If A = 3&1 -1&2 , show that A^2 - 5A + 7I = 0. Hence find A^ -1 . - Vidyadip

Question

$\text{If A} = \begin{bmatrix}3&1\\-1&2\end{bmatrix},$ show that $A^2 - 5A + 7I = 0$. Hence find $A^{-1}$.

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Answer

$\text{Given:}\ \text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$ $\therefore\ \text{A}^2=\text{A.A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$
$\text{L.H.S.}=\text{A}^2-5\text{A}+7\text{I}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$=\begin{bmatrix}8-15&5-5\\-5+5&3-10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$=\begin{bmatrix}-7&0\\0&-7\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$=\begin{bmatrix}-7+7&0+0\\0+0&-7+7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0=\text{R.H.S.}\ \Rightarrow\ \text{A}^2-5\text{A}+7\text{I}=0$
To find: $A^{-1}$, multiplying eq. (1) by $A^{-1}$.
$\Rightarrow\ \text{A}^2\text{A}^{-1}-5\text{A.A}^{-1}+7\text{I}_2\text{A}^{-1}=0.\text{A}^{-1}\ \Rightarrow\ \text{A}-5\text{I}_2+7\text{A}^{-1}=0$
$\Rightarrow\ \ 7\text{A}^{-1}=-\text{A}+5\text{I}_2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}+5\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}+\begin{bmatrix}5&0\\0&5\end{bmatrix}=\begin{bmatrix}2&-1\\1&3\end{bmatrix}$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}$

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