Solve by using matrix method. 4x 3y = 5 6x + 2y = 40

To solve the given system of linear equations using the matrix method (inverse matrix method), we follow these steps:

Given Equations:

$$\{\begin{array}{l}4x-3y=5\text{(1)}\\ 6x+2y=40\text{(2)}\end{array}$$

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

The system can be represented as: $$\left[\begin{array}{cc}4& -3\\ 6& 2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5\\ 40\end{array}\right]$$ Here, $A=\left[\begin{array}{cc}4& -3\\ 6& 2\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\end{array}\right]$, and $B=\left[\begin{array}{c}5\\ 40\end{array}\right]$.

Step 2: Find the determinant of $A$

$$|A|=\left|\begin{array}{cc}4& -3\\ 6& 2\end{array}\right|=(4)(2)-(-3)(6)=8+18=26$$ Since $|A|\ne 0$, the inverse 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: $$A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$$ Applying this: $$A^{-1}=\frac{1}{26}\left[\begin{array}{cc}2& 3\\ -6& 4\end{array}\right]$$

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

$$X=A^{-1}B=\frac{1}{26}\left[\begin{array}{cc}2& 3\\ -6& 4\end{array}\right]\left[\begin{array}{c}5\\ 40\end{array}\right]$$

Calculate the product: $$\left[\begin{array}{c}(2)(5)+(3)(40)\\ (-6)(5)+(4)(40)\end{array}\right]=\left[\begin{array}{c}10+120\\ -30+160\end{array}\right]=\left[\begin{array}{c}130\\ 130\end{array}\right]$$

Now multiply by $\frac{1}{26}$: $$X=\frac{1}{26}\left[\begin{array}{c}130\\ 130\end{array}\right]=\left[\begin{array}{c}\frac{130}{26}\\ \frac{130}{26}\end{array}\right]=\left[\begin{array}{c}5\\ 5\end{array}\right]$$

Final Solution:

$$x=5,y=5$$

Answer: The solution to the system is $\boxed{x=5,y=5}$.