Q.Using matrix method, solve the system of equations $3x + 2y - 2z = 3$, $x + 2y + 3z = 6$, $2x - y + z = 2$.

Concept: Inverse Matrix Method – For a system $AX = B$, the solution is $X = A^{-1}B$, provided $A$ is invertible.

Step 1: Write in matrix form

Step 2: Find $A^{-1}$

First, $\det(A) = 3(2\cdot1 - 3(-1)) - 2(1\cdot1 - 3\cdot2) + (-2)(1(-1) - 2\cdot2)$

$= 3(2+3) - 2(1-6) -2(-1-4) = 3(5) -2(-5) -2(-5) = 15 + 10 + 10 = 35$

Cofactor matrix, then adjugate, then divide by $\det$:

Step 3: Multiply $A^{-1}B$ …

The Inverse Matrix Method is a clean, algebraic way to solve a system of linear equations when the number of equations equals the number of unknowns. The idea is simple: if you can write the system as a single matrix equation $AX = B$, and if the coefficient matrix $A$ is invertible (i.e., its determinant is non-zero), then you can multiply both sides by $A^{-1}$ to get $X = A^{-1}B$. This turns the problem of solving for $x, y, z$ into a single matrix multiplication — elegant and systematic.

Let’s apply it step by step.

1. Write the system in matrix form

The given equations are:

This becomes:

So $AX = B$.

2. Check that $A$ is invertible

We need $\det A \neq 0$. Compute the determinant:

Since $\det A = 35 \neq 0$, $A^{-1}$ exists.

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

Recall:

where $\text{adj}(A)$ is the transpose of the cofactor matrix.

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

• $C_{11} = + \begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix} = 2\cdot 1 - 3\cdot(-1) = 2 + 3 = 5$

• $C_{12} = - \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = -(1\cdot 1 - 3\cdot 2) = -(1 - 6) = 5$

• $C_{13} = + \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = 1\cdot(-1) - 2\cdot 2 = -1 - 4 = -5$

• $C_{21} = - \begin{vmatrix} 2 & -2 \\ -1 & 1 \end{vmatrix} = -(2\cdot 1 - (-2)\cdot(-1)) = -(2 - 2) = 0$

• $C_{22} = + \begin{vmatrix} 3 & -2 \\ 2 & 1 \end{vmatrix} = 3\cdot 1 - (-2)\cdot 2 = 3 + 4 = 7$

• $C_{23} = - \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} = -(3\cdot(-1) - 2\cdot 2) = -(-3 - 4) = 7$

• $C_{31} = + \begin{vmatrix} 2 & -2 \\ 2 & 3 \end{vmatrix} = 2\cdot 3 - (-2)\cdot 2 = 6 + 4 = 10$

• $C_{32} = - \begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} = -(3\cdot 3 - (-2)\cdot 1) = -(9 + 2) = -11$

• $C_{33} = + \begin{vmatrix} 3 & 2 \\ 1 & 2 \end{vmatrix} = 3\cdot 2 - 2\cdot 1 = 6 - 2 = 4$

So the cofactor matrix is: …

Method: Solving a System of Linear Equations by the Matrix (Inverse) Method

Use this method whenever a problem explicitly asks you to "solve using the matrix method" a system with as many equations as unknowns.

Steps

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

Collect the coefficients of $x,y,z$ into a matrix $A$, the unknowns into a column $X$, and the constants into a column $B$, so the whole system becomes the single matrix equation $AX=B$.

Step 2: Confirm $A$ is invertible

Compute $|A|$ by cofactor expansion. If $|A|\neq0$, the system has a unique solution and the inverse method applies directly (if $|A|=0$, this method fails, and consistency must instead be checked via $(\operatorname{adj}A)B$).

Step 3: Find $A^{-1}$ via the adjoint

Compute every cofactor $C_{ij}$, transpose to get $\operatorname{adj}(A)$, then …

Common Mistakes

Mistake 1: Dropping or mis-assigning the cofactor sign $(-1)^{i+j}$

Why it's wrong: with nine $2\times2$ minors to compute, it's easy to forget the alternating $+,-,+$ pattern (e.g. treating $C_{12}$ or $C_{32}$ as positive) — a single sign error silently produces a wrong adjoint and a wrong solution even though every individual $2\times2$ minor was computed correctly.

Mistake 2: Using the cofactor matrix directly as the adjoint, without transposing

Why it's wrong: $\operatorname{adj}(A)$ is the transpose of the cofactor matrix, not the cofactor matrix itself. Skipping the transpose swaps off-diagonal entries (e.g. using $C_{12}$ where $C_{21}$ belongs) and gives a matrix that isn't actually $A^{-1}$ once divided by $\det A$. …