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

The key idea is to solve a system of linear equations using the matrix method, where we write the system as $A \mathbf{x} = \mathbf{b}$ and find $\mathbf{x} = A^{-1} \mathbf{b}$.

Step 1: Write the system in matrix form.

Step 2: Find the inverse of the coefficient matrix. The determinant is $\det(A) = (2)(4) - (-1)(3) = 8 + 3 = 11$. So,

Step 3: Multiply $A^{-1}$ by $\mathbf{b}$. …

We have two equations:

The goal is to find a pair $(x, y)$ that satisfies both simultaneously. The most reliable method here is elimination — we manipulate the equations so that adding them cancels one variable. Why elimination? Because it avoids fractions early and works cleanly with integer coefficients.

1. Choose which variable to eliminate.

   Look at the coefficients: $y$ has $-1$ in equation (1) and $+4$ in equation (2). If we multiply equation (1) by $4$, the $y$ terms become $-4y$ and $+4y$, which cancel when added. That’s a clean move.

2. Multiply equation (1) by 4:

Call this equation (1′).

3. Add equation (1′) to equation (2):

The $-4y$ and $+4y$ cancel perfectly, leaving:

4. Solve for $x$:

5. Substitute $x$ back into one original equation to find $y$. …

Method: Solving a 2×2 System by the Matrix Inverse, With Fractional Values

The matrix method works identically whether the final answer comes out as whole numbers or fractions — the fraction only appears at the very last step, when dividing by the determinant.

Steps

Step 1: Write $AX=B$ from the two equations

Step 2: Compute $\det(A)=ad-bc$ carefully, watching the signs

When $b$ or $c$ is itself negative, $-bc$ can flip from subtraction to addition — write out $ad-bc$ term by term rather than doing it mentally, to avoid a sign slip here.

Step 3: Build $A^{-1}=\dfrac{1}{\det(A)}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$

Keep the $\frac{1}{\det(A)}$ factor attached symbolically rather than dividing early — this keeps the numbers as clean integers for as long as possible. …

Common Mistakes

Mistake 1: Dropping the double-negative when computing $\det(A)$

Why it's wrong: here $\det(A)=(2)(4)-(-1)(3)$; the second term is $-(-1)(3)=+3$, not $-3$, so the correct total is $8+3=11$. A student who forgets the double-negative computes $8-3=5$ instead, and every later step (the inverse, and hence $x,y$) inherits this wrong determinant. Correct approach: expand $-bc$ term by term when $b$ or $c$ is negative, writing $-(-1)(3)$ explicitly rather than simplifying by eye.

Mistake 2: Forgetting to divide by the determinant after building the swap-negate matrix …