A=begin{bmatrix} 1 & 0 & 1 0 & 1 & 1 0 & 1 | vedclass

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(B) આપેલ છે, $A = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0 \end{bmatrix}$.પ્રથમ, $A^2 = A 	imes A = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1

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$-A^{-1}$

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(B) આપેલ છે, $A = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0 \end{bmatrix}$.પ્રથમ, $A^2 = A 	imes A = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \ 0 & 2 & 1 \ 0 & 1 & 1 \end{bmatrix}$ શોધો.હવે, $A^2 - 2A = \begin{bmatrix} 1 & 1 & 1 \ 0 & 2 & 1 \ 0 & 1 & 1 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \ 0 & 2 & 1 \ 0 & 1 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 0 & 2 \ 0 & 2 & 2 \ 0 & 2 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 1 & -1 \ 0 & 0 & -1 \ 0 & -1 & 1 \end{bmatrix}$.આગળ, $|A| = 1(0-1) - 0 + 1(0-0) = -1$ શોધો.સહ-અવયવ શ્રેણિક $C$ માટે $C_{ij} = (-1)^{i+j} M_{ij}$ છે.$C_{11} = -1, C_{12} = 0, C_{13} = 0$.$C_{21} = 1, C_{22} = 0, C_{23} = -1$.$C_{31} = -1, C_{32} = -1, C_{33} = 1$.$adj(A) = C^T = \begin{bmatrix} -1 & 1 & -1 \ 0 & 0 & -1 \ 0 & -1 & 1 \end{bmatrix}$.$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-1} \begin{bmatrix} -1 & 1 & -1 \ 0 & 0 & -1 \ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 1 \ 0 & 0 & 1 \ 0 & 1 & -1 \end{bmatrix}$.$A^2 - 2A = \begin{bmatrix} -1 & 1 & -1 \ 0 & 0 & -1 \ 0 & -1 & 1 \end{bmatrix}$ અને $-A^{-1} = - \begin{bmatrix} 1 & -1 & 1 \ 0 & 0 & 1 \ 0 & 1 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 1 & -1 \ 0 & 0 & -1 \ 0 & -1 & 1 \end{bmatrix}$ ની સરખામણી કરતા.આમ, $A^2 - 2A = -A^{-1}$.

$A=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \Rightarrow A^2-2A=$

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$3$ ક્રમના ચોરસ શ્રેણિક $B$ માટે, જો $B^T=B^{-1}$ અને $|B|=1$ હોય, તો $|B-I|=$

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