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May 23rd

May 23rd

Matrix Operations

Theorem 7
An n × n matrix A is invertible if and only if A is row equivalent to I n, and in this case, any
sequence of elementary row operations that reduces A to In will also transform In to A-1.

Algorithm for finding A-1
Place A and I side-by-side to form an augmented matrix  [ A I ] . Then perform row operations on
this matrix (which will produce identical operations on A and I). So by Theorem 7.
[ A I ] will row reduce to [ I A-1]

or A is not invertible.

EXAMPLE. Find the inverse of
if it exists.

Solution.

Order of multiplication is important→

EXAMPLE Suppose A,B,C, and D are invertible n × n matrices and A = B (D - In) C.
Solve for D in terms of A,B,C and D.

Solution.

2.3 Characterizations of Invertible Matrices

Theorem 8 (The Invertible Matrix Theorem)
Let A be a square n × n matrix. The the fol lowing statements are equivalent (i.e., for a given
A, they are either all true or all false).

a. A is an invertible matrix.
b. A is row equivalent to In.
c. A has n pivot positions.
d. The equation Ax = 0 has only the trivial solution.
e. The columns of A form a linearly independent set.
f. The linear transformation x → Ax is one-to-one.
g. The equation Ax = b has at least one solution for each b in R n.
h. The columns of A span Rn.
i. The linear transformation x → Ax maps Rn onto Rn.
j. There is an n × n matrix C such that CA = In.
k. There is an n × n matrix D such that AD = In.
l. AT is an invertible matrix.

EXAMPLE. Use the Invertible Matrix Theorem to de termine if A is invertible , where

Solution

Circle correct conclusion. Matrix A is / is not invertible.

EXAMPLE. Suppose H is a 5 × 5 matrix and suppose there is a vector v in R5 which is not a
linear combination of the columns of H. What can you say about the number of solutions to
Hx = 0?

Solution Since v in R5 is not a linear combination of the
columns of H, the columns of H do not ___________ R5.
So by the Invertible Matrix Theorem, Hx = 0 has
_________________________________________.

Invertible Linear Transformations

For an invertible matrix A,
A-1Ax = x for all x in Rn
and
AA-1x = x for all x in Rn.

Pictures.
A linear transformation T . Rn → Rn is said to be invertible if there exists a function
S . Rn → Rn such that

S(T(x))= x for all x in Rn
and
T(S(x)) = x for all x in Rn.

Theorem 9
Let T . Rn → Rn be a linear transformation and let A be the standard matrix for T. Then T is
invertible if and only if A is an invertible matrix. In that case, the linear transformation S
given by S(x)= A-1x is the unique function satisfying

S(T(x))= x for all x in Rn
and
T(S(x)) = x for all x in Rn.

Section 2.6 - Leontief Input-Output Model

Economic Model With External Demand
Three sectors in economy. Manufacturing, Agriculture, Services
Production vector
gives total annual production (in dollars).

There is also an open sector which does not produce goods, but con sumes them
according to the final demand vector

When the various sectors produce goods to meet external demands, they re-
quire goods from all sectors for this production. This creates intermediate
demand.

We want to satisfy

Each sector i has a unit consumption vector saying how many dollars of
each sector it requires to produce one dollar 's worth of sector i.

 Inputs Consumed per Unit of Output Purchased from. Manufacturing Agriculture Services Manufacturing 0.50 0.40 0.20 griculture 0.20 0.30 0.10 Services 0.10 0.10 0.30

E.g. if agriculture produces 100 units, it will consume

That is, it will consume ("demand") 40 units from manufacturing, 30 units
from agriculture, and 10 units from services.

A simpler demonstration of the problem with numbers instead of vectors.

Say there's only one sector, agriculture, and there is an external demand of 100
units. Assume that the agriculture sector consumes 0.30 units of agriculture for
every unit it produces.

If agriculture produces 100 units to meet the external demand, this creates an
intermediate demand for (0.3)(100) = 30 units. So agriculture can produce
another 30 units, to meet this intermediate demand.

But producing this 30 units creates intermediate demand for another (0.3)(30)
= 9 units. So agriculture must produce another 9 units to meet this new
intermediate demand. Note that 9 = (0.3)(30) = (0.3)[(0.3)100] = (0.3)2(100).

But producing this extra 9 units creates yet more intermediate demand, of
(0.3)(9) = 2.7 units. So it has to produce yet another 2.7 units. Note that
2.7 = (0.3)(9) = (0.3)[(0.3)2100] = (0.3)3(100)...

This can continue "forever." But notice what will happen. in total, agriculture
will have to produce

If manufacturing wants to produce x1 units, then the intermediate demands
created are units from the three sectors.

Similarly, if agriculture and services want to produce x2 and x3 units respec-
tively, their intermediate demands are and .

Adding them up to get the total intermediate demand created by all three
sectors, we get

where the consumption matrix is

So if they produce ,this creates intermediate demand .

Note that the columns of C do not add up to one. Each column should add
up to a number less than one, otherwise each dollar of production would cost

Now to solve the model. We have

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