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January 23, 2019
MATHEMATICS FOR ECONOMICS II (2018-19)
ECONOMICS, LAW–ECONOMICS, INTERNATIONAL STUDIES–ECONOMICS
SHEET 1. MATRICES, DETERMINANTS AND SYSTEMS
(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.
(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
(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.
(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.
(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)
where we have taken out b − a in the first row.
c): In the determinant ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
we subtract the last row from rows 2 and 3 and we expand it using the first column ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 0 1 1 1
0 − 1 0 1
0 0 − 1 1
1 1 1 0
we add now the first row to the second one and expand using the first column again,
(6) Let A be a square matrix of order n × n such that A
T A = In. 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.
(7) Find the rank of the following matrices:
A =
B =
Solution: The rank
rg A = rg
is the same as
rg
(^) = rg
On the other hand,
rg
= rg
= rg
(8) 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 rg 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,
rg 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, rg B ≥ 2. On the other hand,
rg B = rg
1 x x
2
(^) = rg
1 x x
2
(^) = rg
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,
rg B =
2 , si x = 2 o x = 3;
3 , en los dem´as casos.
Finally, we compute the rank of C.
rg C = rg
x − 1 0 1
0 x − 1 1
1 0 − 1 2
(^) = rg
x − 1 0 1
0 x − 1 1
rg
0 − 1 x 1 − 2 x
0 x − 1 1
(^) = rg
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,
rg C =
2 , if x = 1;
3 , otherwise.
(9) Let A and B be square, invertible matrices of the same order. Solve for X in the following
matrix equations:
(a) X
t · A = B
(b) (X · A)
− 1 = A
− 1 · B
Substract now the third row times
x 2
2 to the second row,
1 0 −x
x 1 − x 2
2 0 0 1
from here we obtain the inverse matrix o A.
Now we compute the inverse matrix of
B =
0 x 3
4 1 −x
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
(11) 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.
Solution: The matrix associated to the system is
( m − 1
1 −m
2 m − 1
whose rank is the same as
rg
1 −m
m − 1
2 m − 1
1
= rg
1 −m
0 m
2 − 1
2 m − 1
1 + m − 2 m
2
Therefore, the rank of this matrix 2 is m
2 6 = 1. That is, if m 6 = 1 and m 6 = −1 then the
system has a unique solution. In this case the system is equivalent to the following one
x − my = 2 m − 1
(m
2 − 1)y = 1 + m − 2 m
2
and the solution is
y =
1 + m − 2 m
2
m^2 − 1
−(m − 1)(1 + 2m)
(m − 1)(m + 1)
− 1 − 2 m
m + 1
x = 2 m − 1 + my = 2m − 1 − m
1 + 2m
m + 1
m + 1
If x = 3 is a solution we must have that
m + 1
this implies that m = − 4 /3.
We study now the case m = 1.
rg
1 −m
0 m
2 − 1
2 m − 1
1 + m − 2 m
2
= rg
= 1 = rg A
and the system has a unique solution. The original system of equations is equivalent to the
following one
x − y = 1
and the set of solutions is {1 + y, y) : y ∈ R}. Taking y = 2, we obtain the solution (3, 2).
For the case m = −1 we have that
rg
1 −m
0 m 2 − 1
2 m − 1
1 + m − 2 m 2
= rg
= 2 6 = rg A
and the system is inconsistent.
(13) Given the system of linear equations
x + ay = 1
ax + z = 1
ay + z = 2
(a) Express it in matrix form;
(b) Write the unknowns vector, the independent term and the associated homogeneous sys-
tem;
Solution: En forma matricial, el sistema queda expresado como AX = B con
A =
1 a 0
a 0 1
0 a 1
B =
X =
x
y
z
(c) Discuss and solve it according to the values of a.
Solution: The augmented matrix is
(A|B) =
1 a 0
a 0 1
0 a 1
First, we swap rows 2 and 3,
rg(A|B) = rg
1 a 0
0 a 1
a 0 1
Now, we subtract row 1 times a from row 3
rg(A|B) = rg
1 a 0
0 a 1
0 −a
2 1
1 − a
We add row 2 times a to row 2,
rg(A|B) = rg
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 rg(A) = rg(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
rg(A|B) = rg
from this we see that rg(A) = rg(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
rg(A|B) = rg
and we see that rg(A) = 2 < rg(A|B) = 3, so the system is inconsistent.
Now, we study the linear system for the value a = 1. In this case,
rg(A|B) = rg
b
0
So, when b 6 = 0 the system is inconsistent, since rg(A) = 1, rg(A|B) = 2.
Finally, if a = 1 and b = 0 then rg(A) = rg(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}.
(16) Discuss and solve the following system, according the values of a and b:
x − 2 y + bz = 3
5 x + 2y = 1
ax + z = 2
Solution: The augmented matrix is
(A|B) =
1 − 2 b
5 2 0
a 0 1
Performing elementary row operations we see that
rg(A|B) = rg
1 − 2 b
0 12 − 5 b
0 2 a 1 − ab
2 − 3 a
We subtract row 2 times a/6 from row 3,
rg(A|B) = rg
1 − 2 b
0 12 − 5 b
ab 6
2 a 3
And we see that if ab 6 = 6 then the unique solution is
z =
12 − 4 a
6 − ab
, y =
12 − 4 a
6 − ab
b, x = 3 +
12 − 4 a
6 − ab
b −
12 − 4 a
6 − ab
b
If ab = 6 we solve for a = 6/b (since b 6 = 0) to obtain
rg(A|B) = rg
1 − 2 b
0 12 − 5 b
0 0 0
2 b− 4 b
So, if b = 2 (and hence a = 3) then rg(A|B) = rg(A) = 2 and the system is consistent and
underdetermined. It is equivalent to the following system
x − 2 y + 2z = 3
12 y − 10 z = − 14
Taking z as the parameter, the set of solutions is
2 − z
5 z − 7
, z) : z ∈ R}
Finally, if b 6 = 2 then the system is inconsistent.
(17) Using Cramer’s method, solve the following system:
−x + y + z = 3
x − y + z = 7
x + y − z = 1
Solution: The determinant of the associated matrix is ∣ ∣ ∣ ∣ ∣ ∣ − 1 1 1
1 − 1 1
1 1 − 1
And the solutions are
x =
= 4 y =
= 2 z =
(18) 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 =
(19) Using Cramers’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 =
