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Lyrx [107]
3 years ago
6

Suppose that A = PDP-1 . Prove that det(A) = det(D)

Mathematics
1 answer:
daser333 [38]3 years ago
5 0

Answer:

Check.

Step-by-step explanation:

To prove it we need to know that  for two matrices A and B we have that:

det(AB) = det(A)*det(B) and det(A^{-1}) = \frac{1}{det(A)}. Now:

A = PDP^{-1}

det(A) = det(PDP^{-1})

det(A) = det(P)*det(D)*det(P^{-1})

det(A) = det(P)*det(D)*\frac{1}{det(P)}

det(A) = det(P)*\frac{1}{det(P)}*det(D)

det(A) = det(D).

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A rectangle has a perimeter of 30 ft. Find a function that models its area A in terms of the length x of one of its sides. What
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Answer:

Step-by-step explanation:

Let the other side of the rectangle be y. The perimeter of the rectangle is expressed as P = 2(x+y)

Given P = 30ft, on substituting P = 30 into the expression;

30 = 2(x+y)

x+y = 15

y = 15-x

Also since the area of the rectangle is xy;

A = xy

Substitute y = 15-x into the area;

A = x(15-x)

A = 15x-x²

The function that models its area A in terms of the length x of one of its sides is A = 15x-x²

The side of length x yields the greatest area when dA/dx = 0

dA/dx = 15-2x

15-2x = 0

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x = -15/-2

x = 7.5 ft

Hence the side length, x that yields the greatest area is 7.5ft.

Since y = 15-x

y = 15-7.5

y = 7.5

Area of the rectangle = 7.5*7.5

Area of the rectangle = 56.25ft²

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Answer:

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