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Playing with Matrix Multiplication
Solutions
Linear Algebra
1. Assume A, B are invertible nxn matrices.
(a) If (AB)2 = I, show that (BA)2 = I.
(b) If (AB)n = I for some n ≥ 1, show (BA)n = I. (Note: This problem is about a
general n, so it is not sufficient to do this for some specific value of n, like n = 5.
However, try experimenting with several small, concrete values of n to figure out
how to model your argument.)
Solution. Since (AB)2 = ABAB = I, left-mulitplying by A−1 gives BAB = A−1 ,
and then right-multiplying by A gives BABA = I, i.e. (BA)2 = I. For general n, the
same trick works.
2. Assume P is invertible and let A = P BP −1 . Solve for B.
Solution. Left-multiply by P −1 and right-multiply by P to get P −1 AP = B.
3. Let A, B, C be invertible matrices such that C −1 (A + X)B −1 = I. Solve for X.
Solution. Left-multiplying by C and right-multiplying by B gives A + X = CB.
Subtracting A gives X = CB − A.
4. Compute
(a) (I − A)(I + A + A2 )
(b) (I − A)(I + A + A2 + · · · + An )
Solution. We have
(I − A)(I + A + A2 ) = I + A + A2 − A − A2 − A3 = I − A3 .
Similarly, for general n the middle terms cancel and we have (I −A)(I +A+· · ·+An ) =
I − An+1 .
5. If A is a n × n matrix such that A3 = 0, find (I − A)−1 . (Hint:Use problem 4.)
Solution. Using 4, we have
(I − A)(I + A + A2 ) = I − A3 = I
where the last equality is because A3 = 0 by assumption. Thus (I −A)−1 = (I +A+A2 ).
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6. If A2 − 2A + I = 0, show A3 = 3A − 2I.
Solution. Multiplying both sides of the equation A2 − 2A + I = 0 by A we have
A3 − 2A2 + A = 0
so A3 = 2A2 − A. The original equation implies A2 = 2A − I, and substituting this
into the previous equation gives the desired result.
7. Consider the matrix
A=
7 6
2 3
.
We want to determine the set V of all 2x2 matrices B that commute with A (i.e for
which AB = BA). For example, the identity matrix I and the matrix A are elements
in V , since AI = IA = A and AA = AA = A2 .
(a) To find all the matrices B that commute with A, first write B with unknown
entries:
x y
B=
.
u v
Multiply out AB and BA and set them equal to each other to find four equations
involving x, y, u, v.
(b) Solve the equations obtained in (a). Use the solution to write the general form of
matrices in V .
(c) Show how to obtain the matrices I and A using the general solution found in (b).
(d) Find a finite set of matrices that span V . That is, every matrix in V can be
written as a linear combination of matrices in your finite set.
(e) Check that the following matrix commutes with A:
4 9
C=
.
3 −2
As a result, C must be an element of V . Show how to obtain C from the general
form of the solution in (d).
Solution.
Letting B = ( ux yv ), we compute
7x + 6u 7y + 6v
AB = B =
2x + 3u 2y + 3v
and BA = B =
2
7x + 2y 6x + 3y
7u + 2v 2u + 3v
.
Setting these matrices equal to one another, we have the system of equations
7x + 6u = 7x + 2y
7y + 6v = 6x + 3y
2x + 3u = 7u + 2v
2y + 3v = 6u + 3v
Solving this system yields y = 3u and x = 2u + v with u and v free. So a general
matrix which commutes with A looks like
2 3
1 0
2u + v 3u
B=
=
u+
.
u
v
1 0
0 1
So the matrices ( 21 30 ) and ( 10 01 ) span the matrices which commute with A. Since A,
I, and C commute with A, they are each of this form. A is obtained by taking u = 2
and v = 3. I is obtained by taking u = 0 and v = 1. C is obtained by taking u = 3
and v = −2. (One can work hard to find these, or just notice that the entries in the
second row determine the coefficients.)
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8. Use linear algebra to find the area of the ellipse x16 + y25 = 1.
Solution. The unit circle has area π, and the transformation with matrix A = ( 40 05 )
maps the unit circle to the ellipse. This can be seen by noticing that the standard
basis vectors e1 and e2 are sent to the (positive) x− and y− intercepts of the ellipse.
Since this matrix has determinant 20, the ellipse has area 20π.
Thanks to Thomas Hagedorn of the TCNJ Math Department for much of this problem set.
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