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...
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
1 answer