Question

# Solve the simultaneous Linear equations using matrix inverse Method. 8x+3y=2 6x+2y=4

Matrices
Solve the simultaneous Linear equations using matrix inverse Method.
8x+3y=2
6x+2y=4

2021-03-03

Step 1
To solve simultaneous linear equations, we have to find to form the equations into three matrices.
8x+3y=2
6x+2y=4
These can be shown in matrix form as,
$$\begin{bmatrix}8 & 3 \\6& 2 \end{bmatrix}\begin{bmatrix}x \\y \end{bmatrix}=\begin{bmatrix}2 \\4 \end{bmatrix}$$
Step 2
So if Ax=B , where A,x and B are matrices.
Then we can multiply it with $$A^{-1}$$ to find x
$$A^{-1} \cdot Ax = A^{-1} \cdot B \Rightarrow x=A^{-1}B$$
So find $$A^{-1} , A=\begin{bmatrix}8 & 3 \\6& 2 \end{bmatrix} , x=\begin{bmatrix}x \\y \end{bmatrix} , B=\begin{bmatrix}2 \\4 \end{bmatrix}$$
$$\begin{bmatrix}8 & 3&|&1&0 \\6& 2&|&0&1 \end{bmatrix} R_1 \Rightarrow \frac{R_1}{8} , \ \ R_2 \Rightarrow R_2-6R_1$$ ,

Applying these row operations we get,
$$\begin{bmatrix}1 & \frac{3}{8}&|&\frac{1}{8}&0 \\0& \frac{-1}{4}&|&\frac{-3}{4}&1 \end{bmatrix} \ \ R_2 \Rightarrow -4R_2 , \ \ R_1 \Rightarrow R_1-\frac{3}{8}R_2$$
$$\begin{bmatrix}1 & 0&|&-1&\frac{3}{2} \\0& 1&|&3&-4 \end{bmatrix} , \text{ So } A^{-1}=\begin{bmatrix}-1&\frac{3}{2} \\3&-4 \end{bmatrix}$$
Step 3
$$A^{-1}AX=A^{-1}B \Rightarrow x=A^{-1}B$$
$$\begin{bmatrix}x \\y \end{bmatrix}=\begin{bmatrix}-1&\frac{3}{2} \\3&-4 \end{bmatrix}\begin{bmatrix}2 \\4 \end{bmatrix}$$
$$\begin{bmatrix}x \\y \end{bmatrix}=\begin{bmatrix}-2+\frac{3}{2}\times4 \\6+-4 \times 4 \end{bmatrix}=\begin{bmatrix}-2+6 \\6-16 \end{bmatrix}=\begin{bmatrix}4 \\-10 \end{bmatrix}$$

x=5 , y=-10
Step 4
So the answers are x = 4 and y = -10