Question

# Solve the given system of equations by matrix equation. 5x-4y=4 3x-2y=3

Equations
Solve the given system of equations by matrix equation.
5x-4y=4
3x-2y=3

2021-02-23

Step 1
Given,
System of equations is
5x-4y=4
3x-2y=3
system of equations in matrix form can be written as:
$$\displaystyle{A}{X}={B}\Rightarrow{X}={A}^{{-{1}}}{B}$$
Step 2
Here,$$A=\begin{bmatrix}5 & -4 \\3 & -2 \end{bmatrix}, B=\begin{bmatrix}4 \\3 \end{bmatrix}, X=\begin{bmatrix}x \\y \end{bmatrix}$$
Formula for $$\displaystyle{A}^{{-{1}}}$$ is:
$$\displaystyle{A}^{{-{1}}}={\frac{{{1}}}{{{\det{{A}}}}}}$$adj A
Now,
$$\displaystyle{\det{{A}}}={5}{\left(-{2}\right)}-{\left(-{4}\right)}{\left({3}\right)}$$
=-10+12
=2
And,
adj $$A=\begin{bmatrix}-2 & 4 \\-3 & 5 \end{bmatrix}$$
So,
$$A^{-1}=\frac{1}{2}\begin{bmatrix}-2 & 4 \\-3 & 5 \end{bmatrix}$$
Step 3
Thus, the solution of the given system of equations is
$$\displaystyle{X}={A}^{{-{1}}}{B}$$
$$\begin{bmatrix}x \\y \end{bmatrix}=\frac{1}{2}\begin{bmatrix}-2 & 4 \\-3 & 5 \end{bmatrix}\begin{bmatrix}4 \\3 \end{bmatrix}$$
$$\begin{bmatrix}x \\y \end{bmatrix}=\frac{1}{2}\begin{bmatrix}-2(4) + 4(3) \\-3(4) + 5(3) \end{bmatrix}$$
$$\begin{bmatrix}x \\y \end{bmatrix}=\frac{1}{2}\begin{bmatrix}-8 +12 \\-12 +15 \end{bmatrix}$$
$$\begin{bmatrix}x \\y \end{bmatrix}= \frac{1}{2}\begin{bmatrix}4 \\3 \end{bmatrix}$$
$$\begin{bmatrix}x \\y \end{bmatrix}= \begin{bmatrix}2 \\1.5 \end{bmatrix}$$
Step 4
Therefore,
The solution of the given system of equations is
x = 2
y = 1.5