Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 23/Solution 1

1st entry: $a_{11}=2$, the determinant of the minor matrix is $-1$, and the sign for this entry is $(-1)^{1+1}=1\longrightarrow -1.$

2nd entry: $a_{12}=1$, the determinant of the minor matrix is $2$, and the sign for this entry is $(-1)^{1+2}=-1\longrightarrow -2.$

2nd entry: $a_{13}=3$, the determinant of the minor matrix is $1$, and the sign for this entry is $(-1)^{1+3}=1\longrightarrow 1.$

Continue with this method, we have the cofactor matrix ${\begin{bmatrix}-1&-2&1\\1&4&-2\\1&1&-1\end{bmatrix}}.$

Transpose the cofactor matrix to find the adjugate/adjoint matrix: ${\begin{bmatrix}-1&1&1\\-2&4&1\\1&-2&-1\end{bmatrix}}.$

We now find $\det(A)$:

$\det(A)=2\times (0-1)-1\times (2-3)+0=-1$.

Therefore,

$A^{-1}=(-1){\begin{bmatrix}-1&1&1\\-2&4&1\\1&-2&-1\end{bmatrix}}={\begin{bmatrix}1&-1&-1\\2&-4&-1\\-1&2&1\end{bmatrix}}.$