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

Find $A^{-1}$. For $A=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix}$, expanding along row 1 gives $|A|=1(-1-2)-2(-2-0)+0=-3+4=1$. The cofactor matrix is $\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}$, so

Relate the system to $A$. With all variables written out, $x-2y+0z=10,\ 2x-y-z=8,\ 0x-2y+z=7$, the coefficient matrix is

$M=\begin{bmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{bmatrix}=A^{T}.$ …

Step 1 — Determinant of $A$

For $A=\begin{bmatrix}1&2&0\\-2&-1&-2\\0&-1&1\end{bmatrix}$, expanding along the first row,

so $A^{-1}$ exists.

Step 2 — Cofactors and adjoint

The adjoint is the transpose of the cofactor matrix, and $|A|=1$, so

Step 3 — Match the system to $A$

Write each equation with all three variables:

The coefficient matrix is

This is the key observation: the system is $A^{T}X=B$, not $AX=B$. Reading off $A^{-1}$ and multiplying by $B$ directly would solve the wrong system.

Step 4 — Solve using $(A^{-1})^{T}$ …

Method: Solving a Linear System with the Inverse Matrix Method — Matching the Coefficient Matrix Carefully

Use this method whenever you're given a matrix $A$ (and asked to find $A^{-1}$), then a separate system of equations to solve "using" it — the key extra step beyond a routine inverse-matrix solve is confirming which matrix actually equals the system's coefficient matrix.

Steps

Step 1: Find $A^{-1}$ using the adjoint

Compute $|A|$, the cofactor matrix, and $\operatorname{adj}(A)$, then assemble $A^{-1}=\dfrac{1}{|A|}\operatorname{adj}(A)$. Do this once, before touching the system.

Step 2: Write the system with every variable shown explicitly

Rewrite each equation so every variable appears with an explicit coefficient, including any that are "missing" (write them with coefficient $0$). This exposes the true coefficient matrix — skipping this step is the most common source of error in this question type.

Step 3: Compare the coefficient matrix against $A$ …

Common Mistakes

Mistake 1: Assuming the given matrix $A$ is automatically the system's coefficient matrix

Why it's wrong: writing the three equations in full ($x-2y+0z=10$, $2x-y-z=8$, $0x-2y+z=7$) gives a coefficient matrix that is actually $A^{T}$, not $A$ — the rows and columns of $A$ have been swapped relative to the equations. Using $X=A^{-1}B$ directly here solves the wrong system entirely. Correct approach: always write out the coefficient matrix explicitly from the equations (filling in $0$s for missing variables) and compare it to $A$ before deciding whether to use $A^{-1}$ or $(A^{-1})^T$.

Mistake 2: Forgetting to include zero coefficients for missing variables …