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(C) Given $A = \begin{bmatrix} 1 & 1 & 3 \ 1 & 7 & 9 \ 2 & 3 & 7 \end{bmatrix}$.We need to find $\operatorname{Tr}(A^2-A)$.First,calculate $A-I$:$A-

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$152$

Vedclass · Answer

(C) Given $A = \begin{bmatrix} 1 & 1 & 3 \ 1 & 7 & 9 \ 2 & 3 & 7 \end{bmatrix}$.We need to find $\operatorname{Tr}(A^2-A)$.First,calculate $A-I$:$A-I = \begin{bmatrix} 1 & 1 & 3 \ 1 & 7 & 9 \ 2 & 3 & 7 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 3 \ 1 & 6 & 9 \ 2 & 3 & 6 \end{bmatrix}$.Now,calculate $A(A-I)$:$A^2-A = \begin{bmatrix} 1 & 1 & 3 \ 1 & 7 & 9 \ 2 & 3 & 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 3 \ 1 & 6 & 9 \ 2 & 3 & 6 \end{bmatrix}$.The diagonal elements of the resulting matrix are:$d_{11} = (1)(0) + (1)(1) + (3)(2) = 0 + 1 + 6 = 7$.$d_{22} = (1)(1) + (7)(6) + (9)(3) = 1 + 42 + 27 = 70$.$d_{33} = (2)(3) + (3)(9) + (7)(6) = 6 + 27 + 42 = 75$.The trace is the sum of the diagonal elements:$\operatorname{Tr}(A^2-A) = 7 + 70 + 75 = 152$.

If $A=\begin{bmatrix} 1 & 1 & 3 \\ 1 & 7 & 9 \\ 2 & 3 & 7 \end{bmatrix}$,then $\operatorname{Tr}(A^2-A) = $

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