Let A=left[begin{array}{lll}2 & 0 & 1 1 & 1 | vedclass

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(A) Given $A = \begin{bmatrix} 2 & 0 & 1 \ 1 & 1 & 0 \ 1 & 0 & 1 \end{bmatrix}$.Since $AB = [AB_1, AB_2, AB_3]$,we have $AB = \begin{bmatrix} 1 & 2

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(A) Given $A = \begin{bmatrix} 2 & 0 & 1 \ 1 & 1 & 0 \ 1 & 0 & 1 \end{bmatrix}$.Since $AB = [AB_1, AB_2, AB_3]$,we have $AB = \begin{bmatrix} 1 & 2 & 3 \ 0 & 3 & 2 \ 0 & 0 & 1 \end{bmatrix}$.We know that $|AB| = |A| |B|$.First,calculate $|A| = 2(1-0) - 0(1-0) + 1(0-1) = 2 - 1 = 1$.Calculate $|AB| = 1(3-0) - 2(0-0) + 3(0-0) = 3$.Since $|A| |B| = |AB|$,we have $1 	imes |B| = 3$,so $\alpha = |B| = 3$.To find $B$,we use $B = A^{-1} (AB)$.$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{1} \begin{bmatrix} 1 & 0 & -1 \ -1 & 1 & 1 \ -1 & 0 & 2 \end{bmatrix}$.$B = \begin{bmatrix} 1 & 0 & -1 \ -1 & 1 & 1 \ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \ 0 & 3 & 2 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 2 \ -1 & 1 & 0 \ -1 & -2 & -1 \end{bmatrix}$.The diagonal elements of $B$ are $1, 1, -1$. Thus,$\beta = 1 + 1 - 1 = 1$.Finally,$\alpha^3 + \beta^3 = 3^3 + 1^3 = 27 + 1 = 28$.

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