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

Question
Forms of linear equations
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
\(\displaystyle\triangle={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{8}&{7}&-{10}\backslash{12}&{3}&-{5}\backslash{15}&-{9}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={1149}\)
since \(\displaystyle\triangle\ne{q}{0}\),there exits a unique solution for the given system.
\(\displaystyle\triangle_{{{x}{1}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{151}&{7}&-{10}\backslash{86}&{3}&-{5}\backslash{187}&-{9}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={1149}\)
\(\displaystyle\triangle_{{{x}{2}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{8}&-{151}&-{10}\backslash{12}&{86}&-{5}\backslash{15}&{187}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}=-{3447}\)
\(\displaystyle\triangle_{{{x}{3}}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{8}&{7}&-{151}\backslash{12}&{3}&{86}\backslash{15}&-{9}&{187}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}=-{3447}\)
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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