Graphics Programs Reference
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which agrees with the formula we had above. (Note the use of
syms
to reset
R
and
P
to
undefined
symbolic quantities.) Thus the column vector [
P
;1]
resulting after
n
payments can be computed by left-multiplying the starting
vector [
A
; 1] by the matrix
B
n
. Assuming
m
>
1, that is, a positive rate of
interest, the calculation
[eigenvectors, diagonalform] = eig(B)
eigenvectors =
[ 1, 1]
[ 0, (m-1)/R]
diagonalform =
[m,0]
[0,1]
shows us that the matrix
B
has eigenvalues
m
and 1, and corresponding
eigenvectors [1; 0] and [1;(
m
−
1)
/
R
]
=
[1;
J
/
R
]. Now we can write the vector
[
A
; 1] as a linear combination of the eigenvectors: [
A
;1]
=
x
[1;0]
+
y
[1;
J
/
R
].
We can solve for the coefficients:
[x, y] = solve('A = x*1 + y*1', '1 = x*0 + y*J/R')
x=
(A*J-R)/J
y=
R/J
and so
[
A
;1]
=
(
A
−
(
R
/
J
))
∗
[1;0]
+
(
R
/
J
)
∗
[1;
J
/
R
]
and
B
n
·
[
A
;1]
=
(
A
−
(
R
/
J
))
∗
m
n
∗
[1;0]
+
(
R
/
J
)
∗
[1;
J
/
R
]
.
Therefore the principal remaining after
n
payments is
P
=
((
A
∗
J
−
R
)
∗
m
n
+
R
)
/
J
=
A
∗
m
n
−
R
∗
(
m
n
−
1)
/
J
.
This is the same result we obtained earlier.
To conclude, let's determine the amount of money
A
one can afford to
borrow as a function of what one can afford to pay as the monthly
payment
R
. We simply solve for
A
in the equation that
P
=
0 after
N
payments.
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