Determinants and Adjoints

[~Sam~]

Homework Statement

Suppose A is a 33 matrix such that det(A)=15.

Then det[A3((adj(A))−1)2]= and det[5A−1(adj(A))]

-1=inverse

Homework Equations

I know the properties of determinants and inverses

The Attempt at a Solution

Problem simplifying to get a number.

 

[Dick]

If you know the properties of determinants and inverses, can't you at least explain what the problem you are having is?

 

[Architeuthis Dux]

First, we can use the property that det(AB) = det(A) * det(B) to simplify the first expression. Since we are given that det(A) = 15, we can rewrite the expression as det(A)^3 * det((adj(A))^-1)^2.

Next, we can use the property that det(A^-1) = 1/det(A) to further simplify the expression. This gives us det(A)^3 * (1/det(adj(A))^2.

Since we are given that det(adj(A))^-1 = 1/det(A), we can substitute this into the expression to get det(A)^3 * (det(A))^2.

Using the property that det(A^n) = (det(A))^n, we can simplify the expression to get det(A)^5.

Finally, since we are given that det(A) = 15, we can substitute this into the expression to get 15^5 = 759375.

For the second expression, we can use the property that det(kA) = k^n * det(A) to simplify the expression. Since we are given that det(A) = 15, we can rewrite the expression as 5^3 * 15^-1 * det(adj(A)).

Using the property that det(A^-1) = 1/det(A), we can further simplify the expression to get 5^3 * 15^-1 * (1/det(A)).

Substituting in the given value of det(A) = 15, we get 5^3 * 15^-1 * (1/15) = 125/15 = 25/3.

Therefore, the final answers for the two expressions are 759375 and 25/3, respectively.