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  • If A = matrix 3 1 -1 2 , show that A^2 - 5A + 7I = O. Hence find A^-1.

Q : 13            If   \small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix}​ , show that  A^2-5A+7I=O. Hence find \small A^-^1

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Given \small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix} then we have to show the relation A^2-5A+7I=0

So, calculating each term;

A^2 = \begin{bmatrix} 3& 1\\ -1& 2 \end{bmatrix}\begin{bmatrix} 3&1 \\ -1& 2 \end{bmatrix} = \begin{bmatrix} 9-1 &3+2 \\ -3-2&-1+4 \end{bmatrix} = \begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix}

therefore  A^2-5A+7I;

=\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+7 &&5-5+0 \\ -5+5+0 && 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 &&0 \\ 0 && 0 \end{bmatrix}

Hence A^2-5A+7I = 0.

\therefore A.A -5A = -7I

\Rightarrow A.A(A^{-1}) - 5AA^{-1} = -7IA^{-1} 

[Post multiplying by A^{-1}, also |A| \neq 0]

\Rightarrow A(AA^{-1}) - 5I = -7A^{-1}

\Rightarrow AI - 5I = -7A^{-1}

\Rightarrow -\frac{1}{7}(AI - 5I)= \frac{1}{7}(5I-A)

\therefore A^{-1} = \frac{1}{7}(5\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}-\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}) = \frac{1}{7}\begin{bmatrix} 2 &-1 \\ 1& 3 \end{bmatrix}

 

Posted by

Divya Prakash Singh

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