Problem

Find the adjoint of the matrix \( A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \).

Answer

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Answer

Finally, the adjoint of \( A \) is the transpose of \( C \), i.e., \( \text{adj}(A) = C^T = \begin{bmatrix} 24 & -30 & 20 \\ 0 & 6 & -5 \\ 0 & 0 & 1 \end{bmatrix} \).

Steps

Step 1 :The adjoint of a matrix is the transpose of the cofactor matrix. So, we first need to find the cofactors of each element in the matrix.

Step 2 :The cofactor of an element \( a_{ij} \) in a matrix is found by removing the i-th row and j-th column, calculating the determinant of the remaining square matrix, and multiplying it by \( (-1)^{i+j} \).

Step 3 :The cofactor matrix \( C \) of \( A \) is given by \( C = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} \), where \( C_{ij} \) is the cofactor of \( a_{ij} \).

Step 4 :For \( A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \), the cofactor matrix \( C \) is \( C = \begin{bmatrix} 24 & 0 & 0 \\ -30 & 6 & 0 \\ 20 & -5 & 1 \end{bmatrix} \).

Step 5 :Finally, the adjoint of \( A \) is the transpose of \( C \), i.e., \( \text{adj}(A) = C^T = \begin{bmatrix} 24 & -30 & 20 \\ 0 & 6 & -5 \\ 0 & 0 & 1 \end{bmatrix} \).

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