Question

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

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asked 2021-02-21
Solve the given system of equations by matrix equation.
5x-4y=4
3x-2y=3

Answers (1)

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

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