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UNIVERSIDAD CARLOS III DE MADRID
MATHEMATICS II
EXERCISES (SOLUTIONS )
CHAPTER 1: Matrices and linear systems
1-1. Compute the following determinants:
a)
b)
c)
Solution: subtract solution to a) is −15, to part b) is −36 and to part c) is 32.
1-2. Use that
a b c
p q r
u v w
= 25, to compute the value of
2 a 2 c 2 b
2 u 2 w 2 v
2 p 2 r 2 q
Solution: In the determinant, ∣ ∣ ∣ ∣ ∣ ∣ 2 a 2 c 2 b
2 u 2 w 2 v
2 p 2 r 2 q
we take out the common number 2 in each row,
3
a c b
u w v
p r q
Now swap columns 2 and 3
3
a b c
u v w
p q r
Finally, swap rows 2 and 3,
3
a b c
p q r
u v w
3 × 25 = 200
1-3. Verify the following identities without expanding the determinants:
a)
1 a
2 a
3
1 b
2 b
3
1 c
2 c
3
bc a a
2
ca b b
2
ab c c
2
b)
1 a b + c
1 b c + a
1 c a + b
Solution: a): In the determinant ∣ ∣ ∣ ∣ ∣ ∣
bc a a
2
ac b b
2
ab c c
2
we multiply and divide by a to obtain
a
a
bc a a
2
ac b b
2
ab c c
2
a
abc a
2 a
3
ac b b
2
ab c c
2
Now, we do the same procedure with b, and we observe that the determinant is the same as
ab
abc a 2 a 3
abc b 2 b 3
ab c c 2
likewise with c,
abc
abc a 2 a 3
abc b 2 b 3
abc c 2 c 3
Taking out abc in the first column, the last expression equals
abc
abc
1 a
2 a
3
1 b
2 b
3
1 c
2 c
3
1 a
2 a
3
1 b
2 b
3
1 c
2 c
3
b): Adding the second column to the third, ∣ ∣ ∣ ∣ ∣ ∣ 1 a b + c
1 b a + c
1 c a + b
we obtain (^) ∣ ∣ ∣ ∣ ∣ ∣
1 a a + b + c
1 b a + b + c
1 c a + b + c
= (a + b + c)
1 a 1
1 b 1
1 c 1
since columns 1 and 3 are the same.
1-4. Solve the following equation, using the properties of the determinants. ∣ ∣ ∣ ∣ ∣ ∣ a b c
a x c
a b x
Solution: We must assume that a 6 = 0. Otherwise, the determinant is 0 for every real number x. Since, ∣ ∣ ∣ ∣ ∣ ∣ a b c
a x c
a b x
= a
1 b c
1 x c
1 b x
the statement is equivalent (assuming a 6 = 0) to ∣ ∣ ∣ ∣ ∣ ∣ 1 b c
1 x c
1 b x
We subtract the first row to the second and third rows to obtain the following equation, ecuaci´on
1 b c
0 x − b 0
0 0 x − c
= (x − b)(x − c)
Hence, the solutions are x = b y x = c.
1-5. Simplify and compute the following expressions.
a)
ab 2 b 2 −bc
a 2 c 3 abc 0
2 ac 5 bc 2 c 2
b)
x x x x
x a a a
x a b b
x a b c
c)
Solution: a): In the determinant ∣ ∣ ∣ ∣ ∣ ∣
ab 2 b
2 −bc
a
2 c 3 abc 0
2 ac 5 bc 2 c
2
we take out a in the first column, b in the second one and c in the third one,
abc
b 2 b −b
ac 3 ac 0
2 c 5 c 2 c
and now we take out b in the first row, a in the second one and c in the third row,
a
2 b
2 c
2
c 3 c 0
2 5 2
Finally, we take out c in the second row
a
2 b
2 c
3
= 3a
2 b
2 c
3
Solution: We have to compute the determinant
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 2 3 4 · · · n + 1
3 4 5 · · · n + 2
4 5 6 · · · n + 3
5 6 7 · · · n + 4
6 7 8 · · · n + 4
. . .
n + 1 n + 2 n + 3 · · · 2 n
If n = 2, this reduces to ∣ ∣ ∣ ∣
Suppose now that n ≥ 3. We subtract the second row to the third one
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 2 3 4 · · · n + 1
3 4 5 · · · n + 2
1 1 1 · · · 1
5 6 7 · · · n + 4
6 7 8 · · · n + 4
. . .
n + 1 n + 2 n + 3 · · · 2 n
Now subtract the first row to the second one ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 2 3 4 · · · n + 1
1 1 1 · · · 1
1 1 1 · · · 1
5 6 7 · · · n + 4
6 7 8 · · · n + 4
. . .
n + 1 n + 2 n + 3 · · · 2 n
since rows 2 and 3 are equal.
1-7. Let A be a square matrix of order n × n such that A t A = I. Show that |A| = ±1.
Solution: Note that 1 = |I| = |A t A| = |A t ||A| = |A| 2
. Hence, |A| 2 = 1 and the only possible values for
the determinant are |A| = 1 o |A| = −1.
1-8. Find the rank of the following matrices:
A =
B =
Solution: The rank
rank A = rank
is the same as
rank
(^) = rank
On the other hand,
rank
(^) = rank
(^) = rank
1-9. Study the rank of the following matrices, depending on the possible values of x.
A =
x 0 x 2 1
1 x 2 x 3 x
0 0 1 0
0 1 x 0
B =
1 x x 2
C =
x − 1 0 1
0 x − 1 1
1 0 − 1 2
Solution: We compute first the rank of A.
1 x 2 x 3
0 1 x
1 x
so rank A ≥ 3 for any value of x. Expanding the determinant of A using the third row
|A| =
x 0 1
1 x
2 x
0 1 0
Now we expand the determinant using row 3
|A| = −
x 1
1 x
= −(x
2 − 1)
so the rank of A is 4 if x
2 6 = 1, that is if x 6 = 1 and x 6 = −1. To sum up,
rank A =
3 , if x = 1 o x = −1;
4 , in all other cases.
Now we compute the rank of B. Note that the minor ∣ ∣ ∣ ∣
does not vanish. Hence, rank B ≥ 2. On the other hand,
rank B = rank
1 x x
2
(^) = rank
1 x x
2
(^) = rank
1 x x
2
0 2 − x 4 − x
2
and this rank is 3, unless 5(2 − x) = 4 − x
2
. This happens if and only if x
2 − 5 x + 6 = 0, that is if
x =
To sum up,
rank B =
2 , si x = 2 o x = 3;
3 , en los dem´as casos.
Finally, we compute the rank of C.
rank C = rank
x − 1 0 1
0 x − 1 1
1 0 − 1 2
(^) = rank
x − 1 0 1
0 x − 1 1
rank
0 − 1 x 1 − 2 x
0 x − 1 1
(^) = rank
0 − 1 x 1 − 2 x
0 0 x 2 − 1 1 + x − 2 x 2
ant this rank es 3 unless x 2 − 1 = 0 and 1 + x − 2 x 2 = 0. The solutions to x 2 − 1 = 0 are x = 1 and x = −1.
The solutions to 2x 2 − x − 1 are
x =
Hence, x = 1 is the only solution to both equations. Therefore,
rank C =
2 , if x = 1;
3 , otherwise.
1-10. Let A and B be square, invertible matrices of the same order. Solve for X in the following equations.
(a) X
t · A = B.
(b) (X · A)
− 1 = A
− 1 · B.
By the usual formula we see that
B
− 1
x 2 − 4 x + 3
x
2
− 12 x − 4 3
4 x 1 −x
We also compute the inverse matrix by noticing that
0 t 3
4 1 −t
now we subtract from the third row the first row times 4,
0 t 3
0 1 4 − t
exchange rows 2 and 3
0 1 4 − t
0 t 3
Add the second row times t to the third row
0 1 4 − t
0 0 t 2 − 4 t + 3
4 t 1 −t
From here we see that |A| = t
2 − 4 t + 3. The roots of this polynomial are
t =
so the inverse matrix exists if and only if t 6 = 1 y t 6 = 3. Assuming this inequality, divide by t 2 − 4 t + 3 the
last row (^)
0 1 4 − t
0 0 1
4 t t^2 − 4 t+
1 t^2 − 4 t+
t t^2 − 4 t+
and add the third row to the first one and third row times t − 4 to the second row,
t 2
t^2 − 4 t+
1 t^2 − 4 t+
−t t^2 − 4 t+ − 12 t^2 − 4 t+
t− 4 t^2 − 4 t+
3 t^2 − 4 t+ 4 t t^2 − 4 t+
1 t^2 − 4 t+
−t t^2 − 4 t+
so
A
− 1
t 2 − 4 t + 3
t
2
− 12 t − 4 3
4 t 1 −t
1-12. Whenever possible, compute the inverse of the following matrices,
A =
B =
C =
Solution: We shall use Gauss’ method to compute the inverse. We begin with the matrix
( A| I) =
divide the first row by 4
now, subtract the third row to the second one
add the third row to the first one times 3,
add the third row to the first one
add the second row times 3/2 to the third row,
multiply the third row by − 2
add the second row to the third one and subtract them to the first one
Thus, the inverse is
Note that |B| = 0, so B does not have an inverse.
We compute the inverse matrix of C using Gauss’ methor
( C| I) =
Multiply the third row by −1 and exchange rows 1 and 3
( C| I) =
add the second row times 2 to the third row
( C| I) =
add the first row times 2 to the first one
( C| I) =
exchange rows 2 and 3
( C| I) =
The third row times −1,
( C| I) =
add the row 3 to row 1,
( C| I) =
Solution: The augmented matrix is
(A|B) =
1 a 0
a 0 1
0 a 1
First, we swap rows 2 and 3,
rank(A|B) = rank
1 a 0
0 a 1
a 0 1
Now, we subtract row 1 times a from row 3
rank(A|B) = rank
1 a 0
0 a 1
0 −a
2 1
1 − a
We add row 2 times a to row 2,
rank(A|B) = rank
1 a 0
0 a 1
0 0 1 + a
1 + a
From this, we see that if a 6 = 0 y a 6 = −1 then rank(A) = rank(A|B) = 3 = the number of unknowns, so the
system has a unique solution. In this case, the original system is equivalent to the following one,
x + ay = 1
ay + z = 2
(1 + a)z = 1 + a
and we obtain the solution z = 1, y = 1/a, x = 0.
If a = −1, then
rank(A|B) = rank
from this we see that rank(A) = rank(A|B) = 2 which is strictly less than the number of unknowns, so the
system is undetermined. Now, the original system is equivalent to the following one,
x − y = 1
−y + z = 2
We take z as the parameter and solve for the variables
y = z − 2 x = 1 + y = z − 1
subtract set of solutions is
{(z − 1 , z − 2 , z) ∈ R
3 : z ∈ R}
Finally, if a = 0, then
rank(A|B) = rank
and we see that rank(A) = 2 < rank(A|B) = 3, so the system is inconsistent.
1-15. Discuss and solve the following system
x + y + z + 2t − w = 1
−x − 2 y + 2w = − 2
x + 2z + 4t = 0
Solution: The augmented matrix A|B) is
(A|B) =
By performing elementary row operations we see that
rank(A|B) = rank
(^) = rank
so rank(A) = rank(A|B) = 2 which is less than the number of unknowns. The system is consistent and
underdetermined. Since there are five unknowns, we obtain 5 − 2 = 3 parameters. We have to solve the
following linear system,
x + y + z + 2t − w = 1
−y + z + 2t + w = − 1
We choose z, w and t as a parameters and solve for y = z + 2t + w + 1, x = 1 − y − z − 2 t − w = − 2 z − 4 t.
The set of solutions is
{(− 2 z − 4 t, z + 2t + w + 1, z, t, w) ∈ R
5 : z, t, w ∈ R}
1-16. Discuss and solve the following system according to the values of the parameters.
x + y + z = 0
ax + y + z = b
2 x + 2y + (a + 1) z = 0
Solution: The augmented matrix is
rank(A|B) = rank
a 1 1
2 2 a + 1
b
0
Performing elementary row operations,
rank(A|B) = rank
0 1 − a 1 − a
0 0 a − 1
b
0
If a 6 = 1, then the system has a unique solution and is equivalent to the following one
x + y + z = 0
y + z =
b
1 − a
(a − 1)z = 0
The solution is
z = 0, y =
b
1 − a
, x =
−b
1 − a
Now, we study the linear system for the value a = 1. In this case,
rank(A|B) = rank
b
0
So, when b 6 = 0 the system is inconsistent, since rank(A) = 1, rank(A|B) = 2.
Finally, if a = 1 and b = 0 then rank(A) = rank(A|B) = 1 and the system is consistent and underdetermined
with 3 − 1 = 2 parameters. The original system is equivalent to the following one
x + y + z = 0
Taking y and z as parameters, the set of solutions is
{(−y − z, y, z) : y, z ∈ R}
To sum up,
a 6 = 1, There is a unique solution: z = 0, y =
b 1 −a , x =
−b 1 −a
a = 1, If
b 6 = 0, inconsistent;
b = 0, undetermined. The set of solutions is the set {(−y − z, y, z) : y, z ∈ R}.
1-19. Using Cramer’s method, solve the following system.
x + y = 12
y + z = 8
x + z = 6
Solution: The determinant of the associated matrix is ∣ ∣ ∣ ∣ ∣ ∣ 1 1 0
0 1 1
1 0 1
And the solutions are
x =
= 5 y =
= 7 z =
1-20. Using Cramer’s method, solve the following system.
x + y − 2 z = 9
2 x − y + 4z = 4
2 x − y + 6z = − 1
Solution: The determinant of the associated matrix is ∣ ∣ ∣ ∣ ∣ ∣ 1 1 − 2
2 − 1 4
2 − 1 6
And the solutions are
x =
= 6 y =
= − 2 z =
1-21. Given the following system of two equations with three unknowns
{ x + 2y + z = 3
ax + (a + 3)y + 3z = 1
(a) Study for what values is of a the system is not compatible.
(b) For each value of the parameter a, for which the system is compatible, write the general solution.
Solution: The rank of the associated matrix is
rank(A|B) = rank
a a + 3 3
= rank
0 3 − a 3 − a
1 − 3 a
We see that if a = 3 then the system has no solutions because rank(A|B) = 2, rank(A) = 1. If a 6 = 3 the
system is consistent and underdetermined. It is equivalent to the following one
x + 2y + z = 3
(3 − a)y + (3 − a)z = 1 − 3 a
Taking z as the parameter, the set of solutions is
7 + 3a
3 − a
1 − 3 a
3 − a
− z, z) : z ∈ R}
1-22. Given the homogeneous system
3 x + 3y − z = 0
− 4 x − 2 y + mz = 0
3 x + 4y + 6z = 0
(a) Compute m so that it has no trivial solutions and
(b) solve it for that value.
Solution: The rank of the associated matrix is
rank
− 4 − 2 m
3 4 6
(^) = rank
− 1 1 m − 1
0 1 7
we swap the first two rows
= rank
− 1 1 m − 1
3 3 − 1
0 1 7
(^) = rank
− 1 1 m − 1
0 6 3 m − 4
0 1 7
and now swap the last two rows
= rank
− 1 1 m − 1
0 1 7
0 6 3 m − 4
(^) = rank
− 1 1 m − 1
0 1 7
0 0 3 m − 46
The system has non-trivial solution if and only if the determinant vanishes. This happens for the value
m = 46/3. In this case, the original system is equivalent to the following one
−x + y +
z = 0
y + 7z = 0
and the set of solutions is
{(22z/ 3 , − 7 z, z) : z ∈ R}
