# How to solve the linear system with Cramer's Rule?

## Consider the linear system: $2 x - 3 y + 4 z = 20 \frac{2}{3}$ $x + 2 y - 3 z + 13.5 = 0$ $- x - 2 y + 5 z = \frac{113}{6}$ Give two reasons for Cramer's Rule being applicable for solving this system. Also use the rule to solve the linear system.

Jan 6, 2018

$P = \left\{\left(\frac{1}{2} , - 3 , \frac{8}{3}\right)\right\}$

#### Explanation:

Give two reasons for Cramer's Rule being applicable for solving this system.

1. $\text{Matrix} \quad A \in {C}^{n \times n}$
2. $D = \det A \ne 0$
3. $P = \left\{\left({D}_{1} / D , \ldots , {D}_{n} / D\right)\right\}$

Assuming $20 \frac{2}{3}$ is improper fraction. Rewriting given to the form:
2x−3y+4z=62/3
x+2y−3z=-27/2
−x−2y+5z=113/6

$\det A = | \left[2 , - 3 , 4\right] , \left[1 , 2 , - 3\right] , \left[- 1 , - 2 , 5\right] | = | \left[\textcolor{red}{0} , - 7 , 10\right] , \left[\textcolor{red}{1} , \textcolor{red}{2} , \textcolor{red}{- 3}\right] , \left[\textcolor{red}{0} , 0 , 2\right] |$
${R}_{1} = {R}_{1} - 2 \times {R}_{2}$
${R}_{3} = {R}_{3} + {R}_{2}$

$\det A = 1 \times {\left(- 1\right)}^{1 + 2} \times | \left[- 7 , 10\right] , \left[0 , 2\right] | = - 1 \left(- 14 - 10 \times 0\right) = \textcolor{red}{14 = D}$

I will be using Sarrus' rule

${D}_{1} = | \left[\frac{62}{3} , - 3 , 4\right] , \left[- \frac{27}{2} , 2 , - 3\right] , \left[\frac{113}{6} , - 2 , 5\right] | = \frac{620}{3} + 27 \times 4 + \frac{113 \cdot 3}{2} - \frac{113 \times 4}{3} - 62 \times 2 - \frac{27 \times 15}{2} = \frac{168}{3} - 16 - \frac{66}{2} = 7$

${D}_{2} = | \left[2 , \frac{62}{3} , 4\right] , \left[1 , - \frac{27}{2} , - 3\right] , \left[- 1 , \frac{113}{6} , 5\right] | = - 27 \times 5 + \frac{113 \times 4}{6} + 62 - 27 \times 2 + 113 - \frac{62 \times 5}{3} = - 14 + \frac{113 \times 4}{6} - \frac{620}{6} = - 42$

${D}_{3} = | \left[2 , - 3 , \frac{62}{3}\right] , \left[1 , 2 , - \frac{27}{2}\right] , \left[- 1 , - 2 , \frac{113}{6}\right] | = \frac{113 \times 2}{3} - \cancel{\frac{62 \times 2}{3}} - \frac{3 \times 27}{2} + \cancel{\frac{62 \times 2}{3}} - 2 \times 27 + \frac{113}{2} = - 54 + \frac{226}{3} + 16 = \frac{226 - 38 \times 3}{3} = \frac{112}{3}$

$x = {D}_{1} / D = \frac{7}{14} = \frac{1}{2}$

$y = {D}_{2} / D = - \frac{42}{14} = - 3$

$z = {D}_{3} / D = \frac{\frac{112}{3}}{14} = \frac{\frac{112}{14}}{3} = \frac{8}{3}$

$P = \left\{\left(\frac{1}{2} , - 3 , \frac{8}{3}\right)\right\}$