Solve system of equations: x + y + z = 6, x - y + z = 2, 2x + y - z = 1 using matrices

Question

Solve the system of equations using matrices (Cramer’s rule or inverse method):

x + y + z = 6

x - y + z = 2

2x + y - z = 1

(CBSE 2024)

Solution — Step by Step

A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 6 \\ 2 \\ 1 \end{pmatrix}

Expand along the first row:

|A| = 1(1 - 1) - 1(-1 - 2) + 1(1 + 2)

= 1(0) - 1(-3) + 1(3) = 0 + 3 + 3 = 6

Since $|A| = 6 \neq 0$ , a unique solution exists.

Cofactor matrix:

$C_{11} = 0$ , $C_{12} = 3$ , $C_{13} = 3$

$C_{21} = 2$ , $C_{22} = -3$ , $C_{23} = 1$

$C_{31} = 2$ , $C_{32} = 0$ , $C_{33} = -2$

\text{adj}(A) = \begin{pmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{pmatrix}

(transpose of the cofactor matrix)

A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{6}\begin{pmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{pmatrix}

X = A^{-1}B = \frac{1}{6}\begin{pmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{pmatrix}\begin{pmatrix} 6 \\ 2 \\ 1 \end{pmatrix}

= \frac{1}{6}\begin{pmatrix} 0 + 4 + 2 \\ 18 - 6 + 0 \\ 18 + 2 - 2 \end{pmatrix} = \frac{1}{6}\begin{pmatrix} 6 \\ 12 \\ 18 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}

Solution: $x = 1, y = 2, z = 3$

Verify: $1 + 2 + 3 = 6$ ✓, $1 - 2 + 3 = 2$ ✓, $2 + 2 - 3 = 1$ ✓

Why This Works

The matrix equation $AX = B$ is the compact form of a linear system. If $A$ is invertible ( $|A| \neq 0$ ), we can multiply both sides by $A^{-1}$ to get $X = A^{-1}B$ — the unique solution.

The adjoint method computes $A^{-1} = \frac{1}{|A|}\text{adj}(A)$ , where $\text{adj}(A)$ is the transpose of the cofactor matrix. Each cofactor is a $2 \times 2$ determinant with appropriate sign.

If $|A| = 0$ , the system either has no solution or infinitely many solutions — the matrix method alone can’t distinguish these cases (you’d need to check the augmented matrix).

Alternative Method — Cramer’s Rule

x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}

where $D = |A| = 6$ , and $D_x, D_y, D_z$ are obtained by replacing the respective column of $A$ with $B$ .

$D_x = \begin{vmatrix} 6 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = 6(0) - 1(-3) + 1(3) = 6$ , so $x = 6/6 = 1$ .

For CBSE boards, the inverse matrix method is preferred (more marks for showing the adjoint). For JEE, Cramer’s rule is faster for 3-variable systems. In either case, always verify the solution by substituting back into ALL three equations — CBSE marking scheme awards marks for verification.

Common Mistake

The most common error in the cofactor method: forgetting to transpose the cofactor matrix to get the adjoint. The cofactor matrix and the adjoint are NOT the same — $\text{adj}(A) = C^T$ . Another frequent mistake is getting signs wrong in cofactors — remember the checkerboard sign pattern: $+, -, +$ / $-, +, -$ / $+, -, +$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Cramer’s Rule

Common Mistake

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