Math 1300: Section 4-6 Matrix Equations and Systems of Linear Equations

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Matrix EquationsMatrix Equations and Systems of Linear Equations

Application

Math 1300 Finite MathematicsSection 4.6 Matrix Equations and Systems of Linear

Equations

Jason Aubrey

Department of MathematicsUniversity of Missouri

Jason Aubrey Math 1300 Finite Mathematics

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Application

Solving a Matrix Equation

Suppose we have an n × n matrix A and an n × 1 columnmatrices B and X . If A is invertible, then we can solve theequation

AX = B

as follows

AX = B

A−1(AX ) = A−1B multiply both sides by A−1 on the left

(A−1A)X = A−1B matrix multiplication is associative

IX = A−1B A−1A = I

X = A−1B


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Application

Using Inverses to Solve Systems of Equations

Example: Use matrix inverse methods to solve the system:

x1 − x2 + x3 = 12x2 − x3 = 1

2x1 + 3x2+0x3 = 1

First, we write this as a matrix equation:1 −1 10 2 −12 3 0

︸︷︷︸

A

x1x2x3

︸︷︷︸

X

=

111

︸︷︷︸

B


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Application

So we have a matrix equation

AX = B

To solve this equation, we find A−1:1 −1 10 2 −12 3 0

∣∣∣∣∣∣1 0 00 1 00 0 1

→ · · · →︸︷︷︸Row Operations

1 0 00 1 00 0 1

∣∣∣∣∣∣3 3 −1−2 −2 1−4 −5 2

Therefore,

A−1 =

3 3 −1−2 −2 1−4 −5 2


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Application

We know that X = A−1B, sox1x2x3

︸︷︷︸

X

=

3 3 −1−2 −2 1−4 −5 2

︸︷︷︸

A−1

111

︸︷︷︸

B

=

5−3−7

So, we conclude that

x1 = 5x2 = −3x3 = −7


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Application

Using Inverse Methods to Solve Systems of Equations

If the number of equations in a system equals the number ofvariables and the coeffiecient matrix has an inverse, then thesystem will always have a unique solution that can be found byusing the inverse of the coefficient matrix to solve thecorresponding matrix equation.

Matrix Equation SolutionAX = B X = A−1B


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Application

Insight. . .There are two cases where inverse methods will not work:

Case 1. If the coefficient matrix is singular.

Case 2. If the number of variables is not the same as thenumber of equations.

In either case, use Guass-Jordan elimination.


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Application

Example: An investment advisor currently has two types ofinvestments available for clients: a conservative investment A thatpays 10% per year and investment B of higher risk that pays 20% peryear. Clients may divide their investments between the two to achieveany total return desired between 10% and 20%. However, the higherthe desired return, the higher the risk. How should each client listedin the table invest to achieve the desired return?

1 2 3 kTotal Investment $20,000 $50,000 $10,000 k1

Annual Return Desired $2,400 $7,500 $1,300 k2

The answer to this problem involves six quantities, two for each client.We will solve this problem for an arbitrary client k with unspecifiedamounts k1 for total investment and k2 for annual return.


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Application

Let

x1 = amount invested in A by a given clientx2 = amount invested in B by a given client

Then we have the following mathematical model:

x1 + x2 = k1 total invested0.1x1 + 0.2x2 = k2 Total annual return desired

Write as a matrix equation:[1 1

0.1 0.2

]︸︷︷︸

A

[x1x2

]︸︷︷︸

X

=

[k1k2

]︸︷︷︸

B


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Application

If A−1 exists, then X = A−1B. So, we find A−1:[1 1

0.1 0.2

∣∣∣∣1 00 1

]→ · · · →︸︷︷︸

Row Operations

[1 00 1

∣∣∣∣ 2 −10−1 10

]

Thus

A−1 =

[2 −10−1 10

]


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Application

Therefore, [x1x2

]︸︷︷︸

X

=

[2 −10−1 10

]︸︷︷︸

A−1

[k1k2

]︸︷︷︸

B

To solve each client’s problem, we replace k1 and k2 withappropriate values from the table and multiply by A−1:[

x1x2

]=

[2 −10−1 10

] [20,0002,400

]︸︷︷︸

Client 1

=

[16,0004,000

]

Solution: x1 = $16,000 in investment A,x2 = $4,000 in investment B


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Application

[x1x2

]=

[2 −10−1 10

] [50,0007,500

]︸︷︷︸

Client 2

=

[25,00025,000

]

Solution: x1 = $25,000 in investment A,x2 = $25,000 in investment B[

x1x2

]=

[2 −10−1 10

] [10,0001,300

]︸︷︷︸

Client 2

=

[7,0003,000

]

Solution: x1 = $7,000 in investment A,x2 = $3,000 in investment B


Documents

Math 1300: Section 4-6 Matrix Equations and Systems of Linear Equations