The element in the first row and third column | vedclass

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(D) Let $A = \begin{bmatrix} 1 & 2 & -3 \ 0 & 1 & 2 \ 0 & 0 & 1 \end{bmatrix}$.First,we find the determinant $|A| = 1(1(1) - 2(0)) - 2(0(1) - 2(0))

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$7$

Vedclass · Answer

(D) Let $A = \begin{bmatrix} 1 & 2 & -3 \ 0 & 1 & 2 \ 0 & 0 & 1 \end{bmatrix}$.First,we find the determinant $|A| = 1(1(1) - 2(0)) - 2(0(1) - 2(0)) + (-3)(0(0) - 1(0)) = 1(1) - 0 + 0 = 1$.Next,we find the cofactor matrix $C$ where $C_{ij} = (-1)^{i+j} M_{ij}$.$C_{11} = +|\begin{smallmatrix} 1 & 2 \ 0 & 1 \end{smallmatrix}| = 1$,$C_{12} = -|\begin{smallmatrix} 0 & 2 \ 0 & 1 \end{smallmatrix}| = 0$,$C_{13} = +|\begin{smallmatrix} 0 & 1 \ 0 & 0 \end{smallmatrix}| = 0$.$C_{21} = -|\begin{smallmatrix} 2 & -3 \ 0 & 1 \end{smallmatrix}| = -2$,$C_{22} = +|\begin{smallmatrix} 1 & -3 \ 0 & 1 \end{smallmatrix}| = 1$,$C_{23} = -|\begin{smallmatrix} 1 & 2 \ 0 & 0 \end{smallmatrix}| = 0$.$C_{31} = +|\begin{smallmatrix} 2 & -3 \ 1 & 2 \end{smallmatrix}| = 4 - (-3) = 7$,$C_{32} = -|\begin{smallmatrix} 1 & -3 \ 0 & 2 \end{smallmatrix}| = -(2 - 0) = -2$,$C_{33} = +|\begin{smallmatrix} 1 & 2 \ 0 & 1 \end{smallmatrix}| = 1$.Thus,$adj(A) = C^T = \begin{bmatrix} 1 & -2 & 7 \ 0 & 1 & -2 \ 0 & 0 & 1 \end{bmatrix}$.Since $A^{-1} = \frac{1}{|A|} adj(A)$ and $|A| = 1$,we have $A^{-1} = \begin{bmatrix} 1 & -2 & 7 \ 0 & 1 & -2 \ 0 & 0 & 1 \end{bmatrix}$.The element in the first row and third column is $7$.

The element in the first row and third column of the inverse of the matrix $\begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}$ is

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