જો A+B=left[begin{array}{lll}2 & 1 & 2 1 & 2 | vedclass

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(A) આપેલ છે કે, $A+B=\left[\begin{array}{lll} 2 & 1 & 2 \ 1 & 2 & 0 \ 0 & 2 & 2\end{array}ight]$ અને $AB=\left[\begin{array}{lll}1 & 2 & 2 \ 1 &

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$\left[\begin{array}{lll}4 & 6 & 6 \ 3 & 4 & 2 \ 1 & 6 & 3\end{array}ight]$

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(A) આપેલ છે કે, $A+B=\left[\begin{array}{lll} 2 & 1 & 2 \ 1 & 2 & 0 \ 0 & 2 & 2\end{array}ight]$ અને $AB=\left[\begin{array}{lll}1 & 2 & 2 \ 1 & 1 & 0 \ 1 & 2 & 1\end{array}ight]$.આપણે $A^2+B(A+B)$ શોધવાનું છે.નોંધો કે $A^2+B(A+B) = A^2+BA+B^2$.આપણે જાણીએ છીએ કે $(A+B)^2 = (A+B)(A+B) = A^2+AB+BA+B^2$.તેથી, $A^2+BA+B^2 = (A+B)^2 - AB$.પ્રથમ, $(A+B)^2 = \left[\begin{array}{lll} 2 & 1 & 2 \ 1 & 2 & 0 \ 0 & 2 & 2\end{array}ight] 	imes \left[\begin{array}{lll} 2 & 1 & 2 \ 1 & 2 & 0 \ 0 & 2 & 2\end{array}ight] = \left[\begin{array}{lll} 5 & 8 & 8 \ 4 & 5 & 2 \ 2 & 8 & 4\end{array}ight]$ ગણીએ.હવે, $A^2+B(A+B) = \left[\begin{array}{lll} 5 & 8 & 8 \ 4 & 5 & 2 \ 2 & 8 & 4\end{array}ight] - \left[\begin{array}{lll} 1 & 2 & 2 \ 1 & 1 & 0 \ 1 & 2 & 1\end{array}ight] = \left[\begin{array}{lll} 4 & 6 & 6 \ 3 & 4 & 2 \ 1 & 6 & 3\end{array}ight]$.

જો $A+B=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2\end{array}\right]$ અને $AB=\left[\begin{array}{lll}1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1\end{array}\right]$ હોય, તો $A^2+B(A+B)=$

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જો $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$ હોય, તો $A^3 - 4A^2 - 6A$ ની કિંમત શોધો.

જો શ્રેણિક $A = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$ હોય,તો:

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