6.6 Inverses
Matrix Algebra
A is a square matrix of order n
I is the identity matrix of order n
AI = IA = A
I 1 = I
If A is invertible, then A1 is unique
AA1 = A1A = I
If A = 0, A may still not be invertible
In general AB = BA
Lecture Notes for Math 1540
6.4 Solving Systems by Reducing Matrices
Three elementary row operations:
1. Ri Rj : interchange rows Ri and Rj .
2. kRi: multiply row Ri by a nonzero constant k.
3. kRi + Rj : add k times row Ri to row Rj (but leave Ri unchanged).
Reduced matrix:
1. Al
6.6 Inverses
Find the inverse of
A=
1 2
3 7
Let C be the inverse of A such that AC = I, where
C=
x z
y w
It follows that
1 2
3 7
Lecture Notes for Math 1540
x z
x + 2y
=
y w
3x + 7y
First
Previous
Next
Last
z + 2w
1 0
=
3z + 7w
0 1
12
6.6 Inverses
We
6.1 Matrices
Denition: A Matrix A is a rectangular array of the form
A11 A12 A1n
A21 A22 A2n
.
.
.
.
.
.
Am1 Am2 Amn
The size of A is m n, where m is the number of rows and n is the number of
columns.
The entry Aij locates in the ith row and the
Name (PRINT): Wang
(Last Name)
Xiangsheng
Student id:
(Given Name)
Mark:
Math 1540M 3.00 W2012 Introductory Mathematics for Economists II
Final exam (total 40 points)
April 10 (Tuesday) 19:00-22:00 ACE 001
Instructions:
1. Please write down your name and
6.7 Leontiefs Input-Output Analysis
Production=Internal Demand+External Demand:
X
= AX + D
240
1200
1200
=
360
1500
1200
500
1500
460
1200
+
200
940
1500
1500
Solving production matrix X from external demand matrix D:
X = AX + D X AX = D (I A)X =
Consumers and Producers Surplus
Demand function: p = f (q).
Supply function: p = g(q).
Equilibrium: p0 = f (q0) = g(q0).
Consumers surplus: CS =
Producers surplus: P S =
Lecture Notes for Math 1540
First
q0
(f (q)
0
q0
(p0
0
Previous
p0)dq.
g(q)dq.
6.1 Matrices
Denition: A Matrix A is a rectangular array of the form
A11 A12 A1n
A21 A22 A2n
.
.
.
.
.
.
Am1 Am2 Amn
The size of A is m n, where m is the number of rows and n is the number of
columns.
The entry Aij locates in the ith row and the
Integrals
An antiderivative of a function f (x) is a function F (x) such that F (x) = f (x).
The indenite integral of f (x) with respect to x is dened by
f (x)dx = F (x) + C,
where is called the integral sign, f (x) is the integrand, x is the variable o
7.6 Articial Variables & 7.7 Minimization
We need to add an articial variable t when the feasible region does not
contain the origin (i.e., 0 is not a solution to the constraints). The objective
function becomes W = Z M t = c1x1 + c2x2 + + cnxn M t with
7.3 Multiple Optimum Solutions
If (x1, y1) and (x2, y2) are two corner points at which an objective function is
optimum, then the objective function will also be optimum at all points (x, y),
where
x = (1 t)x1 + tx2
y = (1 t)y1 + ty2
with 0 t 1. The above
List of Homework
You are expected to do all of the assigned homework. Experience has shown that the only way to learn
math is to do it - math is not a spectator sport! The amount you learn in this course and the grade
you receive will be proportional to t