = kA - 2I$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}=$ $k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} -$ $2 \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}=$ $k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} -$ $\begin{bmatrix}2 &0\\0&2 \end{bmatrix}$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}+$ $\begin{bmatrix}2 &0\\0&2 \end{bmatrix}$ $=k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$\begin{bmatrix}1+2 &-2+0\\4+0&-4+2 \end{bmatrix}$ $=\begin{bmatrix}3k&-2k\\4k&-2k \end{bmatrix}$

$\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ $=\begin{bmatrix}3k&-2k\\4k&-2k \end{bmatrix}$

We have, $3=3k$

$k=\frac{3}{3}=1$

Hence, the value of k is 1.

Posted by

seema garhwal

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