Let T(V)=AV be a linear transformation where A=(3 -2 6 -1 15, 4 3 8 10 -14, 2 -3 4 -4 20)
a.) construct a basis of the kernal T
b.) calculate the basis of the range of T
c.) determine the rank and nullity of T
3 | -2 | 6 | -1 | 15 |
4 | 3 | 8 | 10 | -14 |
2 | -3 | 4 | -4 | 20 |
Divide row1 by 3
1 | -2/3 | 2 | -1/3 | 5 |
4 | 3 | 8 | 10 | -14 |
2 | -3 | 4 | -4 | 20 |
Add (-4 * row1) to row2
1 | -2/3 | 2 | -1/3 | 5 |
0 | 17/3 | 0 | 34/3 | -34 |
2 | -3 | 4 | -4 | 20 |
Add (-2 * row1) to row3
1 | -2/3 | 2 | -1/3 | 5 |
0 | 17/3 | 0 | 34/3 | -34 |
0 | -5/3 | 0 | -10/3 | 10 |
Divide row2 by 17/3
1 | -2/3 | 2 | -1/3 | 5 |
0 | 1 | 0 | 2 | -6 |
0 | -5/3 | 0 | -10/3 | 10 |
Add (5/3 * row2) to row3
1 | -2/3 | 2 | -1/3 | 5 |
0 | 1 | 0 | 2 | -6 |
0 | 0 | 0 | 0 | 0 |
Add (2/3 * row2) to row1
1 | 0 | 2 | 1 | 1 |
0 | 1 | 0 | 2 | -6 |
0 | 0 | 0 | 0 | 0 |
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