Question

Use Cramer's rule to solve the given system of linear equations. x_{1}-x_{2}+4x_{3}=-2 -8x_{1}+3x_{2}+x_{3}=0 2x_{1}-x_{2}+x_{3}=6

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ANSWERED
asked 2021-03-10
Use Cramer's rule to solve the given system of linear equations.
\(\displaystyle{x}_{{{1}}}-{x}_{{{2}}}+{4}{x}_{{{3}}}=-{2}\)
\(\displaystyle-{8}{x}_{{{1}}}+{3}{x}_{{{2}}}+{x}_{{{3}}}={0}\)
\(\displaystyle{2}{x}_{{{1}}}-{x}_{{{2}}}+{x}_{{{3}}}={6}\)

Answers (1)

2021-03-12

Step 1
Solution:
Consider the given system is
\(\displaystyle{x}_{{{1}}}-{x}_{{{2}}}+{4}{x}_{{{3}}}=-{2}\)
\(\displaystyle-{8}{x}_{{{1}}}+{3}{x}_{{{2}}}+{x}_{{{3}}}={0}\)
\(\displaystyle{2}{x}_{{{1}}}-{x}_{{{2}}}+{x}_{{{3}}}={6}\)
We have Cramer's Rule for solving the system of equations
\(\displaystyle{x}={\frac{{{D}_{{{x}}}}}{{{D}}}},{y}={\frac{{{D}_{{{y}}}}}{{{D}}}}\ {\quad\text{and}\quad}\ {z}={\frac{{{D}_{{{z}}}}}{{{D}}}}\)
Where D stands for determinant and rest are to be found as
\(D_{x}=\begin{vmatrix}-2 & -1 & 4 \\0 & 3 & 1 \\ 6 & -1 & 1\end{vmatrix}=-2(3+1)+1(0-6)+4(0-18)=-8-6\) \(-72=-86\)
\(D_{y}=\begin{vmatrix}1 & -2 & 4 \\-8 & 0 & 1 \\ 2 & 6 & 1\end{vmatrix}=1(0-6)+2(-8-2)+4(-48-0)=-6-20\)\(-192=-218\)
\(D_{z}=\begin{vmatrix}1 & -1 & -2 \\-8 & 3 & 0 \\ 2 & -1 & 6\end{vmatrix}=1(18-0)+1(-48-0)-2(8-6)=-30-4\)
\(=-34\)
\(D=\begin{vmatrix}1 & -1 & 4 \\-8 & 3 & 1 \\ 2 & -1 & 1\end{vmatrix}=1(3+1)+1(-8-2)+4(8-6)=4-10+8=2\)
Step 2
Therefore
\(\displaystyle{x}={\frac{{{D}_{{{x}}}}}{{{D}}}}={\frac{{-{86}}}{{{2}}}}=-{43}\)
\(\displaystyle{y}={\frac{{{D}_{{{y}}}}}{{{D}}}}={\frac{{-{218}}}{{{2}}}}=-{109}\)
\(\displaystyle{z}={\frac{{{D}_{{{z}}}}}{{{D}}}}={\frac{{-{34}}}{{{2}}}}=-{17}\)

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