If A, B, and C are 3×3 matrices; and det(A) = −3, det(B) = −2, and det(C) = −1 then compute:
det(2A^−1C^−1A^TC^2B^2) = 0
since
|A^-1| = 1/|A|
|A^T| = |A|
|A^2| = |A|^2
|2A| = 2^3|A| for 3x3 A
Now, also det is associative, so we can break up the product into separate determinants, giving us
|2A^-1 C^-1 A^T C^2 B^2|
= 8 * 1/|A| * 1/|C| * |A| * |C|^2 * |B|^2
= 8 * 1/-3 * 1/-1 * -3 * 1 * 4
= -32
Not sure how you can set that to zero, but maybe there's something you left out...
Well, first things first, we need to find the determinants of A^(-1), C^(-1), A^T, C^2, and B^2. Let's break it down step by step:
1. Determinant of A^(-1):
We know that det(A) = -3, so det(A^(-1)) = 1/det(A) = 1/(-3) = -1/3.
2. Determinant of C^(-1):
Similarly, det(C^(-1)) = 1/det(C) = 1/(-1) = -1.
3. Determinant of A^T:
Since A is a 3x3 matrix and A^T is its transpose, the determinant of A^T will be the same as det(A), which is -3.
4. Determinant of C^2:
To find det(C^2), we square the determinant of C. So det(C^2) = (det(C))^2 = (-1)^2 = 1.
5. Determinant of B^2:
Similarly, det(B^2) = (det(B))^2 = (-2)^2 = 4.
Now, let's put it all together and compute det(2A^(-1)C^(-1)A^TC^2B^2):
det(2A^(-1)C^(-1)A^TC^2B^2) = 2 * det(A^(-1)) * det(C^(-1)) * det(A^T) * det(C^2) * det(B^2)
Substituting the determinants we found earlier:
det(2A^(-1)C^(-1)A^TC^2B^2) = 2 * (-1/3) * (-1) * (-3) * 1 * 4
Simplifying:
det(2A^(-1)C^(-1)A^TC^2B^2) = 2 * 1/3 * 3 * 1 * 4
= 8/3 * 4
= 32/3
So, the determinant of 2A^(-1)C^(-1)A^TC^2B^2 is 32/3, which is not equal to zero.
Hope that clears things up! Well, unless you were looking for the determinant to be a "zero", but I can't make that happen. Sorry!
To compute det(2A^−1C^−1A^TC^2B^2), we need to simplify the expression step by step. Here is the step-by-step solution:
Step 1: Inverse of A and C matrices.
Let's find the inverse of A (A^−1) and C (C^−1) matrices.
Step 2: Transpose of A matrix.
Next, let's find the transpose of the A matrix (A^T).
Step 3: Compute (A^−1C^−1A^T).
Multiply A^−1, C^−1, and A^T matrices together.
Step 4: Compute (C^2B^2).
Multiply C^2 and B^2 matrices together.
Step 5: Combine (A^−1C^−1A^T) and (C^2B^2).
Multiply (A^−1C^−1A^T) and (C^2B^2) matrices together.
Step 6: Multiply the result by 2.
Multiply the resulting matrix from Step 5 by 2.
Step 7: Compute the determinant.
Find the determinant of the matrix obtained from Step 6.
Step 8: Determine the final result.
Check if the determinant obtained from Step 7 is equal to 0.
Let's compute these steps one by one.
To compute det(2A^(-1)C^(-1)A^TC^2B^2), we can use the following properties of determinants:
1. det(AB) = det(A) * det(B) (multiplicative property)
2. det(A^(-1)) = 1/det(A) (property of the inverse)
3. det(A^T) = det(A) (property of the transpose)
Using these properties, let's break down the expression step by step:
1. First, apply the multiplicative property to separate the determinants:
det(2A^(-1)C^(-1)A^TC^2B^2) = 2 * det(A^(-1)) * det(C^(-1)) * det(A^T) * det(C^2) * det(B^2)
2. Apply property 2 to det(A^(-1)) and det(C^(-1)):
2 * det(A^(-1)) * det(C^(-1)) * det(A^T) * det(C^2) * det(B^2) = 2 * (1/det(A)) * (1/det(C)) * det(A^T) * det(C^2) * det(B^2)
3. Apply property 3 to det(A^T):
2 * (1/det(A)) * (1/det(C)) * det(A^T) * det(C^2) * det(B^2) = 2 * (1/det(A)) * (1/det(C)) * det(A) * det(C^2) * det(B^2)
4. Combine terms and simplify:
2 * (1/det(A)) * (1/det(C)) * det(A) * det(C^2) * det(B^2) = 2 * (1/det(C)) * det(C^2) * det(B^2)
= 2 * (1/det(C)) * (det(C) * det(C)) * det(B^2)
= 2 * (1/det(C)) * (-1 * -1) * det(B^2)
= 2 * (1/-1) * (-1 * -1) * det(B^2) (substituting det(C) = -1)
5. Simplify further:
2 * (1/-1) * (-1 * -1) * det(B^2) = -2 * 1 * det(B^2) = -2 * det(B^2)
Since we know det(B) = -2, we can substitute it in the expression:
-2 * det(B^2) = -2 * (-2 * -2) = -2 * 4 = -8
So, det(2A^(-1)C^(-1)A^TC^2B^2) = -8.
Therefore, the answer is -8.