3x+2y+z=8
2x+y+3z=7
5x-3y+4z=3
3x + 2y + z = 8
2x + y + 3z = 7
5x - 3y + 4z = 3
The augmented matrix is
[3 2 1 | 8]
[2 1 3 | 7]
[5 -3 4 | 3]
We need to get 0's in the three lower left
positions, that is, in the positions below
the upper left to lower-right diagonal:
To get a 0 where the 2 in row 2 column 1 is,
we multiply row 1 temporarily by -2 and add it
to 3 times row 2. This is easy to do mentally
if you write -2 to the left of row 1 and 3 left
of row 2:
-2[3 2 1 | 8]
3[2 1 3 | 7]
[5 -3 4 | 3]
[3 2 1 | 8]
[0 -1 7 | 5]
[5 -3 4 | 3]
To get a 0 where the 5 in row 3 column 1 is,
we multiply row 1 temporarily by -5 and add it
to 3 times row 3. This is easy to do mentally
if you write -5 to the left of row 1 and 3 left
of row 3:
-5[3 2 1 | 8]
[0 -1 7 | 5]
3[5 -3 4 | 3]
[3 2 1 | 8]
[0 -1 7 | 5]
[0 -19 7 | -31]
To get a 0 where the -19 in row 3 column 2 is,
we multiply row 2 temporarily by 19 and add it
to -1 times row 3. This is easy to do mentally
if you write 19 to the left of row 2 and 1 left
of row 3:
[3 2 1 | 8]
19[0 -1 7 | 5]
-1[0 -19 7 | -31]
[3 2 1 | 8]
[0 -1 7 | 5]
[0 0 126 | 126]
Now that we have the three zeros,
the above is the abbreviation of this system:
3x + 2y + 1z = 8
0x - 1y + 7z = 5
0x + 0y + 126z = 126
or
3x + 2y + z = 8
-y + 7z = 5
126z = 126
The bottom equation,
126z = 126
z = 1
Substitute (1) for z in
-y + 7z = 5
-y + 7(1) = 5
-y + 7 = 5
-y = -2
y = 2
Substitute (1) for z and (2) for y in
3x + 2y + z = 8
3x + 2(2) + (1) = 8
3x + 4 + 1 = 8
3x + 5 = 8
3x = 3
x = 1
Solution:
(x, y, z) = (1, 2, 1)
Edwin
