Solve the following equations using Matrix method: x + 3y = 13 x + y = 7

Solving the simultaneous equations by the matrix method

The given linear equations are

$$\begin{array}{rl}x+3y& =13\text{(1)}\\ x+y& =7\text{(2)}\end{array}$$

1. Write in matrix form

$$\underset{A}{\underbrace{\left[\begin{array}{cc}1& 3\\ 1& 1\end{array}\right]}}\underset{X}{\underbrace{\left[\begin{array}{c}x\\ y\end{array}\right]}}=\underset{B}{\underbrace{\left[\begin{array}{c}13\\ 7\end{array}\right]}}$$

Thus $AX=B$.

2. Find the inverse of the coefficient matrix $A$

Determinant

$$\det (A)=(1)(1)-(1)(3)=1-3=-2\ne 0$$

Since the determinant is non‑zero, $A$ is invertible.

Adjoint matrix

$$\mathrm{adj}(A)=\left[\begin{array}{cc}1& -3\\ -1& 1\end{array}\right]$$

Inverse

$$A^{-1}=\frac{1}{\det (A)}\mathrm{adj}(A)=\frac{-1}{2}\left[\begin{array}{cc}1& -3\\ -1& 1\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{2}& \frac{3}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]$$

3. Multiply the inverse with the constant matrix

$$X=A^{-1}B=\left[\begin{array}{cc}-\frac{1}{2}& \frac{3}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}13\\ 7\end{array}\right]$$

$$\begin{array}{rl}x& =\left(-\frac{1}{2}\right)13+\left(\frac{3}{2}\right)7=-6.5+10.5=4.0\\ y& =\left(\frac{1}{2}\right)13+\left(-\frac{1}{2}\right)7=6.5-3.5=3.0\end{array}$$

4. Final solution

$$\boxed{x=4,y=3}$$

Result: $x=4$ and $y=3$.