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asked 2021-03-10

Use Cramer's rule to solve the given system of linear equations.

\(\displaystyle{x}_{{{1}}}-{x}_{{{2}}}+{4}{x}_{{{3}}}=-{2}\)

\(\displaystyle-{8}{x}_{{{1}}}+{3}{x}_{{{2}}}+{x}_{{{3}}}={0}\)

\(\displaystyle{2}{x}_{{{1}}}-{x}_{{{2}}}+{x}_{{{3}}}={6}\)

\(\displaystyle{x}_{{{1}}}-{x}_{{{2}}}+{4}{x}_{{{3}}}=-{2}\)

\(\displaystyle-{8}{x}_{{{1}}}+{3}{x}_{{{2}}}+{x}_{{{3}}}={0}\)

\(\displaystyle{2}{x}_{{{1}}}-{x}_{{{2}}}+{x}_{{{3}}}={6}\)

asked 2020-12-25

Use cramer's rule to solve system of linear equations

\(13x-6y=17\)

\(26x-12y=8\)

asked 2021-01-27

\(x_1'=2x_1+x_2\)

\(x_2'=4x_1-x_2\)

Convert this system to a second order differential equations and solve this second order differential equations

asked 2021-03-06

Cramer’s Rule to solve (if possible) the system of linear equations.

\(\displaystyle-{8}{x}_{{1}}+{7}{x}_{{2}}{\mid}-{10}{x}_{{3}}=-{151}\)

\(\displaystyle{12}{x}_{{1}}+{3}{x}_{{2}}-{5}{x}_{{3}}={86}\)

\(\displaystyle{15}{x}_{{1}}-{9}{x}_{{2}}+{2}{x}_{{3}}={187}\)

\(\displaystyle-{8}{x}_{{1}}+{7}{x}_{{2}}{\mid}-{10}{x}_{{3}}=-{151}\)

\(\displaystyle{12}{x}_{{1}}+{3}{x}_{{2}}-{5}{x}_{{3}}={86}\)

\(\displaystyle{15}{x}_{{1}}-{9}{x}_{{2}}+{2}{x}_{{3}}={187}\)

asked 2021-05-19

Use the given inverse of the coefficient matrix to solve the following system

\(5x_1+3x_2=6\)

\(-6x_1-3x_2=-2\)

\(A^{-1}=\begin{bmatrix}-1 & -1 \\2 & \frac{5}{3} \end{bmatrix}\)

\(5x_1+3x_2=6\)

\(-6x_1-3x_2=-2\)

\(A^{-1}=\begin{bmatrix}-1 & -1 \\2 & \frac{5}{3} \end{bmatrix}\)

asked 2021-04-06

Use Cramer’s Rule to solve (if possible) the system of linear equations.

13x-6y=17

26x-12y=8

13x-6y=17

26x-12y=8

asked 2021-02-04

\(\begin{cases}20x+8y=11\\12x-24y=21\end{cases}\)

asked 2021-03-05

Use cramer's rule to solve

\(x_1+2x_2=5\)

\(-x_1+x_2=1\)

\(x_1+2x_2=5\)

\(-x_1+x_2=1\)

asked 2020-11-30

Use Cramer’s Rule to solve (if possible) the system of linear equations.

2x-y=-10

3x+2y=-1

2x-y=-10

3x+2y=-1

asked 2021-02-21

Cramer’s Rule to solve (if possible) the system of linear equations.

\(\displaystyle{\frac{{{5}}}{{{6}}}}{x}_{{1}}-{x}_{{2}}=-{20}\)

\(\displaystyle{\frac{{{3}}}{{{4}}}}{x}_{{1}}-{\frac{{{7}}}{{{2}}}}{x}_{{2}}=-{51}\)

\(\displaystyle{\frac{{{5}}}{{{6}}}}{x}_{{1}}-{x}_{{2}}=-{20}\)

\(\displaystyle{\frac{{{3}}}{{{4}}}}{x}_{{1}}-{\frac{{{7}}}{{{2}}}}{x}_{{2}}=-{51}\)