Elementary Row Operations For Matrices Page 3
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1
2
33) Finish the Matrix
Now that the first column is finished, we can work on the next column applying the same
procedures.
We need a 1 in the second row second position. The easiest way to accomplish this is to use the
second type of row operation. Multiply the second row by because
1
0
-3
1
1
0
-3
1
R
0
8
16
0
0
1
2
0
R
2
2
0
14
-10
10
0
14
-10
10
The next column to be worked out.
Continue repeating these steps as needed until the matrix is completed.
1
0
-3
1
1
0
-3
1
1
0
-3
1
0
1
2
0
0
1
2
0
0
1
2
0
0
14
-10
10
0
0
-38
10
0
0
1
-
R
R
-14 R
+ R
- R
2
3
3
3
3
1
0
-3
1
1
0
0
0
1
0
0
1
0
0
0
1
0
0
1
-
-
R
R
-2 R
+ R
3 R
+ R
3
2
2
3
1
1
Recall that the first column represents the x, the second y, the third z. We can rewrite the matrix
back to a system of linear equations.
1
0
0
1x + 0y + 0z =
0
1
0
0x + 1y + 0z =
0
0
1
-
0x + 0y + 1z = -
So, the solution to this system of the linear equations is x =
, y =
, and z = -
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