The inverse of begin{bmatrix} 3 & 5 & 7 2 & | vedclass

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(D) Let $A = \begin{bmatrix} 3 & 5 & 7 \ 2 & -3 & 1 \ 1 & 1 & 2 \end{bmatrix}$.First,calculate the determinant $|A|$:$|A| = 3((-3)(2) - (1)(1)) - 5(

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(D) Let $A = \begin{bmatrix} 3 & 5 & 7 \ 2 & -3 & 1 \ 1 & 1 & 2 \end{bmatrix}$.First,calculate the determinant $|A|$:$|A| = 3((-3)(2) - (1)(1)) - 5((2)(2) - (1)(1)) + 7((2)(1) - (-3)(1))$$|A| = 3(-6 - 1) - 5(4 - 1) + 7(2 + 3)$$|A| = 3(-7) - 5(3) + 7(5) = -21 - 15 + 35 = -1$.Next,find the matrix of cofactors $C_{ij}$:$C_{11} = +((-3)(2) - (1)(1)) = -7$$C_{12} = -((2)(2) - (1)(1)) = -3$$C_{13} = +((2)(1) - (-3)(1)) = 5$$C_{21} = -((5)(2) - (7)(1)) = -3$$C_{22} = +((3)(2) - (7)(1)) = -1$$C_{23} = -((3)(1) - (5)(1)) = 2$$C_{31} = +((5)(1) - (7)(-3)) = 26$$C_{32} = -((3)(1) - (7)(2)) = 11$$C_{33} = +((3)(-3) - (5)(2)) = -19$$Adj(A) = \begin{bmatrix} -7 & -3 & 26 \ -3 & -1 & 11 \ 5 & 2 & -19 \end{bmatrix}^T = \begin{bmatrix} -7 & -3 & 5 \ -3 & -1 & 2 \ 26 & 11 & -19 \end{bmatrix}$.Finally,$A^{-1} = \frac{1}{|A|} Adj(A) = \frac{1}{-1} \begin{bmatrix} -7 & -3 & 5 \ -3 & -1 & 2 \ 26 & 11 & -19 \end{bmatrix} = \begin{bmatrix} 7 & 3 & -5 \ 3 & 1 & -2 \ -26 & -11 & 19 \end{bmatrix}$.Since this result does not match any of the given options,the correct answer is $(d)$.

The inverse of $\begin{bmatrix} 3 & 5 & 7 \\ 2 & -3 & 1 \\ 1 & 1 & 2 \end{bmatrix}$ is

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