
Patrick B. answered 07/26/20
Math and computer tutor/teacher
1 1 3 1 6
2 3 7 5 21
3 -2 4 -1 -5
1 5 7 2 20
4 -3 5 -2 -9
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-2*row1 + row2;
-3*row1 + row3;
-1*row1 + row4;
-4*row1 + row5;
1 1 3 1 6
0 1 1 3 9
0 -5 -5 -4 -23
0 4 4 1 14
0 -7 -7 -6 -33
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5*row2 + row3
-4*row2 + row4
7*row2 + row5
1 1 3 1 6
0 1 1 3 9
0 0 0 11 22
0 0 0 -11 -22
0 0 0 15 30
the last three rows give w = 2
the second row becomes y + z + 3w = 9
y + z + 3(2) = 9
y + z + 6 = 9
y + z = 3
y = 3-z
the first row becomes x + y + 3z + w = 6
x + y + 3z + 2 = 6
x + y + z + 2z + 2 = 6
x + (3) + 2z + 2 = 6
x + 2z + 5 = 6
x + 2z = 1
x = 1 - 2z
So there are infinitely many solutions..
In terms of free variable z, they are
{x=1-2z, y = 3-z, z, 2}
For example, if z=0, then the particular solution is {1,3,0,2}
For example, if z=1, then the particular solution is {-1,2,1,2}