QUESTION 6.2 Adjoint matrix of a $3 \times 3$ matrix Choose one $\cdot 4$ points Calculate the adjoint matrix of \[ M=\left(\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 2 & 1 & 4 \end{array}\right), \operatorname{adj}(M) \] You are given the following cofactors: \[ c_{11}=7, c_{12}=2, c_{13}=-4, c_{21}=-5 \] You will only need to find : \[ c_{31}=?, c_{32}=?, c_{33}=\text { ? } \]
Solution
Step 1 ：Define the matrix M as \(M = \left(\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 2 & 1 & 4 \end{array}\right)\)
Step 2 ：Define a function to calculate the determinant of a 2x2 matrix
Step 3 ：Calculate the cofactors \(c_{31}\), \(c_{32}\), \(c_{33}\) using the formula \(c_{ij} = (-1)^{i+j} \times \text{det}(M_{ij})\) where \(M_{ij}\) is the matrix obtained by removing the i-th row and j-th column from M
Step 4 ：Calculate \(c_{31} = (-1)^{3+1} \times \text{det}(M_{1:3, 2:4}) = -4\)
Step 5 ：Calculate \(c_{32} = (-1)^{3+2} \times \text{det}(M_{1:3, \{1, 3\}}) = -1\)
Step 6 ：Calculate \(c_{33} = (-1)^{3+3} \times \text{det}(M_{1:3, 1:3}) = 2\)
Step 7 ：The cofactors \(c_{31}\), \(c_{32}\), and \(c_{33}\) are \(-4\), \(-1\), and \(2\) respectively. So, \(c_{31} = \boxed{-4}\), \(c_{32} = \boxed{-1}\), and \(c_{33} = \boxed{2}\)
