Math 166 Week-in-Review - © S. Nite 11/10/2012
WIR #9
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Week in Review #9 (F1-F.4, 4.3-4.4, 5.1-5.3)
Simple Interest
I = Prt, where I is the interest, P is the principal, r is the interest rate, and t is the time in years.
A = P(1 + rt), where A is the accumulated amount, P is the principal, r is the interest rate, and t is the time
in years.
Discounts
The discount, D on a discounted loan of M dollars at a sample annual interest rate of r for t years is D =
Mrt
Where D = discount (interest paid at time of loan)
M = maturity value (amount borrowed)
r = discount rate (annual simple interest rate)
t = length of loan.
The proceeds P of the loan is the actual amount the borrower receives when the loan is made and is given
by P = M – D.
For discounted loans, the effective interest rate is reff =
r
, where r is the interest rate and t is the
1 − rt
number of years.
Compound Interest
n
For regular loans, ref
r

=  1 +  − 1 , where re
n

is the effective interest rate, r is the annual
interest rate, and n is the number of compounding periods per year.

A = P 1 +

nt
r
 , where A is the accumulated amount, P is the principal, r is the annual interest rate, n is
n
the number of compounding periods per year, and t is the time in years for compound interest. Use TVM
solver in the calculator.
A = Pe rt , where A is the accumulated amount, P is the principal, r is the annual interest rate
compounded continuously, and t is the time in years.
Present Value and Future Value of Loans, Annuities, and Sinking Funds
Use TVM solver, enter known information and press ALPHA – Enter with cursor on PV (or FV) to solve
for present value (or future value). Note: PMT = 0 if there is a one-time deposit/loan; PV or FV = 0 if
there are regular payments with nothing in the beginning.
A good estimate for doubling time is 72 divided by the interest rate.
The Gauss elimination method involves a sequence of operations on a system of linear equations to
obtain equivalent systems until a solution is found.
Row-Reduced Form of a Matrix
1. Each row consisting entirely of zeros lies below any other row having nonzero entries.
2. The first nonzero entry in each row is 1 (called a leading 1).
3. In any two successive (nonzero) rows, the leading 1 in the lower row lies to the right of the
leading 1 in the upper row.
4. If a column contains a leading 1, then the other entries in that column are zeros.
A column in a coefficient matrix is in unit form if one of the entries in the column is a 1 and the other
entries are zeros.
The Gauss-Jordan Elimination Method
1. Write the augmented matrix corresponding to the linear system.
2. Interchange rows (Operation 1), if necessary, to obtain an augmented matrix in which the first
entry in the first row is nonzero. Then pivot the matrix about this entry.
Math 166 Week-in-Review - © S. Nite 11/10/2012
WIR #9
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3. Interchange the second row with any row below it, if necessary, to obtain an augmented matrix in
which the second entry in the second row is nonzero. Pivot the matrix about this entry.
4. Continue until the final matrix is in row-reduced form.
If there is a row in the augmented matrix containing all zeros to the left of the vertical line and a nonzero
entry to the right of the line, then the system of equations has no solution.
Theorem
a. If the number of equations is greater than or equal to the number of variables in a linear system,
then one of the following is true:
i.
The system has no solution.
ii.
The system has exactly one solution.
iii.
The system has infinitely many solutions.
b. If there are fewer equations than variables in a linear system, then the system either has no
solution or it has infinitely many solutions.
Using Matrices to Represent Data
A matrix is an ordered rectangular array of numbers. A matrix with m rows and n columns has size m x
n. The entry in the ith row and jth column is denoted by a ij .
If A is an m x n matrix with elements a ij , then the transpose of A is the n x m matrix AT with elements
a ji .
Scalar Multiplication
If A is a matrix and c is a real number, then the scalar product cA is the matrix obtained by multiplying
each entry of A by c.
Find the matrix X satisfying the matrix equation 2X + B = 3A.
Matrix Multiplication in General
If A is a matrix of size m x n and B is a matrix of size n x p, then the matrix product of A and B, AB is
defined and is a matrix of size m x p.
The identity matrix of size n (n rows and n columns) is given by
The identity matrix has the property that I n A = A for any n x r matrix A, and BI n = B for any s x n
matrix B. In particular, if A is a square matrix of size n, then I n A = AI n = A .
The Inverse of a Square Matrix
Let A be a square matrix of size n. A square matrix A −1 of size n such that A −1 A = AA −1 = I n is called
the inverse of A.
Solving Systems of Equations with Inverse Matrices
If AX = B is a linear system of n equations in n unknowns and if A-1 exists,
then X = A-1B is the unique solution of the system.
Math 166 Week-in-Review - © S. Nite 11/10/2012
WIR #9
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1. Tommy needs $5,300 to pay off a bill. He has taken out a discount loan that has a monthly
discount rate of 5.6%. The loan must be paid back in 10 months. How much will Tommy pay
back (i.e., what is the maturity value)?
2. Find the effective yield on a discount loan with a discount rate of 3.5% for 5 months. Round
your answer to two decimal places.
3. A principal of $14,500 earns 3.6% per year simple interest. How long will it take for the future
value to become $26,700? Round the answer to two decimal places.
4. Find the present value of $60,000 due in 3 years at at rate of 7% per year, compounded
quarterly.
5. Find the accumulated amount if $23,000 is invested at 3.1% per year, compounded monthly
for 10 years.
6. Find the amount needed to deposit into an account today to yield semi-annual pension
payments of $13,000 for the next 15 years if the account earns 3.2% per year, compounded semiannually.
7. A couple is saving for their daughter’s education at Texas A&M University. They want to
have $85,000 at the end of 16 years. (a) How much should they put each month into a savings
account that earns an annual rate of 5% compounded monthly? (b) How much interest would
they earn over the life of the account? (c) Determine the fund balance after 10 years.
8. A business creates a sinking fund in order to have $270,000 to replace machinery in 12 years.
(a) How much should be deposited into the account at the end of each quarter if the annual
interest rate is 5% compounded quarterly? (b) How much interest would they earn over the life
of the account? (c) Determine the balance after 5 years and 7 years.
9. The Juarez’s have $26,000 for a down payment on a house. They plan to spend $1500 in
monthly payments on the house. If the mortgage rate is 6.2% per year compounded monthly for
the 30-year mortgage, what price house can they afford?
10. A $136,000 loan is to be amortized in equal monthly payments over 30 years with interest
rate of 4% per year compounded monthly. Complete the first three rows of an amortization table.
11. Six years ago, Jenny financed $185,000 to purchase a house. The term of the mortgage was
30 years with a 6.5%/year interest rate, compounded monthly on the unpaid balance. The interest
rate has dropped to 4.1%/year compounded monthly, and Jenny plans to refinance her house.
(a) What is Jenny’s current monthly mortgage payment?
(b) What is Jenny’s current outstanding principal?
(c) If Jenny decides to refinance by securing a 30-year mortgage loan for the outstanding
principal at 4.1%/year compounded monthly, what will be her monthly mortgage payment?
(d) How much interest will Jenny save by refinancing?
Math 166 Week-in-Review - © S. Nite 11/10/2012
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12. Jing plans to retire in 30 years and wants to start a retirement account to provide $5,000 per
month for 20 years when she retires. She can get at 4% annual interest rate, compounded
monthly. If she starts with $200 and makes monthly payments from now until retirement, how
much must she deposit each month.
13. Determine which of the following matrices are in row-reduced form. If not, why not?
1 0 0 0 
0 1 0 − 2 
1 0 − 7 
1 0 0 2 
1 − 2 0 0








a) 1 0 0 3 b) 1 0 0 − 3 c) 0 − 2 0  d) 0 1 0 − 5 e) 0 0 1 3
0 1 0 2
0 0 1 2 
0 0
0 0 1 0 
0 0 2 0
0 
14. Pivot the matrix about the circled element.
1
4
1 5
0 − 2 2 − 3


0 3 − 2 − 5
15. Solve the following systems and interpret the solution.
(a) -2x + 3y – 4z = 7
x - 2y + 5 = 4z
-5y + 4z – 6 = 2x
(c) 2x + 4y – 6 = z
-x – 3y + z = -2
3y + 4z – 3x = -4
(b) x – y + z = 20
2x + 3y – z = -15
-3x + 3y - 20 = 3z
(d) -2x + 3y + 5 = z
2x – y – z = 4
4x – 10 + 2z = 6y
16. Set up, solve, and interpret the solution:
Julio has $25,000 to invest in three stocks A, B, and C. He wants to invest three times as much in
stock C as in stocks A and B combined. He estimates a return of 2% on stock A, 4% on stock B,
and 6% on stock C. How much should he invest in each stock to have a return of $1750 per year
on the total investment?
Math 166 Week-in-Review - © S. Nite 11/10/2012
WIR #9
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17. Set up, solve, and interpret the solution:
A shop makes necklaces, earrings, and bracelets. The table shows the number of minutes each
item spends in each area. If there are 32 hours available in the stringing area, 43 hours available
in the painting area, and 38 hours available in the finishing area, how many of each item can be
made if all the available hours are to be used?
Necklace
Earring pair
Bracelet
Stringing
minutes
10
5
8
Painting
minutes
12
8
10
Finishing
minutes
8
6
12
 2 3
− 1 0 c 
, find AB.
18. Given A =  − 1 0 and B = 
2 a − 2

− 2 5
 2 3
− 1 0 c 
19. Given A =  − 1 0 and B = 
, find BA.
2 a − 2

− 2 5
4 5 
0 2 − 1 
20. Given A = 3 − 2 and B = 
, find 2AT - 3B.

3 1 − 2
0 1 
− 2 1 
0 2 
 2 3
21. Given A = 
, B=
, and C = 


 , find matrix K such that KA + B = C.
 0 − 1
3 1 
1 0 
22. Set up the matrix equation and solve the system by inverse matrices.
3x – 2y + 5 = 2z
-2z + x + 2y = -7
y – 2x = -14