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Matrices — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMatrices3 Marks
Question
If $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right] \& \ I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] ;$ Find $k,$ So that $\mathrm{A}^2=\mathrm{kA} - 2l.$
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Answer

$A^2 = A.A$
$ = \left[ {\begin{array}{*{20}{c}} 3&{ - 2} \\ 4&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&{ - 2} \\ 4&{ - 2} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {9 - 8}&{ - 6 + 4} \\ {12 - 8}&{ - 8 + 4} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ 4&{ - 4} \end{array}} \right]$
$A^2 = kA - 2I.$
$\Rightarrow$$\left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ 4&{ - 4} \end{array}} \right] = k\left[ {\begin{array}{*{20}{c}} 3&{ - 2} \\ 4&{ - 2} \end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$
$\Rightarrow$$\left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ 4&{ - 4} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3k}&{ - 2k} \\ {4k}&{ - 2k} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&0 \\ 0&2 \end{array}} \right]$
$\Rightarrow$$\left[ {\begin{array}{*{20}{c}} 3&{ - 2} \\ 4&{ - 2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3k}&{ - 2k} \\ {4k}&{ - 2k} \end{array}} \right]$
$\Rightarrow 3k = 3$
$\Rightarrow k = 1$

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