Q.Solve the following system of linear equations using the matrix method: $2x + 3y + 3z = 5$ $x - 2y + z = -4$ $3x - y - 2z = 3$

Write the system as $AX=B$:

Determinant: $\det(A)=2(5)-3(-5)+3(5)=10+15+15=40\neq0$.

Adjoint and inverse:

$\operatorname{adj}(A)=\begin{pmatrix}5 & 3 & 9\\ 5 & -13 & 1\\ 5 & 11 & -7\end{pmatrix},\qquad A^{-1}=\frac{1}{40}\begin{pmatrix}5 & 3 & 9\\ 5 & -13 & 1\\ 5 & 11 & -7\end{pmatrix}.$ …

The core idea here is that a system of linear equations can be compactly represented as a single matrix equation. Instead of juggling three equations, we let a matrix hold all the coefficients, a column vector hold the variables, and another column vector hold the constants. The problem then reduces to finding the inverse of the coefficient matrix — a powerful, systematic approach that works for any number of equations.

We have:

This can be written as $AX = B$, where:

If $A$ is invertible, then $X = A^{-1}B$. So our job is to find $A^{-1}$ and multiply.

1. Find the determinant of $A$. We expand along the first row:

Compute each $2 \times 2$ determinant:

• First: $(-2)(-2) - (1)(-1) = 4 + 1 = 5$
• Second: $(1)(-2) - (1)(3) = -2 - 3 = -5$
• Third: $(1)(-1) - (-2)(3) = -1 + 6 = 5$

So:

Since $\det(A) \neq 0$, the matrix is invertible.

2. Find the matrix of cofactors.

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

   • $C_{11} = + \begin{vmatrix} -2 & 1 \\ -1 & -2 \end{vmatrix} = 5$

   • $C_{12} = - \begin{vmatrix} 1 & 1 \\ 3 & -2 \end{vmatrix} = -(-5) = 5$

   • $C_{13} = + \begin{vmatrix} 1 & -2 \\ 3 & -1 \end{vmatrix} = 5$

   • $C_{21} = - \begin{vmatrix} 3 & 3 \\ -1 & -2 \end{vmatrix} = -[ (3)(-2) - (3)(-1) ] = -[ -6 + 3 ] = 3$

   • $C_{22} = + \begin{vmatrix} 2 & 3 \\ 3 & -2 \end{vmatrix} = (2)(-2) - (3)(3) = -4 - 9 = -13$

   • $C_{23} = - \begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} = -[ (2)(-1) - (3)(3) ] = -[ -2 - 9 ] = 11$

   • $C_{31} = + \begin{vmatrix} 3 & 3 \\ -2 & 1 \end{vmatrix} = (3)(1) - (3)(-2) = 3 + 6 = 9$

   • $C_{32} = - \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} = -[ (2)(1) - (3)(1) ] = -[ 2 - 3 ] = 1$

   • $C_{33} = + \begin{vmatrix} 2 & 3 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (3)(1) = -4 - 3 = -7$

   So the cofactor matrix is:

3. Find the adjoint (transpose of the cofactor matrix). …

Method: Solving $AX=B$ Efficiently by Choosing the Best Expansion Row

The arithmetic load of the matrix method depends heavily on which row or column you expand along — choosing well is itself part of the technique.

Steps

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

Form the coefficient matrix $A$, the variable column $X$, and the constants column $B$ directly from the three equations, keeping the variable order ($x$, $y$, $z$) consistent across all three rows.

Step 2: Scan all three rows/columns before expanding $\det(A)$

Look for the row or column with the most zeros or the smallest numbers — expanding along that one, rather than automatically using the first row, cuts down the number of $2\times2$ minors you need.

Step 3: Confirm $\det(A)\neq 0$, then build $\operatorname{adj}(A)$ …

Common Mistakes

Mistake 1: Automatically expanding along the first row instead of scanning for the simplest one

Why it's wrong: expanding along a row/column with large or awkward numbers multiplies out the arithmetic unnecessarily and increases the chance of a slip, compared to a row/column with a zero or a small number. Correct approach: glance at all three rows and columns first, and expand along whichever has the fewest or smallest entries.

Mistake 2: Sign errors when building the cofactor matrix

Why it's wrong: the $(-1)^{i+j}$ pattern alternates by position, and misapplying it to even one cofactor propagates into a wrong adjoint and a wrong final answer. Correct approach: write out the sign pattern explicitly before computing any cofactor, and apply it entry by entry. …