Category: Finite Math

How Do I Solve A System Of Equations That Results In Two Equations And Three Variables?

When using Left to Right Elimination or Gauss’s Methods, a system of three equation in three variables often results in a unique solution. In other words, a single value for each of the three variables solves each of the three equations in the system. However when some systems are put into row echelon form, all of the constants and variables in one of the equations drop out. In a sense, one of the equations has disappeared and now you are left with a system of two equations in three variables.

Don’t confuse this with an inconsistent system. In an inconsistent system, the row operations lead to one of the equations being false…something like 0 = 6. This type of system has no solutions.

Let’s examine a system of three equations in three variables in which one of the equations drops out. Take the system

$latex \displaystyle \begin{matrix}
x&+2y&-4z &=5 \\
2x&+3y&-z &=3 \\
3x&+5y&-5z &=8 \\
\end{matrix}$

If we use left to right elimination or matrix elimination to solve this system, we get the equivalent system

$latex \displaystyle \begin{matrix}
&x&+2y&-4z &=5 \\
&&y&-7z &=7 \\
& & &0 &=0 \\
\end{matrix}$

This new system contains only two equations since 0 = 0 does not describe a relationship between the variables. This new system has many solutions. We can show this by solving the second equation for y to give

$latex \displaystyle y=7z+7$

and substituting it into the first equation:

$latex \displaystyle x+2\left( 7z+7 \right)-4z=5$

Now solve this equation for x,

$latex \displaystyle \begin{matrix}
x+14z+14-4z=5 \\
x+10z+14=5 \\
x=-10z-9 \\
\end{matrix}$

The solution to the system is

$latex \displaystyle \begin{matrix}
x&=&-10z-9 \\
y&=&7z+7 \\
\end{matrix}$

This may also be written as an ordered triple $latex \left( -10z-9,7z+7,z \right)$.  We have solved for x and y, but z is not specified. We can pick it to be anything. For instance, if  z = 0, then x = -0 and y = 7. This means that the ordered triple (-9, 7, 0) solves the original system of equations. If z = -1, then x = 1 and y = 0, So (1, 0, -1) also solves the system.

How Do You Find The Inverse Of A 2 x 2 Matrix?

Suppose the given square matrix is called A. To find the inverse of any matrix, we write the matrix in a larger matrix along side an identity matrix of the same size,

$latex \displaystyle \left[ \left. A\, \right|\,I \right]$

Now use row operations to rewrite this matrix so that the identity appears on the left side. The inverse of the original matrix will be on the right side of the transformed matrix,

$latex \displaystyle \left[ \left. I\, \right|\,{{A}^{-1}} \right]$

For instance, suppose we want to find the inverse of

$latex \displaystyle A=\left[ \begin{matrix}
2 & 2 \\
2 & 1 \\
\end{matrix} \right]$

Start with

$latex \displaystyle \left[ \left. \begin{matrix}
2 & 2 \\
2 & 1 \\
\end{matrix}\, \right|\,\begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]$

$latex \displaystyle \frac{1}{2}{{R}_{1}}\to {{R}_{1}}$

$latex \displaystyle \left[ \left. \begin{matrix}
1 & 1 \\
2 & 1 \\
\end{matrix} \right|\begin{matrix}
\frac{1}{2} & 0 \\
0 & 1 \\
\end{matrix} \right]$

$latex \displaystyle -2{{R}_{1}}+{{R}_{2}}\to {{R}_{2}}$

$latex \displaystyle \left[ \left. \begin{matrix}
1 & 1 \\
0 & -1 \\
\end{matrix} \right|\begin{matrix}
\frac{1}{2} & 0 \\
-1 & 1 \\
\end{matrix} \right]$

$latex \displaystyle -1{{R}_{2}}\to {{R}_{2}}$

$latex \displaystyle \left[ \left. \begin{matrix}
1 & 1 \\
0 & 1 \\
\end{matrix} \right|\begin{matrix}
\frac{1}{2} & 0 \\
1 & -1 \\
\end{matrix} \right]$

$latex \displaystyle -1{{R}_{2}}+{{R}_{1}}\to {{R}_{1}}$

$latex \displaystyle \left[ \left. \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right|\begin{matrix}
-\frac{1}{2} & 1 \\
1 & -1 \\
\end{matrix} \right]$

Let’s apply this strategy to finding a few more inverses.

Problem 1 Find the inverse of

$latex \displaystyle \left[ \begin{matrix}
2 & 4 \\
2 & 5 \\
\end{matrix} \right]$

m152_inv_matrix_2

m152_inv_matrix_2b

Problem 2 Find the inverse of

$latex \displaystyle \left[ \begin{matrix}
1 & 3 \\
2 & 7 \\
\end{matrix} \right]$

m152_inv_matrix_3

How Do You Find the Annual Interest Rate From the Compound Interest Formula?

Suppose 5000 dollars is deposited in an account that earns compound interest that is done annually. If there is 7000 dollars in the account after 2 years, what is the annual interest rate?

The easiest way to approach this problem is to use the compound interest formula,

compound_01

This formula applies when interest is earned on an annual basis and the interest is earned once a year.

Let’s look at the quantities in the problem statement:

5000 dollars is deposited in an account > P = 5000

If there is 7000 dollars in the account after 2 years > A = 7000 and n = 2

Putting these values into the formula above gives us

We need to find the annual interest rate r. Since the r is hidden in the parentheses, we start by isolating the parentheses.

To get at the r, we need to remove the square on the parentheses.

compound_04

Using a calculator to do the square root, we get r ≈ 0.183 or 18.3%.

Note that instead of taking the square root to get rid of the square, we could raise both sides to the 1/2 power:

Then we would be able to solve for r,

and get the same value for r since square rooting and raising to the 1.2 power yield the same result.

Now what if the interest is earned over six years instead of two years? Instead of a square on the parentheses we now have a sixth power. To solve for r in this equation, we follow similar steps as before.

compound_03

The root can be computed in a graphing calculator using the MATH button, by raising to the 1/6 power (^(1/6)), or put into WolframAlpha:

wolframalpha_root

Either method gives r ≈ 0.577 or 5.77%. Notice that the annual interest is lower when it is earned over a longer period of time.

Compare the Mean and Median Using a Spreadsheet

During the week of 6/7/2012 through 6/14/2012, eight homes were sold in Paradise Valley, Arizona in the area code 85253. The sales prices for these homes are listed below.

900,000
535,000
182,500
1,550,000
2,250,000
1,525,000
490,000
1,525,000

a. Find the mean sales price.

Solution Use the AVERAGE command to compute the mean of the data.

  1. Enter the data from the table into the different cells in a spreadsheet.

stats_excel005

2. Click on cell A11. This is where we will place the mean of the data. Type =AVERAGE( as shown to the right. The command will be shown in the cell as well as the function bar. To indicate the location of the data, type A2:A9. You can also click in cell A2, hold the left mouse button down and drag the cursor to cell A9. Type ) to complete the command.

Press Enter to compute the mean. In cell B11, type Mean to identify the type of central tendency.

stats_excel006

b. Find the median sales price.

Solution Use the MEDIAN command to compute the median of the data.

  1. In cell A12, type =MEDIAN(A2:A9).

stats_excel007

3. Press Enter to compute the median.

4. In cell B12, type Median to identify the measure of central tendency.

stats_excel008