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mth 164
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questions for ch. 11-13
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I do not understand how to do a problem like this....Find the x^ 5 term in the expansion of ( x + 2 y^2 ) ^ 15.
and how to Set a system as a matrix multiplication of the form A B = C. Find the inverse A^-1 of A and use it to solve the system.
-7 x + -5y = 21 .
-20 x + -18 y = -36 .
@& By the binomial expansion
(a + b)^n = C(n, 0) a^n + C(n, 1) a^(n-1) b + C(n, 2) a^(n-2) b^2 + ... + C(n, n-2) a^2 b^(n - 2) + C(n, n-1) a b^(n - 1) + b^n.
If a = x and b = 2 y^2, with n = 15, we get the expansion of (x + 2 y^2)^15.
The fifth term of the expansion would be
C(15, 5) x^5 ( 2y^2 ) ^(15 - 5)
= C(15, 5) x^5 ( 2 y^2) ^ 10
= C(15, 5) x^5 * 2^10 * (y^2)^5
= C(15, 5) * 1024 x^5 y^10.
C(15, 5) = 15 ! / (5 ! * (15 - 5)! ) = 15 ! / (5 ! * 10 !) = 15 * 14 * 13 * 12 * 11 / (5 !) = 3 * 7 * 13 * 11 = etc..
The system is written in matrix form as
[ [-7, -5], [-20, -18] ] * [x; y] = [21; -36]
This is
A X = Y
with
A = [ [-7, -5], [[-20, -18] ]
If we row-reduce the augmented matrix
[ [-7, -5, 1, 0], [[-20, -18, 0, 1] ]
to the form
[ [1, 0, a, b], [0, 1, c, d]]
then the matrix
[ [a, b], [c, d] ]
is the inverse of A.
The system is then solved:
A^-1 * A X = A^-1 * Y
so that
X = A^-1 * Y = [ [a, b], [c, d] ] * [ 21; -18 ].
This is illustrated better in the associated QA and in the text, where the matrices can be shown in standard format.
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