Q.If $A = \begin{pmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{pmatrix}$, find $A^{-1}$ and use it to solve the following system of equations: $x - 2y = 10$, $\ 2x - y - z = 8$, $\ -2y + z = 7$.

Concept: Inverse Matrix Method – For a system $AX = B$, if $A$ is invertible, then $X = A^{-1}B$.

Step 1 – Find $A^{-1}$

First compute $\det(A)$:

Since $\det(A)=1 \neq 0$, $A^{-1}$ exists. Find the adjugate (cofactor matrix transpose):

Cofactors:

$C_{11}=(-1-2)=-3$, $C_{12}=-(2-0)=-2$, $C_{13}=(-4-0)=-4$

$C_{21}=-(-2-0)=2$, $C_{22}=(1-0)=1$, $C_{23}=-(-2-0)=2$

$C_{31}=(2-0)=2$, $C_{32}=-(-1-0)=1$, $C_{33}=(-1+4)=3$

So adjugate matrix (transpose of cofactor matrix) is:

Since $\det(A)=1$, $A^{-1} = \text{adj}(A)$:

Step 2 – Write system in matrix form …

The core idea: a system of linear equations can be written as a single matrix equation $A\mathbf{x} = \mathbf{b}$. If $A$ is invertible, multiplying both sides by $A^{-1}$ gives $\mathbf{x} = A^{-1}\mathbf{b}$ — a clean, one-shot solution. No substitution, no elimination; just compute the inverse and multiply.

Here the coefficient matrix $A$ is exactly the $3 \times 3$ matrix given. The constant vector $\mathbf{b}$ comes from the right-hand sides: $10$, $8$, $7$. So the system is:

We need $A^{-1}$ first.

1. Find the determinant of $A$

For a $3 \times 3$ matrix, expand along a row with zeros to save work. Row 1 has a zero in column 3, so expand along row 1:

Compute each $2 \times 2$ determinant:

• First: $(-1)(1) - (-1)(-2) = -1 - 2 = -3$
• Second: $(2)(1) - (-1)(0) = 2 - 0 = 2$

So:

2. Find the matrix of cofactors

For each entry $a_{ij}$, the cofactor is $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor (determinant of the matrix after removing row $i$, column $j$).

Let’s compute systematically:

• $C_{11} = + \det\begin{pmatrix} -1 & -1 \\ -2 & 1 \end{pmatrix} = -3$

• $C_{12} = - \det\begin{pmatrix} 2 & -1 \\ 0 & 1 \end{pmatrix} = - (2) = -2$

• $C_{13} = + \det\begin{pmatrix} 2 & -1 \\ 0 & -2 \end{pmatrix} = (-4 - 0) = -4$

• $C_{21} = - \det\begin{pmatrix} -2 & 0 \\ -2 & 1 \end{pmatrix} = - [(-2)(1) - (0)(-2)] = -(-2) = 2$

• $C_{22} = + \det\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 1$

• $C_{23} = - \det\begin{pmatrix} 1 & -2 \\ 0 & -2 \end{pmatrix} = - [(-2) - 0] = 2$

• $C_{31} = + \det\begin{pmatrix} -2 & 0 \\ -1 & -1 \end{pmatrix} = (2 - 0) = 2$

• $C_{32} = - \det\begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix} = - [(-1) - 0] = 1$

• $C_{33} = + \det\begin{pmatrix} 1 & -2 \\ 2 & -1 \end{pmatrix} = (-1 + 4) = 3$

So the cofactor matrix is:

3. Transpose to get the adjugate

The adjugate (or adjoint) is the transpose of the cofactor matrix:

4. Multiply by $1/\det A$ to get $A^{-1}$

Since $\det A = 1$: …