91Ó°ÊÓ

First, calculate the determinant of matrix \( A \). For a 3x3 matrix \( A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \), the determinant is given by \( |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \).Substitute the values from the given matrix \( A = \begin{bmatrix} 2 & 1 & 1 \ 3 & 2 & 2 \ 1 & 1 & 2 \end{bmatrix} \).Calculate: \[|A| = 2(2 \cdot 2 - 2 \cdot 1) - 1(3 \cdot 2 - 1 \cdot 1) + 1(3 \cdot 1 - 1 \cdot 2)\]\[|A| = 2(4 - 2) - 1(6 - 1) + 1(3 - 2)\]\[|A| = 2(2) - 1(5) + 1(1)\]\[|A| = 4 - 5 + 1 = 0\]So \(|A| = 0\).

Now we need to find the cofactor of each element in matrix \( A \) by using the formula for the cofactor of element \( a_{ij} \) as \( C_{ij} = (-1)^{i+j}M_{ij} \), where \( M_{ij} \) is the minor of \( a_{ij} \) obtained by deleting the \( i \)-th row and \( j \)-th column.Let's calculate for the first row:- \( C_{11} = (-1)^{2} \cdot (2\cdot2 - 2\cdot1) = 2 \)- \( C_{12} = (-1)^{3} \cdot (3\cdot2 - 2\cdot1) = -4 \)- \( C_{13} = (-1)^{4} \cdot (3\cdot1 - 2\cdot1) = 1 \)For the second row:- \( C_{21} = (-1)^{3} \cdot (1\cdot2 - 2\cdot1) = -2 \)- \( C_{22} = (-1)^{4} \cdot (2\cdot2 - 1\cdot1) = 3 \)- \( C_{23} = (-1)^{5} \cdot (2\cdot1 - 1\cdot1) = -1 \)For the third row:- \( C_{31} = (-1)^{4} \cdot (1\cdot2 - 2\cdot2) = -2 \)- \( C_{32} = (-1)^{5} \cdot (2\cdot2 - 1\cdot2) = 2 \)- \( C_{33} = (-1)^{6} \cdot (2\cdot1 - 1\cdot3) = 1 \)Thus, the cofactors matrix is:\[C = \begin{bmatrix} 2 & -4 & 1 \ -2 & 3 & -1 \ -2 & 2 & 1 \end{bmatrix}.\]

The adjugate of a matrix, \( \text{adj}(A) \), is the transpose of the cofactor matrix. Transpose the cofactor matrix obtained in the previous step:\[\text{adj}(A) = C^T = \begin{bmatrix} 2 & -2 & -2 \ -4 & 3 & 2 \ 1 & -1 & 1 \end{bmatrix}.\]

As determined in Step 1, \(|A| = 0\). Thus, we need to check if \( A \times \text{adj}(A) = \text{adj}(A) \times A = 0 \times I \), meaning it should result in the zero matrix. Calculate these two product results:1. Compute \( A \times \text{adj}(A) \):\[A = \begin{bmatrix} 2 & 1 & 1 \ 3 & 2 & 2 \ 1 & 1 & 2 \end{bmatrix},\quad \text{adj}(A) = \begin{bmatrix} 2 & -2 & -2 \ -4 & 3 & 2 \ 1 & -1 & 1 \end{bmatrix}.\]Carrying out the multiplication should result in the zero matrix:\[A \times \text{adj}(A) = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}.\]2. Compute \( \text{adj}(A) \times A \), it should similarly result in the zero matrix.\[\text{adj}(A) \times A = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}.\]Both multiplications confirm the identity \( A(\text{adj}A) = (\text{adj}A)A = |A|I = 0 \times I \).