Question 1184020
 Given the following systems of equations

{{{2x + y+ z = 12}}}
{{{6x + 5 y - 3z = 6}}}
{{{4x - y + 3z = 5}}}

Solve for the {{{x}}},{{{y }}}and {{{z}}} using:

(i) Gaussian elimination

{{{matrix(3,4,2,	1,	1,	12,
6,	5,	-3,	6,
4,	-1,	3,	5)}}}

R1 / 2 → R1 (divide the 1 row by 2)



{{{matrix(3,4,1,	0.5,	0.5,	6,
6,	5,	-3,	6,
4,	-1,	3,	5)}}}

R2 - 6 R1 → R2 (multiply 1 row by 6 and subtract it from 2 row); R3 - 4 R1 → R3 (multiply 1 row by 4 and subtract it from 3 row)

{{{matrix(3,4,1,	0.5,	0.5,	6,
0,	2,	-6,	-30,
0,	-3,	1,	-19)}}}

R2 / 2 → R2 (divide the 2 row by 2)

{{{matrix(3,4,1,	0.5,	0.5,	6,
0,	1,	-3,	-15,
0,	-3,	1,	-19)}}}

R1 - 0.5 R2 → R1 (multiply 2 row by 0.5 and subtract it from 1 row); R3 + 3 R2 → R3 (multiply 2 row by 3 and add it to 3 row)

{{{matrix(3,4,1,	0,	2,	13.5,
0,	1,	-3,	-15,
0,	0,	-8,	-64)}}}

R3 / -8 → R3 (divide the 3 row by -8)

{{{matrix(3,4,1,	0,	2,	13.5,
0,	1,	-3,	-15,
0,	0,	1,	8)}}}

R1 - 2 R3 → R1 (multiply 3 row by 2 and subtract it from 1 row); R2 + 3 R3 → R2 (multiply 3 row by 3 and add it to 2 row)

{{{matrix(3,4,1,	0,	0,	-2.5,
0,	1,	0,	9,
0,	0,	1,	8)}}}

{{{x = -2.5}}}
{{{y = 9}}}
{{{z = 8}}}



(ii) Gauss- Jordan elimination

{{{matrix(3,4,
2,	1,	1,	12,
6,	5,	-3,	6,
4,	-1,	3,	5)}}}

Make the pivot in the 1st column by dividing the 1st row by 2

{{{matrix(3,4,1,	1/2,	1/2,	6,
6,	5,	-3,	6,
4,	-1,	3,	5)}}}


Eliminate the 1st column


{{{matrix(3,4,1,	1/2,	1/2,	6,
0,	2,	-6,	-30,
0,	-3,	1,	-19)}}}

Make the pivot in the 2nd column by dividing the 2nd row by 2

{{{matrix(3,4,1,	1/2,	1/2,	6,
0,	1,	-3,	-15,
0,	-3,	1,	-19)}}}


Eliminate the 2nd column

{{{matrix(3,4,1,	0,	2,	27/2,
0,	1,	-3,	-15,
0,	0,	-8,	-64)}}}

Make the pivot in the 3rd column by dividing the 3rd row by -8


{{{matrix(3,4,1,	0,	2,	27/2,
0,	1,	-3,	-15,
0,	0,	1,	8)}}}

Eliminate the 3rd column

{{{matrix(3,4,
1,	0,	0,	-5/2,
0,	1,	0,	9,
0,	0,	1,	8)}}}

{{{x = -2.5}}}
{{{y = 9}}}
{{{z = 8}}}





(iv) Cramer`s rule

{{{matrix(3,4,
2,	1,	1,	12,
6,	5,	-3,	6,
4,	-1,	3,	5)}}}

Write down the main matrix and find its determinant

{{{matrix(3,3,
2,	1,	1,	
6,	5,	-3,
4,	-1,	3)}}}

Δ = {{{2*5*3+6*(-1)*1+4*1(-3)-(2(-1)*(-3))-(6*1*3)-(4*5*1)}}}
Δ ={{{30-6-12 -6-18-20}}}
Δ ={{{-32}}}

Replace the 1st column of the main matrix with the solution vector and find its determinant

{{{matrix(3,3,12,	1,	1,
6,	5,	-3,
5,	-1,	3)}}}

Δ1 = {{{80}}}

Replace the 2nd column of the main matrix with the solution vector and find its determinant

{{{matrix(3,3,2,	12,	1,
6,	6,	-3,
4,	5,	3)}}}

Δ2 ={{{ -288}}}

Replace the 3rd column of the main matrix with the solution vector and find its determinant

{{{matrix(3,3,2,	1,	12,
6,	5,	6,
4,	-1,	5)}}}

Δ3 = {{{-256}}}

{{{x }}}= Δ1 / Δ = {{{80 / (-32) = -5/2}}}
{{{y }}}= Δ2 / Δ = {{{(-288) / (-32) = 9}}}
{{{z}}} = Δ3 / Δ ={{{ (-256) / (-32) = 8}}}

leaving to you to do:

(iii) Use Sarrus` rule to find determinant of the equation above