College Algebra 9th txtbk

480 CHAPTER 7 SYSTEMS OF EQUATIONS AND MATRICES
(A) X A 1 B
x 1 3 3 1 3 8
£ x 2 § £ 2 2 1§£
1 § £ 4 §
x 3 4 5 2 4 9
The solution is, x 1 8, x 2 4, and x 3 9
(B) X A 1 B
x 1 3 3 1 5 6
£ x 2 § £ 2 2 1§£
2 § £ 3 §
x 3 4 5 2 3 4
The solution is, x 1 6, x 2 3, and x 3 4
MATCHED PROBLEM 6 Use matrix inverse methods to solve each of the following systems (see Matched Problem 5):
(A) 3x 1 x 2 x 3 3
x 1 x 2 3
x 1 x 3 2
(B) 3x 1 x 2 x 3 5
x 1 x 2 1
x 1
x 3 4
As Examples 5 and 6 illustrate, inverse methods are very convenient for hand calculations
because once the inverse is found, it can be used to solve any new system formed by
changing only the constant terms. Since most graphing calculators can compute the inverse
of a matrix, this method also adapts readily to graphing calculator solutions. However, if your
graphing calculator also has a built-in procedure for finding the reduced form of an augmented
coefficient matrix, then it is just as convenient to use Gauss–Jordan elimination. Furthermore,
Gauss–Jordan elimination can be used in all cases and, as noted previously, matrix
inverse methods cannot always be used.
The application in Example 7 illustrates the usefulness of matrix inverses.
EXAMPLE 7 Investment Allocation
An investment adviser currently has two types of investments available for clients: an investment
A that pays 4% per year and an investment B of higher risk that pays 8% per year. Clients
may divide their investments between the two to achieve any total return desired between 4
and 8%. However, the higher the desired return, the higher the risk. How should each client
listed in the table invest to achieve the indicated return?
Client
1 2 3 k
Total investment $20,000 $50,000 $10,000 k 1
Annual return desired $1,200 $3,750 $500 k 2
(6%) (7.5%) (5%)
SOLUTION
We will first solve the problem for an arbitrary client k using inverses, and then apply the
result to the three specific clients.
Let
x 1 Amount invested in A
x 2 Amount invested in B