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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices3 Marks
Question
If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$ show that $\mathrm{A}^2-5 \mathrm{~A}+7 \mathrm{I}=0$
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Answer

Given: $A = \left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right]$
$\therefore  A^2 – 5A + 7I   = \left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {9 - 1}&{3 + 2} \\ { - 3 - 2}&{ - 1 + 4} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {15}&5 \\ { - 5}&{10} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 7&0 \\ 0&7 \end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}} 8&5 \\ { - 5}&3 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {15}&5 \\ { - 5}&{10} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 7&0 \\ 0&7 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {8 - 15}&{5 - 5} \\ { - 5 + 5}&{3 - 10} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 7&0 \\ 0&7 \end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}} { - 7}&0 \\ 0&{ - 7} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 7&0 \\ 0&7 \end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}} { - 7 + 7}&{0 + 7} \\ {0 + 7}&{ - 7 + 7} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$
$\therefore $ L.H.S= R.H.S. Hence Proved.

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