Math 141: Business Mathematics I
Fall 2015
§2.3 Solving Systems of Linear Equations II
Instructor: Yeong-Chyuan Chung
Outline
• The Gauss-Jordan elimination method - systems of linear equations with no solution
or infinitely many solutions
General Systems of Linear Equations
Recall that a system of linear equations is simply a list of linear equations. In §2.2, we had
examples of systems consisting of 2, 3, or 4 equations, and the number of equations was the
same as the number of variables. The systems we saw there also happened to have unique
solutions. In general, there is no restriction on the number of equations we can have in a
system. The number of equations can also be different from the number of variables. Just
as we saw in §2.1, systems of linear equations do not always have unique solutions as they
can also have no solution or infinitely many solutions.
When we had two linear equations in two variables in §2.1, we could think of the equations
geometrically in terms of the lines they represent, and that was sufficient to tell us whether
the system had a unique solution, no solution, or infinitely many solutions. With more
equations and more variables, this option becomes unavailable. However, the Gauss-Jordan
elimination method that we saw in §2.2 can be used in all these situations.
The Gauss-Jordan Elimination Method (continued)
Recall that the Gauss-Jordan elimination method is a way of replacing the original system of
linear equations by ones that are simpler but have the same solution(s). It involves performing row operations on matrices corresponding to the systems, and converting the matrices
to row-reduced form. Also recall that the ”rref” command on the graphing calculator can
be used to find the row-reduced form of a matrix.
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§2.3 Solving Systems of Linear Equations II
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Solve the following systems:
Example (Exercise 29 in the text).
x − 2y + 3z = 4
2x + 3y − z = 2
−3z + 2y + x = −6
Example (Exercise 23 in the text).
x + 2y + z = −2
−2x − 3y − z = 1
2x + 4y + 2z = −4
Example (Exercise 27 in the text).
x − 2z + y = −3
2x − y + 3z = 7
x − 2y + 5z = 0
§2.3 Solving Systems of Linear Equations II
Solve the following systems:
Example (Exercise 15 in the text).
2x − y = 3
x + 2y = 4
2x + 3y = 7
Example (Exercise 31 in the text).
4x + y − z = 4
8x + 2y − 2z = 8
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§2.3 Solving Systems of Linear Equations II
Example (Exercise 41 in the text). Mr. and Mrs. Garcia have a total of $100,000 to be
invested in stocks, bonds, and a money market account. The stocks have a rate of return
of 6%/year, while the bonds and the money market account pay 4%/year and 2%/year,
respectively. The Garcias have stipulated that the amount invested in stocks should be equal
to the sum of the amount invested in bonds and 3 times the amount invested in the money
market account. How should the Garcias allocate their resources if they require an annual
income of $5,000 from their investments? Give two specific options.