8.2. CONSTRUCTING THE MATRIX OF A LINEAR TRANSFORMATION Example 8.2.14 Let
1 A= 2 4
2 1 5
143
0 2 2
3 1 7
Find ker (A). Equivalently, find the solution space to the system of equations Ax = 0. This asks you to find {x : Ax = 0} . In other words you are asked to solve the system, T Ax = 0. Let x = (x, y, z, w) . Then this amounts to solving
1 2 3 2 1 1 4 5 7
x 0 0 y = 0 2 z 2 0 w
This is the linear system x + 2y + 3z = 0 2x + y + z + 2w = 0 4x + 5y + 7z + 2w = 0 and you know how to solve this using row operations, (Gauss Elimination). Set up the augmented matrix, 1 2 3 0 | 0 2 1 1 2 | 0 4 5 7 2 | 0 Then row reduce to obtain the row reduced echelon form,
1
0
0
− 13
4 3
1
5 3
− 23
0
0
0
0
| 0
| 0 . | 0
This yields x = 13 z − 43 w and y = 23 w − 53 z. Thus ker (A) consists of vectors of the form,
1 3z 2 3w
− 43 w
1 3 − 53
− 53 z = z 1 z 0 w
− 43
2 +w 3 0 1
.
Example 8.2.15 The general solution of a linear system of equations is just the set of all solutions. Find the general solution to the linear system, x 1 2 3 0 9 y 7 2 1 1 2 z = 4 5 7 2 25 w given that
¡
1 1
2
1
¢T
=
¡
x
y
z
w
¢T
is one solution.