Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 23E from Chapter 4.8 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 23E
Chapter:
Problem:
0
Step-by-Step Solution
Given Information
We are given with the unpaid balance after $k$ th monthly payment ($y _ { k }$) and the first payment is given as\[y _ { 1 } = 10,000 + ( 0.01 ) 10,000 - 450\]We have to find the difference equation for the above case, create a table for the same and find the k for last payment and total money paid
Step-1:
Let the difference equation has the form: \[y _ { k + 1 } - a y _ { k } = b\]The coefficient $a$ can be found from the interest rate \[\begin{aligned} a & = \left( 1 + \dfrac { r } { 100 } \right) \\ & = \left( 1 + \dfrac { 1 } { 100 } \right) \\ & = 1.01 \end{aligned}\]The coefficient $b$ is the monthly payment: \[b = - 450\]The initial amount is: \[y _ { 0 } = 10000\]So the difference equation is:
\[y _ { k + 1 } = 1.01 y _ { k } - 450 , \quad y _ { 0 } = 10,000\]
Step-2:
The table is computed using the MATLAB code;
>> P = 450; y=10000; m=0; Table=[0 ; y];
>>while y>450
y=1.01*y-P;
m=m+1
Table=[Table ; y];
end>>
>>m,y
Step-2:
We can see from Table above, the last payment is 114.88 and total amount paid is; \[\begin{array}{l}T = 450 \times 25 + 114.88\\\\T = 11364.88\end{array}\]
We are given with the unpaid balance after $k$ th monthly payment ($y _ { k }$) and the first payment is given as\[y _ { 1 } = 10,000 + ( 0.01 ) 10,000 - 450\]We have to find the difference equation for the above case, create a table for the same and find the k for last payment and total money paid
Step-1:
Let the difference equation has the form: \[y _ { k + 1 } - a y _ { k } = b\]The coefficient $a$ can be found from the interest rate \[\begin{aligned} a & = \left( 1 + \dfrac { r } { 100 } \right) \\ & = \left( 1 + \dfrac { 1 } { 100 } \right) \\ & = 1.01 \end{aligned}\]The coefficient $b$ is the monthly payment: \[b = - 450\]The initial amount is: \[y _ { 0 } = 10000\]So the difference equation is:
\[y _ { k + 1 } = 1.01 y _ { k } - 450 , \quad y _ { 0 } = 10,000\]
Step-2:
The table is computed using the MATLAB code;
>> P = 450; y=10000; m=0; Table=[0 ; y];
>>while y>450
y=1.01*y-P;
m=m+1
Table=[Table ; y];
end>>
>>m,y
| m | Y | m | Y | m | Y |
|---|---|---|---|---|---|
| 0 | 10000 | 11 | 6338.226 | 21 | 2293.349 |
| 1 | 9650 | 12 | 5951.608 | 22 | 1866.282 |
| 2 | 9296.5 | 13 | 5561.124 | 23 | 1434.945 |
| 3 | 8939.465 | 14 | 5166.735 | 24 | 999.2944 |
| 4 | 8578.86 | 15 | 4768.403 | 25 | 559.2873 |
| 5 | 8214.648 | 16 | 4366.087 | 26 | 114.8802 |
| 6 | 7846.795 | 17 | 3959.747 | 27 | 0 |
| 7 | 7475.263 | 18 | 3549.345 | ||
| 8 | 7100.015 | 19 | 3134.838 | ||
| 9 | 6721.015 | 20 | 2716.187 | ||
| 10 | 6338.226 | 21 | 2293.349 |
Step-2:
We can see from Table above, the last payment is 114.88 and total amount paid is; \[\begin{array}{l}T = 450 \times 25 + 114.88\\\\T = 11364.88\end{array}\]