Q.Solve the following system of linear equations using the matrix method: $x - y + z = 4$ $2x + y - 3z = 0$ $x + y + z = 2$

Write the system as $AX=B$:

Determinant: $\det(A)=1(1+3)+1(2+3)+1(2-1)=4+5+1=10\neq0$, so $A^{-1}$ exists.

Adjoint: the cofactors of $A$ give

$\operatorname{adj}(A)=\begin{pmatrix}4 & 2 & 2\\ -5 & 0 & 5\\ 1 & -2 & 3\end{pmatrix},\qquad A^{-1}=\frac{1}{10}\begin{pmatrix}4 & 2 & 2\\ -5 & 0 & 5\\ 1 & -2 & 3\end{pmatrix}.$ …

The core idea here is that a system of linear equations can be written as a single matrix equation: $AX = B$, where $A$ is the coefficient matrix, $X$ is the column of variables, and $B$ is the column of constants. If $A$ is invertible (its determinant is non-zero), we can multiply both sides by $A^{-1}$ to get $X = A^{-1}B$. This is elegant because it turns solving three equations into a single matrix multiplication — once you have the inverse, you have all variables at once.

Let’s set it up.

1. Write the system in matrix form.

   The equations are:

   $x - y + z = 4$

   $2x + y - 3z = 0$

   $x + y + z = 2$

   So:

2. Check if $A$ is invertible — find $\det(A)$. Compute the determinant:

Since $\det(A) = 10 \neq 0$, $A$ is invertible. Good.

3. Find $A^{-1}$ using the adjoint method.

   We need the matrix of cofactors, then its transpose (the adjoint), then divide by $\det(A)$.

   First, find all cofactors $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor (determinant of the submatrix after removing row $i$, column $j$).

   • $C_{11} = + \begin{vmatrix} 1 & -3 \\ 1 & 1 \end{vmatrix} = 1 \cdot 1 - (-3) \cdot 1 = 1 + 3 = 4$
   • $C_{12} = - \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = -(2 \cdot 1 - (-3) \cdot 1) = -(2 + 3) = -5$
   • $C_{13} = + \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = 2 \cdot 1 - 1 \cdot 1 = 2 - 1 = 1$
   • $C_{21} = - \begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix} = -((-1) \cdot 1 - 1 \cdot 1) = -(-1 - 1) = -(-2) = 2$
   • $C_{22} = + \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = 1 \cdot 1 - 1 \cdot 1 = 0$
   • $C_{23} = - \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = -(1 \cdot 1 - (-1) \cdot 1) = -(1 + 1) = -2$
   • $C_{31} = + \begin{vmatrix} -1 & 1 \\ 1 & -3 \end{vmatrix} = (-1) \cdot (-3) - 1 \cdot 1 = 3 - 1 = 2$
   • $C_{32} = - \begin{vmatrix} 1 & 1 \\ 2 & -3 \end{vmatrix} = -(1 \cdot (-3) - 1 \cdot 2) = -(-3 - 2) = -(-5) = 5$
   • $C_{33} = + \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = 1 \cdot 1 - (-1) \cdot 2 = 1 + 2 = 3$

   So the cofactor matrix is:

   $\text{Cof}(A) = \begin{pmatrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{pmatrix}$ …

Method: Choosing Between the Inverse Method and Cramer's Rule

Both techniques solve $AX=B$ once $\det(A) \neq 0$; picking the more efficient one for the numbers in front of you saves time in an exam.

Steps

Step 1: Set up $AX=B$ and compute $\det(A)$

Form the coefficient matrix, the variable column and the constant column exactly as usual, then expand $\det(A)$ along the row/column with the fewest or simplest entries.

Step 2: Decide which finishing technique to use

If you only need one variable, or the matrix has awkward numbers, Cramer's rule (replace one column of $A$ with $B$ and divide by $\det A$) is often faster:

$x=\frac{\det A_x}{\det A}, \quad y=\frac{\det A_y}{\det A}, \quad z=\frac{\det A_z}{\det A}$ …

Common Mistakes

Mistake 1: Forgetting to replace the correct column when using Cramer's rule

Why it's wrong: to find $x$ you replace the FIRST column of $A$ with $B$; students often replace the wrong column or replace a row instead of a column, which computes a completely different determinant. Correct approach: keep track of which column corresponds to which variable and replace only that one column.

Mistake 2: Sign slip in a $3\times3$ cofactor expansion …