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

We solve using the matrix method: write $AX=B$ and find $X=A^{-1}B$.

$\det(A)=(4)(-5)-(-3)(3)=-20+9=-11$.

For a $2\times2$ matrix, $\operatorname{adj}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$, so $\operatorname{adj}(A)=\begin{pmatrix}-5 & 3\\ -3 & 4\end{pmatrix}$ and $A^{-1}=\dfrac{1}{-11}\begin{pmatrix}-5 & 3\\ -3 & 4\end{pmatrix}$. …

The stem asks for the matrix method, so instead of eliminating variables algebraically we package the system as a single matrix equation and invert the coefficient matrix.

The two equations are:

Step 1: Write as $AX=B$.

Step 2: Find $\det(A)$.

Since $\det(A)\neq0$, $A$ is invertible and the system has a unique solution.

Step 3: Find $\operatorname{adj}(A)$.

For a $2\times2$ matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, the adjoint is $\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$. Here:

Step 4: Find $A^{-1}$.

$A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A)=\frac{1}{-11}\begin{pmatrix}-5 & 3\\ -3 & 4\end{pmatrix}.$ …

Method: Solving a 2×2 System When the Determinant is Negative

A negative determinant does not change the method at all — $A^{-1}$ still exists whenever $\det(A)\neq0$ — but it does require extra care when distributing the sign through the inverse.

Steps

Step 1: Write $AX=B$

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

If this comes out negative, that is completely fine — it still means $A$ is invertible and a unique solution exists. The sign of the determinant has no bearing on consistency.

Step 3: Build $A^{-1}=\dfrac{1}{\det(A)}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$, keeping the negative denominator explicit …

Common Mistakes

Mistake 1: Silently dropping the negative sign of the determinant

Why it's wrong: here $\det(A)=(4)(-5)-(-3)(3)=-20+9=-11$, a negative value. Using $\frac{1}{11}$ instead of $\frac{1}{-11}$ when forming $A^{-1}$ flips the sign of every entry in the final solution, turning the correct $x=-\frac{6}{11},\,y=-\frac{19}{11}$ into the wrong $x=\frac{6}{11},\,y=\frac{19}{11}$. Correct approach: carry the negative sign explicitly through $\frac{1}{\det(A)}$ rather than working with its absolute value and fixing signs 'at the end' from memory.

Mistake 2: A sign error subtracting equation (4) from equation (3) during elimination …