Matrices — Class 12 Maths Solution

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Given that, $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\{ - 1}&2\end{array}} \right]$

$\Rightarrow$ ${A^2} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\{ - 1}&2\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\{ - 1}&2\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{9 - 1}&{3 + 2}\\{ - 3 - 2}&{ - 1 + 4}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}8&5\\{ - 5}&3\end{array}} \right]$

and $5A = 5\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&1\\{ - 1}&2\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{15}&5\\{ - 5}&{10}\end{array}} \right]$

Now, substituting the values, we have
${A^2} - 5A + 7I = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}8&5\\{ - 5}&3\end{array}} \right] - \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{15}&5\\{ - 5}&{10}\end{array}} \right] + \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}7&0\\0&7\end{array}} \right]$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 7}&0\\0&{ - 7}\end{array}} \right] + \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}7&0\\0&7\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0\\0&0\end{array}} \right] = O$

Hence, proved.