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

Concept: Matrix Equation Solving — rewriting the system as $AX = B$, then solving via $X = A^{-1}B$.

First, write the system in matrix form:

Find the inverse of $A$. The determinant is $\det(A) = (5)(2) - (2)(3) = 10 - 6 = 4$.

So

Multiply $A^{-1}$ by $B$: …

Why the matrix method works

A system of linear equations can be written compactly as $A\mathbf{x} = \mathbf{b}$, where $A$ is the coefficient matrix, $\mathbf{x}$ is the column of variables, and $\mathbf{b}$ is the column of constants. If $A$ is invertible (i.e., its determinant is non-zero), we can multiply both sides by $A^{-1}$ to get $\mathbf{x} = A^{-1}\mathbf{b}$. This turns solving into a single matrix multiplication — clean, systematic, and free of substitution or elimination guesswork.

For a $2\times 2$ system, the inverse has a simple closed form, so the entire process is quick and reliable.

Step-by-step solution

1. Write the system in matrix form

The given equations are:

So:

The system is $A\mathbf{x} = \mathbf{b}$.

2. Compute the determinant of $A$

For a $2\times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $\det(A) = ad - bc$.

Since $\det(A) \neq 0$, $A$ is invertible and a unique solution exists.

3. Find the inverse of $A$

For a $2\times 2$ matrix, the inverse is:

Here $a=5,\; b=2,\; c=3,\; d=2$, so:

$A^{-1} = \frac{1}{4} \begin{pmatrix} 2 & -2 \\ -3 & 5 \end{pmatrix}$ …

Method: Solving a 2×2 System by the Matrix Inverse — Standard Case

This is the base-case version of the matrix method for two equations in two unknowns: write the system as $AX=B$, invert $A$ using the $2\times2$ shortcut, then multiply.

Steps

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

Step 2: Compute $\det(A)=ad-bc$

Double-check you multiplied the correct diagonal pairs — $a$ with $d$, and $b$ with $c$ — not $a$ with $c$.

Step 3: Form the inverse

$A^{-1}=\frac{1}{\det(A)}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ …

Common Mistakes

Mistake 1: Swapping the wrong pair of entries when forming $A^{-1}$

Why it's wrong: for $A=\begin{pmatrix}5&2\\3&2\end{pmatrix}$, the shortcut swaps the diagonal entries ($5\leftrightarrow2$ in position, giving $2,\,2$ visually similar here) and negates the OFF-diagonal $2$ and $3$. A student who instead negates the diagonal entries gets $\begin{pmatrix}-5&2\\3&-2\end{pmatrix}$ instead of the correct $\begin{pmatrix}2&-2\\-3&5\end{pmatrix}$ — an easy mix-up here since two of the four entries are already the same number ($2$), making a misplaced sign harder to notice. Correct approach: write the general pattern $\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ next to the specific matrix and substitute entry-by-entry rather than editing the original matrix by eye.

Mistake 2: Computing $\det(A)$ as $ac-bd$ instead of $ad-bc$ …