Econ 1012: Linear Algebra for Economists
Answers to Exercise 0
AAUSC, Department of Economics
March 16, 2019


1 0 0
1. A =  0 1 0 
0 0 1
(This is the 3 × 3 identity matrix).
0 1
1 −1
0 + 1 1 + (−1)
1 0
2. A + B =
+
=
=
,
2 3
5
2
2+5
3+2
7 5
0 3
0 1
.
=
3A = 3
6 9
2 3
3. u = 3 and v = 2. (Equating the elements in row 1 and column 3 gives u = 3. Then, equating
those in row 2 and column 3 gives u − v = 5 and so v = 2. The other elements then need to
be checked, but this is obvious.)
1 0 4
0 + 1 1 + (−1) −1 + 5
1 −1 5
0 1 −1
,
=
=
+
4. A + B =
2 4 16
2+0
3+1
7+9
0
1 9
2 3
7
−1 2 −6
0 − 1 1 − (−1) −1 − 5
1 −1 5
0 1 −1
,
=
=
−
A−B =
2 2 −2
2−0
3−1
7−9
0
1 9
2 3
7
3 −3 15
0 5 −5
1 −1 5
0 1 −1
−
=
−3
5A − 3B = 5
0
3 27
10 15 35
0
1 9
2 3
7
0 − 3 5 − (−3) −5 − 15
−3 8 −20
=
=
.
10 − 0 15 − 3
35 − 27
10 12
8
0 −2 −1 4
0 · (−1) + (−2) · 1
5. (a) AB =
=
3
1
1 5
3 · (−1) + 1 · 1
BA =
−1 4 0 −2
(−1) · 0 + 4 · 3
=
1 5 3
1
1·0+5·3
8
(b) AB =
1
26
=
6
0 · 4 + (−2) · 5
−2 −10
=
,
3·4+1·5
−2
17
(−1) · (−2) + 4 · 1
12 6
=
.
1 · (−2) + 5 · 1
15 3


2 −2
3 −2 
8 · 2 + 3 · 4 + (−2) · 1

4
3 =
0
4
1·2+0·4+4·1
1 −5
3
,
−22
1
8 · (−2) + 3 · 3 + (−2) · (−5)
1 · (−2) + 0 · 3 + 4 · (−5)


2 · 8 + (−2) · 1
2 −2 8 3 −2



3
= 4·8+3·1
BA = 4
1 0
4
1 · 8 + (−5) · 1
1 −5


14 6 −12
4 .
=  35 12
3 3 −22

2 · 3 + (−2) · 0
4·3+3·0
1 · 3 + (−5) · 0

2 · (−2) + (−2) · 4
4 · (−2) + 3 · 4 
1 · (−2) + (−5) · 4



 

0 0
·
0
0
·
(−2)
0
·
3
0
0
0
4 −6 ,
(c) AB =  −2  0 −2 3 =  (−2) · 0 (−2) · (−2) (−2) · 3  =  0
4
4·0
4 · (−2)
4·3
0 −8 12


0
BA = 0 −2 3  −2  = [0 · 0 + (−2) · (−2) + 4 · 3] = [16], a 1 × 1 matrix.
4
(d) AB is not defined.



3 · (−1) + 1 · 2
3 1 −1
0
=  (−1) · (−1) + 1 · 2
BA =  −1 1 
2 4
0 · (−1) + 2 · 2
0 2
 

3·0+1·4
−1 4
(−1) · 0 + 1 · 4  =  3 4 .
0·0+2·4
4 8

  
ax + dy + ez
a d e x
6. We start by performing the multiplication  d b f  y  =  dx + by + f z . Next,
ex + f y + cz
e f c z


ax + dy + ez
x y z  dx + by + f z  = [ax2 + by 2 + cz 2 + 2dxy + 2exz + 2f yz], which is a 1 × 1 matrix.
ex + f y + cz

0 3 1
3 4
0 2
3 −1
0
0
= A0 + B 0 ,
=
, (A + B) =
, B =
7.
=
4 7
1 7
2 2
2
5
0 0 4 10
4 10
−6
2
−6 −4
0
0
0
= B 0 A0 ,
=
= αA , (AB) =
=
(αA) =
10 8
10 8
−4 −10
2 −10
0 −2 10
−2 4
0
0
= (BA)0 .
=
and A B =
4 14
10 14
A0
8. The firm’s profit is given by


11,
000
15, 000
pq − wz = 10 12
− 10 10 8 15, 000
27, 000
15, 000
= 10(15, 000) + 12(27, 000) − [10(11, 000) + 10(15, 000) + 8(15, 000)]
= 150, 000 + 324, 000 − (110, 000 + 150, 000 + 120, 000)
= 94, 000
2