If A = begin{bmatrix} 3 & 2 0 & 1 end{bmatri | vedclass

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(A) Given $A = \begin{bmatrix} 3 & 2 \ 0 & 1 \end{bmatrix}$.First,find the determinant: $|A| = (3)(1) - (2)(0) = 3$.Next,find the adjoint matrix: $Ad

Vedclass · Accepted Answer

$\frac{1}{27} \begin{bmatrix} 1 & -26 \ 0 & 27 \end{bmatrix}$

Vedclass · Answer

(A) Given $A = \begin{bmatrix} 3 & 2 \ 0 & 1 \end{bmatrix}$.First,find the determinant: $|A| = (3)(1) - (2)(0) = 3$.Next,find the adjoint matrix: $Adj(A) = \begin{bmatrix} 1 & -2 \ 0 & 3 \end{bmatrix}$.Thus,$A^{-1} = \frac{1}{|A|} Adj(A) = \frac{1}{3} \begin{bmatrix} 1 & -2 \ 0 & 3 \end{bmatrix}$.Now,calculate $(A^{-1})^2 = \left( \frac{1}{3} \begin{bmatrix} 1 & -2 \ 0 & 3 \end{bmatrix} ight) \left( \frac{1}{3} \begin{bmatrix} 1 & -2 \ 0 & 3 \end{bmatrix} ight) = \frac{1}{9} \begin{bmatrix} 1 & -8 \ 0 & 9 \end{bmatrix}$.Finally,$(A^{-1})^3 = (A^{-1})^2 \cdot A^{-1} = \frac{1}{9} \begin{bmatrix} 1 & -8 \ 0 & 9 \end{bmatrix} \cdot \frac{1}{3} \begin{bmatrix} 1 & -2 \ 0 & 3 \end{bmatrix} = \frac{1}{27} \begin{bmatrix} 1 & -2 - 24 \ 0 & 27 \end{bmatrix} = \frac{1}{27} \begin{bmatrix} 1 & -26 \ 0 & 27 \end{bmatrix}$.

If $A = \begin{bmatrix} 3 & 2 \\ 0 & 1 \end{bmatrix}$,then $(A^{-1})^3$ is equal to:

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