ECON30020 Mathematical Economics Tutorial 9. Matrices. The Inverse Matrices.
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ECON30020 Mathematical Economics
Tutorial 9. Matrices. The Inverse Matrices.
Question 1. Matrix Operations
Compute, for the following matrices A and B
A = 、 and B = ,
(a) AB
(b) BA
(c) (AB)T
(d) (BA)T
and their traces. For AB and (AB)T also compute the inverses.
Solution.
(a) AB = 、 = 、 = 、 (b) BA = 、 = = (c) (AB)T = 、
╱ 38 28 5 、
(d) (BA)T = 36 21 10
( |
Further, tT(AB) = 53 + 1 = 54, tT(BA) = 38 + 21 △ 5 = 54, which is also the trace of their transposes. Note that tT(AB) = tT(BA) and tT(C) = tT(CT).
The inverse matrices are calculated by using the following expression, derived in lectures: swap the elements on the main diagonal, and multiply by (-1) the elements on the other diagonal; divide by det(A).
A− 1 =
、 =
△a12
Therefore,
(AB)− 1 = z △22(1)
、 = z △22(1)
、 = △ z △22(1)
△49、
and
((AB)T )− 1 = △
Note that (CT)− 1 = (C − 1 )T ; here C = AB .
1 △49
、.
Question 2. Inverses
For the matrix
AA− 1 = I2 ,
where I2 = z 0(1) 1(0) 、, following the steps in L9, p.9 to p.11. Then verify that A− 1 is of the
form given there. What is the determinant of A?
Solution. Let A− 1 = , then from I = AA− 1 , we have
α21 α22
z0(1) 1(0)、 = z1(4) 2(3)、 = 、 .
This gives us four equations that could be split into two systems (two equations for each column):
í
ìα11 + 2α21 = 0
and
í
ìα12 + 2α22 = 1
Solving the first system for α 11 and α21 :
Let’s use the substitution method. From the second equation α 11 = △2α21 , substituting in the first equation: 4( △2α21 ) + 3α21 = 1 gives us α21 = -1/5 . Substituting back into the expression for α 11 , we obtain α 11 = △2α21 = 2/5 .
Solving the second system for α 12 and α22 :
Let’s solve it by elimination. Multiplying the second equation by 4 and substracting from the first equation gives us 3α22 △ 4 A 2α22 = △4. Thus, α22 = 4/5 . Substituting into the first equation: 4α12 + 3 A 4/5 = 0, we obtain α 12 = -3/5
Therefore,
A− 1 =
、 = z △ 1(2)
△4(3) 、.
The determinant of A is det(A) = 4 A 2 △ 1 A 3 = 5 and we have indeed
z 、
Question 3. Representing systems of linear equations in matrix form
Consider the system of three linear equations and three unknowns x1 , x2 and x3
2x1 △ 3x2 + 4 = 7x3 (2)
6x1 + 5x2 = 0. (3)
Rewrite this system in the form
Ax = b,
╱ x1 、 ╱ b1 、
where x = x2 and b = b2 is a vector of coefficients and A is a 3 3 3 matrix of
ì x3 | ì b3 |
coefficients. What are A and b?
Solution. Rewriting the system to get all xi’s onto the left hand and all constants onto the right hand sides gives
3x1 + 4x2 + 8x3 = 5 (4) 2x1 △ 3x2 △ 7x3 = △4 (5)
6x1 + 5x2 + 0x3 = 0. (6)
We can write it in the matrix form as
╱3 4 8 、╱x1 、 ╱ 5 、
2 △3 △7 x2 = △4 .
ì6 5 0 |ìx3 | ì 0 |
↓ 乂↓ 乂 ↓ 乂
A 北 b
╱ 3 Hence A = 2 ì |
4 △3 5 |
8 、 ╱ 5 、 △7 and b = △4 0 0 |
Question 4. The Cofactor Method
Obtain the inverses of the following matrices by using the cofactor method:
┌ 1 2 (a) A = '(')0 1 ' |
3┐ 2 '(') ' |
┌ 1 (b) B = '(') 0 ' |
2 1 2 |
△1┐ 0 '(') ' |
Solution.
(a) First, it is necessary to calculate the minors for every element of the matrix A, which
are as follows:
M11 = det z0(1) M12 = det z0(0) M13 = det z0(0)
z
M22 = det z0(1)
1(2)、 = 1
1(2)、 = 0
0(1)、 = 0
、
1(3)、 = 1
M23 = det z0(1) M31 = det z 1(2) M32 = det z0(1)
z
0(2)、 = 0
2(3)、 = 1
2(3)、 = 2
、
Second, we calculate the matrix of cofactors, where cofactors are given by Cij = ( △ 1)i+j Mij . We then transpose it to get the adjoint matrix:
┌ 1 C = '(')△2 ' |
0 1 △2 |
0┐ 0 '(') ' |
transpose to get =去 |
┌ 1 adj(A) = CT = '(')0 ' |
△2 1 0 |
1 ┐ △2 '(') ' |
The determinant of A is simply det(A) = 1 3 1 3 1 = 1 (the product of the elements on the main diagonal), because it is an upper-triangular matrix. Hence, the inverse of matrix A is
1 ┌ 1 A− 1 = det(A)adj(A) = '(')0 |
△2 1 0 |
1 ┐ △2 '(') ' |
(b) This repeats the same process as in part (a).
M11 = det z 2(1)
z
M13 = det z △(0)5 M21 = det z 2(2)
z
3(0)、 = 3
3(0)、 = 0
2(1)、 = 5
、
3(△)1、 = △2
M23 = det 、 = 12
z 、
M32 = det z0(1) 1(2)、 = 0
M33 = det 、 = 1
Forming the matrix of cofactors and transposing
┌ 3 C = '(')△8 ' |
0 △2 0 |
5 ┐ △12 '(') ' |
transposes to =去 |
give |
┌3 adj(B) = '(')0 ' |
△8 △2 △12 |
1┐ 0 '(') 1 |
2022-06-20