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

To solve the given system of linear equations using the Matrix Method (Inverse Method), we follow these steps:

Given Equations:

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

Step 1: Write the system in matrix form $AX=B$

The system can be expressed as: $$\left[\begin{array}{cc}1& 3\\ 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}13\\ 7\end{array}\right]$$ Here,

• $A=\left[\begin{array}{cc}1& 3\\ 1& 1\end{array}\right]$ (coefficient matrix),
• $X=\left[\begin{array}{c}x\\ y\end{array}\right]$ (variable matrix),
• $B=\left[\begin{array}{c}13\\ 7\end{array}\right]$ (constant matrix).

Step 2: Find the determinant of $A$

$$|A|=\left|\begin{array}{cc}1& 3\\ 1& 1\end{array}\right|=(1)(1)-(3)(1)=1-3=-2$$ Since $|A|\ne 0$, the inverse of $A$ exists.

Step 3: Find the inverse of $A$

The inverse of a $2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is given by: $$A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$$ Applying this to $A$: $$A^{-1}=\frac{1}{-2}\left[\begin{array}{cc}1& -3\\ -1& 1\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{2}& \frac{3}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]$$

Step 4: Multiply $A^{-1}$ with $B$ to find $X$

$$X=A^{-1}B=\left[\begin{array}{cc}-\frac{1}{2}& \frac{3}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]\left[\begin{array}{c}13\\ 7\end{array}\right]$$ Perform the matrix multiplication: $$x=\left(-\frac{1}{2}\right)(13)+\left(\frac{3}{2}\right)(7)=-\frac{13}{2}+\frac{21}{2}=\frac{8}{2}=4$$ $$y=\left(\frac{1}{2}\right)(13)+\left(-\frac{1}{2}\right)(7)=\frac{13}{2}-\frac{7}{2}=\frac{6}{2}=3$$

Final Solution:

The solution to the system is: $$\boxed{x=4,y=3}$$