Q.Solve the following system of equations by matrix method. $3x - 2y + 3z = 8$, $2x + y - z = 1$, $4x - 3y + 2z = 4$.

Write the system as $AX=B$ and use $X=A^{-1}B$.

Determinant. $\det A=3(-1)+2(8)+3(-10)=-3+16-30=-17\neq0$.

Adjoint. The cofactor matrix is $\begin{bmatrix}-1&-8&-10\\-5&-6&1\\-1&9&7\end{bmatrix}$, so

$\operatorname{adj}A=\begin{bmatrix}-1&-5&-1\\-8&-6&9\\-10&1&7\end{bmatrix},\qquad A^{-1}=\frac{1}{-17}\operatorname{adj}A.$ …

Set up

If $\det A\neq0$ the unique solution is $X=A^{-1}B$.

Step 1 — Determinant

Expanding along row 1,

Since $\det A=-17\neq0$, a unique solution exists.

Step 2 — Cofactors

Cofactor matrix: $\begin{bmatrix}-1&-8&-10\\-5&-6&1\\-1&9&7\end{bmatrix}$.

Step 3 — Adjoint and inverse

Transposing, …

Method: Solving a 3-Variable System by the Matrix (Adjoint) Method

This method solves a system of three linear equations in three unknowns by writing it as $AX=B$ and computing $X=A^{-1}B$ using the adjoint, since no simple $2\times2$-style shortcut exists at this size.

Steps

Step 1: Write the system as $AX=B$

Collect coefficients into a $3\times3$ matrix $A$, unknowns into $X=(x,y,z)^T$, constants into $B$.

Step 2: Compute $\det(A)$ by cofactor expansion

Expand along whichever row or column has the most convenient entries (zeros or small numbers). If $\det(A)=0$, the matrix method fails here — check consistency by another route instead.

Step 3: Compute all nine cofactors $C_{ij}$

Delete each row/column pair, evaluate the resulting $2\times2$ minor, and attach the checkerboard sign $(-1)^{i+j}$.

Step 4: Transpose the cofactor matrix to get $\operatorname{adj}(A)$

Step 5: Form $A^{-1}$ and multiply by $B$

$X=A^{-1}B=\frac{1}{\det(A)}\operatorname{adj}(A)\cdot B$ …

Common Mistakes

Mistake 1: A sign or arithmetic slip in one of the nine cofactors, carried through to a wrong inverse

Why it's wrong: computing $\det(A)$ and all nine cofactors for a $3\times3$ system involves many small $2\times2$ determinants — a single sign error (e.g. mishandling a negative entry like $-2$ or $-1$ in the coefficient matrix) silently produces a wrong adjoint and a wrong final solution, with no obvious warning sign. Correct approach: after finding $X$, substitute back into all three original equations — a genuine arithmetic slip almost always fails at least one of the three checks.

Mistake 2: Dividing by $\det(A)$ too early, creating messy fractions that then get mis-simplified …