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

Question

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

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Answer 1 · 2018-01-06T21:57:10

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

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

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

$detA=|[2,-3,4],[1,2,-3],[-1,-2,5]|=|[color(red)0,-7,10],[color(red)1,color(red)2,color(red)(-3)],[color(red)0,0,2]|$
$R_1=R_1-2xxR_2$
$R_3=R_3+R_2$

$detA=1xx(-1)^(1+2)xx|[-7,10],[0,2]|=-1(-14-10xx0)=color(red)(14=D)$

I will be using Sarrus' rule

$D_1=|[62/3,-3,4],[-27/2,2,-3],[113/6,-2,5]|=620/3+27xx4+(113*3)/2-(113xx4)/3-62xx2-(27xx15)/2=(168)/3-16-66/2=7$

$D_2=|[2,62/3,4],[1,-27/2,-3],[-1,113/6,5]|=-27xx5+(113xx4)/6+62-27xx2+113-(62xx5)/3=-14+(113xx4)/6-(620)/6=-42$

$D_3=|[2,-3,62/3],[1,2,-27/2],[-1,-2,113/6]|=(113xx2)/3-cancel((62xx2)/3)-(3xx27)/2+cancel((62xx2)/3)-2xx27+113/2=-54+226/3+16=(226-38xx3)/3=112/3$

$x=D_1/D=7/14=1/2$

$y=D_2/D=-42/14=-3$

$z=D_3/D=(112/3)/14=(112/14)/3=8/3$

$P={(1/2, -3, 8/3)}$