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

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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