જો $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ હોય,તો સાબિત કરો કે $A^{2} - 5A + 7I = 0$.

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(A) આપેલ છે કે $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$.
પ્રથમ,$A^{2} = A \cdot A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ ની ગણતરી કરીએ.
$A^{2} = \begin{bmatrix} 3(3) + 1(-1) & 3(1) + 1(2) \\ -1(3) + 2(-1) & -1(1) + 2(2) \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$.
હવે,પદાવલિ $L.H.S. = A^{2} - 5A + 7I$ ની કિંમત શોધીએ.
$L.H.S. = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
$L.H.S. = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$.
$L.H.S. = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5-(-5)+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$.
આમ,$A^{2} - 5A + 7I = 0$ સાબિત થાય છે.

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