LINEAR ALGEBRA — PRACTICE EXAM 2
(1) Invertible matrices.
Suppose A,B, and Xare matrices that satisfy the relation AX −A=B, where
A=
0 1 0
1
3−2
3
1
3
−1 2 0
and B=
−1 3 4
1 2 0
4 5 6
.
Solve for X.
answer: Solving for Xsymbolically, we see that
AX =B+A⇒X=A−1(B+A) = A−1B+I.
We not compute A−1:
0 1 0 1 0 0
1
3−2
3
1
30 1 0
−1 2 0 0 0 1
∼
0 1 0 1 0 0
1−21030
−1 2 0 0 0 1
∼
0 1 0 1 0 0
1−21030
0 0 1 0 3 1
∼
1−21030
0 1 0 1 0 0
0 0 1 0 3 1
∼
1−2 0 0 0 −1
0 1 0 1 0 0
0 0 1 0 3 1
∼
10020−1
0 1 0 1 0 0
0 0 1 0 3 1
,
hence
A−1=
2 0 −1
1 0 0
0 3 1
.
Now solving for X:
X=A−1B+I
=
2 0 −1
1 0 0
0 3 1
−1 3 4
1 2 0
4 5 6
+
100
010
001
=
−612
−134
7 11 6
+
100
010
001
=
−512
−144
7 11 7
(2) Column space and null space of a matrix.
Let
A=
2−1 1 −6 8
1−2−4 3 −2
−7 8 10 3 −10
4−5−7 0 4
.
Find a basis for Col(A) and Nul(A).
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