Q.Solve the following system of linear equations using the matrix method: $5x + 2y = 4$ $7x + 3y = 5$

Concept: Matrix Equation Solving — rewriting the system as $A\mathbf{x} = B$ and solving via $\mathbf{x} = A^{-1}B$.

First, write in matrix form:

Find the inverse of $A$. The determinant is:

Since $\det(A) = 1$, the inverse is simply the adjugate:

Multiply $A^{-1}$ by $B$: …

The matrix method is just a compact way of solving linear systems. Instead of juggling equations, we let a single matrix equation do the work. The key idea: if we can find the inverse of the coefficient matrix, we can "divide" both sides by it — multiplying by the inverse isolates the variable vector.

For a system like

we write it as $A\mathbf{x} = \mathbf{b}$, where

If $A$ is invertible (its determinant is non-zero), then $\mathbf{x} = A^{-1}\mathbf{b}$. That's the whole logic — find the inverse, multiply, done.

1. Write the system in matrix form

So $A = \begin{pmatrix} 5 & 2 \\ 7 & 3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}$.

2. Check that $A$ is invertible Compute the determinant:

Since $\det(A) \neq 0$, the inverse exists. This also tells us the system has a unique solution.

3. Find the inverse of $A$ For a $2\times2$ matrix, the inverse formula is:

Here $a=5$, $b=2$, $c=7$, $d=3$, and $\det(A)=1$, so: …

Method: Solving a 2×2 System by the Matrix Inverse Shortcut

For a $2\times2$ system, the inverse of the coefficient matrix has a direct shortcut formula, avoiding cofactor computation altogether.

Steps

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

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

A non-zero determinant guarantees $A^{-1}$ exists and the solution is unique — you may proceed.

Step 3: Build $A^{-1}$ using the swap-and-negate shortcut

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

Common Mistakes

Mistake 1: Negating the wrong pair of entries when building $A^{-1}$

Why it's wrong: the shortcut swaps the DIAGONAL entries ($a\leftrightarrow d$) and negates the OFF-DIAGONAL entries ($b,c\to -b,-c$). For $A=\begin{pmatrix}5&2\\7&3\end{pmatrix}$, this gives $\begin{pmatrix}3&-2\\-7&5\end{pmatrix}$; mixing up which pair gets negated — e.g. negating the diagonal instead — produces $\begin{pmatrix}-3&2\\7&-5\end{pmatrix}$, a completely wrong inverse. Correct approach: always swap the top-left and bottom-right entries first, then attach a minus sign to the two entries that did NOT move.

Mistake 2: Multiplying in the wrong order, $BA^{-1}$ instead of $A^{-1}B$ …