Question

Cramer’s Rule to solve (if possible) the system of linear equations. -8x_1+7x_2|-10x_3=-151 12x_1+3x_2-5x_3=86 15x_1-9x_2+2x_3=187

Forms of linear equations
ANSWERED
asked 2021-03-06
Cramer’s Rule to solve (if possible) the system of linear equations.
\(\displaystyle-{8}{x}_{{1}}+{7}{x}_{{2}}{\mid}-{10}{x}_{{3}}=-{151}\)
\(\displaystyle{12}{x}_{{1}}+{3}{x}_{{2}}-{5}{x}_{{3}}={86}\)
\(\displaystyle{15}{x}_{{1}}-{9}{x}_{{2}}+{2}{x}_{{3}}={187}\)

Answers (1)

2021-03-07

Given system of equations
\(\displaystyle-{8}{x}_{{1}}+{7}{x}_{{2}}{\mid}-{10}{x}_{{3}}=-{151}\)
\(\displaystyle{12}{x}_{{1}}+{3}{x}_{{2}}-{5}{x}_{{3}}={86}\)
\(\displaystyle{15}{x}_{{1}}-{9}{x}_{{2}}+{2}{x}_{{3}}={187}\)
This implies that
\(\triangle=\begin{bmatrix}-8 & 7 & -10 \\12 & 3 & -5 \\15 & -9 & 2\end{bmatrix}=1149\)
since \(\displaystyle\triangle\ne{q}{0}\),there exits a unique solution for the given system.
\(\triangle_{x1}=\begin{bmatrix}-151 & 7 & -10 \\86 & 3 & -5 \\187 & -9 & 2\end{bmatrix}=1149\)
\(\triangle_{x2}=\begin{bmatrix}-8 & -151 & -10 \\12 & 86 & -5 \\15 & 187 & 2\end{bmatrix}=-3447\)
\(\triangle_{x3}=\begin{bmatrix}-8 & 7 & -151 \\12 & 3 & 86 \\15 & -9 & 187\end{bmatrix}=-3447\)
By cramer's rule
\(\displaystyle{x}_{{1}}={\frac{{\triangle{x}_{{1}}}}{{\triangle}}},{x}_{{2}}={\frac{{\triangle{x}_{{2}}}}{{\triangle}}}{\quad\text{and}\quad}{x}_{{3}}={\frac{{\triangle{x}_{{3}}}}{{\triangle}}}\)
\(\displaystyle{x}_{{1}}={\frac{{{11490}}}{{{1149}}}},,{x}_{{2}}={\frac{{\triangle-{3447}}}{{{1149}}}}{\quad\text{and}\quad}{x}_{{3}}={\frac{{\triangle{5745}}}{{{1149}}}}\)
\(\displaystyle{x}_{{1}}={10},{x}_{{2}}=-{3}{\quad\text{and}\quad}{x}_{{3}}={5}\)
Thus,the solution is \(\displaystyle{\left({x}_{{1}},{x}_{{2}},{x}_{{3}}\right)}={\left({10},-{3},{5}\right)}\)

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