Q.If matrix $A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 3 & -3 \\ -2 & -4 & -4 \end{bmatrix}$, then find $A^{-1}$. OR Solve the system of equations by matrix method: $2x + 3y + 3z = 5$; $x - 2y + z = -4$; $3x - y - 2z = 3$.

$\det A=-8\ne0$, so $A^{-1}=\dfrac{1}{\det A}\,\text{adj}(A)$. Computing the cofactors and transposing gives $A^{-1}=\dfrac18\begin{bmatrix}24&8&12\\ -10&-2&-6\\ -2&-2&-2\end{bmatrix}$.

Concept. $A^{-1}=\dfrac{1}{|A|}\,\text{adj}(A)$, where $\text{adj}(A)$ is the transpose of the cofactor matrix. This requires $|A|\ne 0$. (Main part; the OR alternative is solved separately below.)

Determinant. For $A=\begin{bmatrix}1&1&3\\ 1&3&-3\\ -2&-4&-4\end{bmatrix}$:

$|A|=1(3\cdot(-4)-(-3)(-4))-1(1\cdot(-4)-(-3)(-2))+3(1\cdot(-4)-3\cdot(-2))$

$=1(-12-12)-1(-4-6)+3(-4+6)=-24+10+6=-8.$

Cofactors $C_{ij}$.

$C_{11}=-24,\ C_{12}=10,\ C_{13}=2,$

$C_{21}=-8,\ C_{22}=2,\ C_{23}=2,$

$C_{31}=-12,\ C_{32}=6,\ C_{33}=2.$

Adjoint (transpose of the cofactor matrix):

$\text{adj}(A)=\begin{bmatrix}-24&-8&-12\\ 10&2&6\\ 2&2&2\end{bmatrix}.$

Inverse. …