Question 1122649
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<pre>
The matrix form is


{{{(matrix(4,4, 2,-1,0,1,  0,3,-1,-1, 1,0,1,-3, 1,1,2,0))}}} . {{{(matrix(4,1, x1,x2,x3,x4))}}} = {{{(matrix(4,1, 9,-9,-2,0))}}}.


The rule is simple: where the unknown is missed (omitted) in the current equation, you put zero coefficient to the matrix.


To solve it, I used free of charge online solver

https://matrix.reshish.com/gaussSolution.php


The solution process and the results are shown below:



Your matrix

       X1	X2	X3	X4	b
1	2	-1	0	1	9
2	0	3	-1	-1	-9
3	1	0	1	-3	-2
4	1	1	2	0	0


Find the pivot in the 1st column and swap the 3rd and the 1st rows

       X1	X2	X3	X4	b
1	1	0	1	-3	-2
2	0	3	-1	-1	-9
3	2	-1	0	1	9
4	1	1	2	0	0


Multiply the 1st row by 2

       X1	X2	X3	X4	b
1	2	0	2	-6	-4
2	0	3	-1	-1	-9
3	2	-1	0	1	9
4	1	1	2	0	0


Subtract the 1st row from the 3rd row and restore it

       X1	X2	X3	X4	b
1	1	0	1	-3	-2
2	0	3	-1	-1	-9
3	0	-1	-2	7	13
4	1	1	2	0	0


Subtract the 1st row from the 4th row

       X1	X2	X3	X4	b
1	1	0	1	-3	-2
2	0	3	-1	-1	-9
3	0	-1	-2	7	13
4	0	1	1	3	2


Find the pivot in the 2nd column (inversing the sign in the whole row) and swap the 3rd and the 2nd rows

       X1	X2	X3	X4	b
1	1	0	1	-3	-2
2	0	1	2	-7	-13
3	0	3	-1	-1	-9
4	0	1	1	3	2


Multiply the 2nd row by 3

       X1	X2	X3	X4	b
1	1	0	1	-3	-2
2	0	3	6	-21	-39
3	0	3	-1	-1	-9
4	0	1	1	3	2


Subtract the 2nd row from the 3rd row and restore it

       X1	X2	X3	X4	b
1	1	0	1	-3	-2
2	0	1	2	-7	-13
3	0	0	-7	20	30
4	0	1	1	3	2


Subtract the 2nd row from the 4th row

      X1	X2	X3	X4	b
1	1	0	1	-3	-2
2	0	1	2	-7	-13
3	0	0	-7	20	30
4	0	0	-1	10	15


Find the pivot in the 3rd column (inversing the sign in the whole row) and swap the 4th and the 3rd rows

       X1	X2	X3	X4	b
1	1	0	1	-3	-2
2	0	1	2	-7	-13
3	0	0	1	-10	-15
4	0	0	-7	20	30


Subtract the 3rd row from the 1st row

       X1	X2	X3	X4	b
1	1	0	0	7	13
2	0	1	2	-7	-13
3	0	0	1	-10	-15
4	0	0	-7	20	30


Multiply the 3rd row by 2

       X1	X2	X3	X4	b
1	1	0	0	7	13
2	0	1	2	-7	-13
3	0	0	2	-20	-30
4	0	0	-7	20	30


Subtract the 3rd row from the 2nd row and restore it

       X1	X2	X3	X4	b
1	1	0	0	7	13
2	0	1	0	13	17
3	0	0	1	-10	-15
4	0	0	-7	20	30


Multiply the 3rd row by -7

       X1	X2	X3	X4	b
1	1	0	0	7	13
2	0	1	0	13	17
3	0	0	-7	70	105
4	0	0	-7	20	30


Subtract the 3rd row from the 4th row and restore it

       X1	X2	X3	X4	b
1	1	0	0	7	13
2	0	1	0	13	17
3	0	0	1	-10	-15
4	0	0	0	-50	-75


Make the pivot in the 4th column by dividing the 4th row by -50

       X1	X2	X3	X4	b
1	1	0	0	7	13
2	0	1	0	13	17
3	0	0	1	-10	-15
4	0	0	0	1	3/2


Multiply the 4th row by 7

       X1	X2	X3	X4	b
1	1	0	0	7	13
2	0	1	0	13	17
3	0	0	1	-10	-15
4	0	0	0	7	21/2


Subtract the 4th row from the 1st row and restore it

       X1	X2	X3	X4	b
1	1	0	0	0	5/2
2	0	1	0	13	17
3	0	0	1	-10	-15
4	0	0	0	1	3/2


Multiply the 4th row by 13

       X1	X2	X3	X4	b
1	1	0	0	0	5/2
2	0	1	0	13	17
3	0	0	1	-10	-15
4	0	0	0	13	39/2


Subtract the 4th row from the 2nd row and restore it

       X1	X2	X3	X4	b
1	1	0	0	0	5/2
2	0	1	0	0	-5/2
3	0	0	1	-10	-15
4	0	0	0	1	3/2


Multiply the 4th row by -10

       X1	X2	X3	X4	b
1	1	0	0	0	5/2
2	0	1	0	0	-5/2
3	0	0	1	-10	-15
4	0	0	0	-10	-15


Subtract the 4th row from the 3rd row and restore it

       X1	X2	X3	X4	b
1	1	0	0	0	5/2
2	0	1	0	0	-5/2
3	0	0	1	0	0
4	0	0	0	1	3/2


Solution set:

x1 = 5/2
x2 = -5/2
x3 = 0
x4 = 3/2

</pre>

Solved.