Q.Solve the following system of linear equations using the matrix method: $2x + y + z = 1$ $x - 2y - z = \dfrac{3}{2}$ $3y - 5z = 9$

Concept: Matrix Equation Solving – Represent the system as $AX = B$, then solve using $X = A^{-1}B$.

Step 1 – Write in matrix form

Step 2 – Find $A^{-1}$

First, $\det(A) = 2(10+3) - 1(-5-0) + 1(3-0) = 2(13) + 5 + 3 = 34$.

Cofactor matrix:

Adjugate $= C^T$, so

Step 3 – Multiply $A^{-1}B$ …

The matrix method turns a system of linear equations into a single compact 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 (determinant non-zero), we can multiply both sides by $A^{-1}$ to get $X = A^{-1}B$. This gives the solution directly, provided we can compute the inverse correctly.

Let’s set it up.

1. Write the system in matrix form.

   The equations are:

   $2x + y + z = 1$

   $x - 2y - z = \frac{3}{2}$

   $3y - 5z = 9$

   Notice the third equation has no $x$ term — that’s fine; we just put a 0 in the coefficient matrix. So:

2. Check if $A$ is invertible — compute $\det(A)$. Expand along the first row:

Compute each:

• First: $(-2)(-5) - (-1)(3) = 10 + 3 = 13$
• Second: $(1)(-5) - (-1)(0) = -5 - 0 = -5$
• Third: $(1)(3) - (-2)(0) = 3 - 0 = 3$

So $\det(A) = 2(13) - 1(-5) + 1(3) = 26 + 5 + 3 = 34$.

Since $\det(A) \neq 0$, $A^{-1}$ exists.

3. Find the adjoint of $A$.

   The adjoint is the transpose of the cofactor matrix. Compute each cofactor $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor (determinant after removing row $i$, column $j$).

   • $C_{11} = + \det\begin{pmatrix} -2 & -1 \\ 3 & -5 \end{pmatrix} = 13$
   • $C_{12} = - \det\begin{pmatrix} 1 & -1 \\ 0 & -5 \end{pmatrix} = -(-5) = 5$
   • $C_{13} = + \det\begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix} = 3$
   • $C_{21} = - \det\begin{pmatrix} 1 & 1 \\ 3 & -5 \end{pmatrix} = -[ (1)(-5) - (1)(3) ] = -(-5 - 3) = 8$
   • $C_{22} = + \det\begin{pmatrix} 2 & 1 \\ 0 & -5 \end{pmatrix} = (2)(-5) - (1)(0) = -10$
   • $C_{23} = - \det\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} = -[ (2)(3) - (1)(0) ] = -6$
   • $C_{31} = + \det\begin{pmatrix} 1 & 1 \\ -2 & -1 \end{pmatrix} = (1)(-1) - (1)(-2) = -1 + 2 = 1$
   • $C_{32} = - \det\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} = -[ (2)(-1) - (1)(1) ] = -(-2 - 1) = 3$
   • $C_{33} = + \det\begin{pmatrix} 2 & 1 \\ 1 & -2 \end{pmatrix} = (2)(-2) - (1)(1) = -4 - 1 = -5$

   So the cofactor matrix is:

The adjoint (transpose) is: …

Method: Solving a Linear System Using $X = A^{-1}B$

This method solves any system of three linear equations in three unknowns by writing it as a single matrix equation and inverting the coefficient matrix.

Steps

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

Read off the coefficients of $x$, $y$, $z$ from each equation (in order) to form the rows of $A$; if a variable is missing from an equation, its coefficient is $0$ — do not skip forming that entry. Stack the constants on the right into $B$.

Step 2: Compute $\det(A)$ and confirm it is non-zero

Expand along any row or column (whichever has more zeros, to save work). If $\det(A) \neq 0$, $A^{-1}$ exists and the system has a unique solution — proceed. If $\det(A) = 0$, this method fails and you must instead test for consistency using $(\operatorname{adj}A)B$. …

Common Mistakes

Mistake 1: Using the cofactor matrix directly instead of its transpose as the adjoint

Why it's wrong: the adjoint is defined as the transpose of the cofactor matrix, not the cofactor matrix itself — using it un-transposed silently swaps every off-diagonal entry and gives a wrong inverse. Correct approach: always write out the cofactor matrix first, then explicitly transpose it before dividing by $\det(A)$.

Mistake 2: Sign errors in the cofactors …