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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES3 Marks
Question
If $\text{A} = \begin{bmatrix}3 & 1 \\-1 & 2\end{bmatrix},$ show that $A^2 - 5A + 7I = 0$.
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Answer

Given: $\text{A} = \begin{bmatrix}3 & 1 \\-1 & 2\end{bmatrix},$$\therefore \ \text{A}^{2}-5\text{A}+7\text{I}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$= \begin{bmatrix}9-1&3 + 2\\-3-2&-1 + 4\end{bmatrix}-\begin{bmatrix}15 & 5 \\-5 & 10\end{bmatrix} + \begin{bmatrix}7 & 0 \\0 & 7 \end{bmatrix} $
$= \begin{bmatrix}8 & 5 \\-5 & 3 \end{bmatrix} - \begin{bmatrix}15 & 5\\-5& 10 \end{bmatrix}+ \begin{bmatrix}7 & 0 \\0 & 7 \end{bmatrix}$
$=\begin{bmatrix}8-15&5-5\\-5+5&3-10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$=\begin{bmatrix}-7&0\\0&-7\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}-7+7&0+7\\0+7&-7+7\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0 = \text{R.H.S.}$

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