1. No category

Matrix Multiplication and Inverses
3
4 7
and B =
. Find AB and BA (if they make sense).
2
5 1
7 −11
−1 26
Solution. AB =
and BA =
. Notice that it is not true that AB = BA.
14
9
−9 17
−2
1. Let A =
1
2. Let A =
1
2
and B =
3
4
. Find AB and BA (if they make sense).
Solution. AB = [11] and BA =
3
4
6
.
8


4
1  and B = 0 2 1 . Find AB and BA (if they make sense).
0
Solution. BA = −6 2 , but AB is not defined.
1
3. Let A =  −3
0
4. If A is an n × m matrix, what is In A? How about AIm ? (Recall that In denotes the n × n identity
matrix).
Solution. We know that In A is the matrix of the composition S ◦ T , where S : Rn → Rn is defined
by S(~x) = In ~x and T : Rm → Rn is defined by T (~x) = A~x. This composition is
(S ◦ T )(~x) = S(T (~x))
= S(A~x) since T (~x) = A~x
= A~x since S sends every vector to itself
So, the matrix of S ◦ T is A, which means that In A = A .
Here’s another way of thinking about things. Let’s think about what happens when we multiply In A
by a vector ~x:
(In A)~x = In (A~x)
= A~x since In times any vector is the vector itself
So, multiplying any vector ~x by In A is the same as multiplying it by A. Therefore, In A must be equal
to A.
Similarly, if we multiply IAm by any vector ~x, we have (AIm )~x = A(Im ~x) = A~x since Im ~x = ~x. So,
multiplying any vector ~x by Im A is the same as multiplying it by A, which means that AIm = A .
−3/5
5. The matrix of reflection about the line y = 2x in R is A =
4/5
2
4/5
. Find A2 .
3/5
Solution. Of course, you can calculate this, but it’s much simpler to think about this in terms of
composition: A2 is the product AA, which means “do A, then do A again”. If we reflect and reflect
again, we’ll have done nothing: every vector ends up exactly where it started. So, A2 must be the
1 0
identity matrix
.
0 1
1
6. Recall that a n × m matrix A is invertible if the system Ax = b has a unique solution for all b in Rn .
If A is invertible, what can you say about rref A? Is the converse true?
Solution. By definition, an n × m matrix A is invertible if A~x = ~y has exactly one solution for every ~y
in Rn . Let’s imagine solving
the linear system A~x = ~y using Gauss-Jordan; we do this by row-reducing
the augmented matrix A ~y . If A~x = ~y has exactly one solution, then the system has no free
variables, so every column of rref A must have a leading 1.
On the
of 0s, then we can pick some ~y so that
other hand, if rref(A) has a row consisting entirely
rref A ~y has a row of the form 0 · · · 0 1 . This would mean that the system A~x = ~y is
inconsistent. Since we are told this doesn’t happen, we can conclude that rref(A) does not have a row
consisting entirely of 0s. That is, every row of rref(A) must have a leading 1. The only way that every
row and every column can have a leading 1 is if A is a square matrix (n × n) and rref A = In .
So far, we’ve figured out that, if A is invertible, then rref A = In . Conversely, if rref A = In , then it
n
is
certainly
the case that A~x = ~y has exactly one solution for every ~y in R : when you row reduce
A ~y , you’ll end up with In ∗ , which has exactly one solution.
3
7. (a) The matrix
7
2
is invertible; find its inverse.
5
Solution. When we want to find the inverse of a matrix A, we essentially want to solve the
system A~x = ~y to write ~x in terms of ~y .(1) As usual, we solve this using Gauss-Jordan, i.e., by
y
row-reducing the augmented matrix A ~y . If we write ~y = 1 , then we have:
y2
y
÷3
3 2 y1
1 23 31
→
−7(I)
7 5 y2
7 5 y2
y1
2
1 3
3
→
÷ 13
0 13 − 7y31 + y2
y1
1 23
− 23 (II)
3
→
0 1 −7y1 + 3y2
1 0
5y1 − 2y2
→
0 1 −7y1 + 3y2
So,
~x =
5y1 − 2y2
5
=
−7y1 + 3y2
−7
which shows that
A
−1
5
=
−7
−2 y1
,
3 y2
−2
.
3
Note: From now on, we will write this calculation in a slightly different way; the entries of
the right most column in the augmented matrix are all linear combinations of y1 and y2 , so we
can instead
keep track
of the coefficients of y1 and y2 . That is, instead of row-reducing the
3 2 y1
, we can think of y1 and y2 as 1y1 + 0y2 and 0y1 + 1y2 and row-reduce
matrix
7 5 y2
3 2 1 0
. If we row-reduce, then we end up with
7 5 0 1
1 0
5 −2
.
0 1 −7
3
(1) More
precisely, we want to write ~
x as B~
y for some matrix B; the matrix B is the inverse of A.
2
The right side of this is the inverse A−1
. So, the general method here is: to find
the inverse of an
n × n matrix, row-reduce the matrix A In to end up with In A−1 .
= 7
.
= 13
3 2 p
7
3 2
Solution. We can rewrite the system as
=
. Since we’ve seen that
is
7 5 q
13
7 5
−1 p
3 2
7
invertible, the system has a unique solution,
=
. Using our answer to (a), this
q
7 5
13
5 −2
7
9
p
9
.
is equal to
=
. So, the system has a unique solution,
=
−7
3 13
−10
q
−10
(b) Solve the linear system
3p
7p
+ 2q
+ 5q
8. Let A be an invertible n × n matrix. What is A−1 A? How about AA−1 ?
Solution. If T (~x) = A~x, then we know T −1 (T (~x)) = ~x for all ~x ∈ Rn . In matrix form, this tells us
that A−1 A~x = ~x = In ~x for all ~x ∈ Rn , so A−1 A = In . Similarly, T (T −1 (~y )) = ~y for all ~y ∈ Rn , so
AA−1 = In .
9. If A and B are invertible n × n matrices, express (AB)−1 in terms of A−1 and B −1 .
Solution. Intuitively, we know that AB means “do B, then do A”. To undo this, we need to first
undo A (by doing A−1 ), then undo B (by doing B −1 ), so the inverse of AB is B −1 A−1 .
(A useful way of remembering this is to think, “I put on my socks, then my shoes. When I take them
off, I take off my shoes, then my socks.”)

1
10. Let A = 0
0
1
1
0

1
1.
1
(a) A is invertible; find its inverse.
Solution.
the augmented matrix
As explained in the solution of
#7(a), if we row reduce
A I3 , we will end up with I3 A−1 . So, let’s row reduce A I3 :




1 1 1 1 0 0
−(II)
1 0 0 1 −1 0
 0 1 1 0 1 0 
1 0  −(III)
→ 0 1 1 0
0 0 1 0 0 1
0 0 1 0
0 1


1 0 0 1 −1
0
1 −1 
→ 0 1 0 0
0 0 1 0
0
1
So, A−1

1
= 0
0

2
(b) If B = 0
0
0
3
0

−1
0
1 −1.
0
1

0
0, find all 3 × 3 matrices X which satisfy the equation AXA−1 = B.
5
Solution. If we multiply the equation AXA−1 = B by A−1 on the left and A on the right, we
3
get
A−1 (AXA−1 )A = A−1 BA
(We’ve colored the A−1 and A just to make clear that we did the same thing to both sides of the
equation.) Since matrix multiplication is associative, we can re-arrange the parentheses in this
expression (as long as we don’t change the order of the matrices):
(A−1 A)X(A−1 A) = A−1 BA
Since A−1 A = In , this simplifies to just
X = A−1 BA

2
Since we already found A−1 , it is straightforward to compute this, and we find that it is 0
0

−1 −1
3 −2 .
0
5
v1
w1
11. (a) Just for fun. Show that the area of the parallelogram defined by two vectors ~v =
and w
~=
v2
w2
in R2 is |v1 w2 − v2 w1 |.
~v
w
~
Solution. There are lots of ways to do this, but here’s one. If we let θ be the angle between ~v
and w,
~ then the area of the parallelogram is k~v kkwk
~ sin θ. After all, half of the parallelogram is a
triangle (shown in yellow below), and if we think of the vector w
~ as being the base of the triangle,
the height of the triangle (shown in blue below) is k~v k sin θ:
~v
w
~
θ
On the other hand, we know that we can express the angle θ using the dot product formula
4
~v · w
~ = k~v kkwk
~ cos θ. So, we can manipulate this:
area of parallelogram = k~v kkwk
~ sin θ
p
= k~v kkwk
~
1 − cos2 θ
p
~ 2 (1 − cos2 θ)
= k~v k2 kwk
p
= k~v k2 kwk
~ 2 − (k~v kkwk
~ cos θ)2
p
= k~v k2 kwk
~ 2 − (~v · w)
~ 2
q
= (v12 + v22 )(w12 + w22 ) − (v1 w1 + v2 w2 )2
q
= v12 w12 + v12 w22 + v22 w12 + v22 w22 − (v12 w12 + 2v1 w1 v2 w2 + v22 w22 )
q
= v12 w22 − 2v1 w2 v2 w2 + v22 w12
p
= (v1 w2 − v2 w1 )2
= |v1 w2 − v2 w1 |
(b) Can you use the result of (a) to find a simple way to determine whether a 2 × 2 matrix
a
c
b
is
d
invertible?
Solution. We have seen that vectors ~v1 , . . . , ~vn form a basis of Rn ⇐⇒ the matrix ~v1
is invertible. So,
a b
a
b
is invertible ⇐⇒
,
form a basis of R2
c d
c
d
· · · ~vn
Geometrically, we have an easy way to tell whether two vectors form a basis of R2 (see #6(d) on
the worksheet “How much data do you need to determine a linear transformation?”):
a
b
,
are not collinear (that is, they are not in the same line)
c
d
a
b
⇐⇒ the area of the parallelogram defined by
,
is not 0
c
d
⇐⇒
⇐⇒ |ad − bc| =
6 0 by (a)
⇐⇒ ad − bc 6= 0
5