જો A = begin{bmatrix} 2 & 1 7 & 4 end{bmatri | vedclass

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(B) આપેલ છે કે $A = \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix}$.પ્રથમ,$A^2 = A 	imes A = \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix} \begin{bmatri

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$\frac{1}{4} \begin{bmatrix} -3 & 1 \ 7 & -1 \end{bmatrix}$

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(B) આપેલ છે કે $A = \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix}$.પ્રથમ,$A^2 = A 	imes A = \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix} \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix} = \begin{bmatrix} 4+7 & 2+4 \ 14+28 & 7+16 \end{bmatrix} = \begin{bmatrix} 11 & 6 \ 42 & 23 \end{bmatrix}$.ત્યારબાદ,$5A = 5 \begin{bmatrix} 2 & 1 \ 7 & 4 \end{bmatrix} = \begin{bmatrix} 10 & 5 \ 35 & 20 \end{bmatrix}$.તેથી,$A^2 - 5A = \begin{bmatrix} 11 & 6 \ 42 & 23 \end{bmatrix} - \begin{bmatrix} 10 & 5 \ 35 & 20 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 7 & 3 \end{bmatrix}$.ધારો કે $B = A^2 - 5A = \begin{bmatrix} 1 & 1 \ 7 & 3 \end{bmatrix}$.નિશ્ચાયક $|B| = (1)(3) - (1)(7) = 3 - 7 = -4$.વ્યસ્ત શ્રેણિક $B^{-1} = \frac{1}{|B|} 	ext{adj}(B) = \frac{1}{-4} \begin{bmatrix} 3 & -1 \ -7 & 1 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -3 & 1 \ 7 & -1 \end{bmatrix}$.

જો $A = \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$ હોય,તો $(A^2 - 5A)^{-1}$ શું થાય?

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