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

Equations
Use Cramer's rule to solve the given system of linear equations.
$$\displaystyle{x}_{{{1}}}-{x}_{{{2}}}+{4}{x}_{{{3}}}=-{2}$$
$$\displaystyle-{8}{x}_{{{1}}}+{3}{x}_{{{2}}}+{x}_{{{3}}}={0}$$
$$\displaystyle{2}{x}_{{{1}}}-{x}_{{{2}}}+{x}_{{{3}}}={6}$$

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
$$\displaystyle{x}={\frac{{{D}_{{{x}}}}}{{{D}}}}={\frac{{-{86}}}{{{2}}}}=-{43}$$
$$\displaystyle{y}={\frac{{{D}_{{{y}}}}}{{{D}}}}={\frac{{-{218}}}{{{2}}}}=-{109}$$
$$\displaystyle{z}={\frac{{{D}_{{{z}}}}}{{{D}}}}={\frac{{-{34}}}{{{2}}}}=-{17}$$