Sign in
Please select an account to continue using cracku.in
↓ →
Let A, B, C be three 4x4 matrices such that det A = 5, det B = -3, and det $$C = \frac{1}{2}$$. Then the det $$(2AB^{-1}C^{3}B^{T})$$ is ........
Correct Answer: 10
For A, B square matrix of n x n :
1. det(AB) = det(A) x det(B)
2. det($$B^T$$) = det(B)
3. det($$B^{-1}$$) = $$\dfrac{\ 1}{\det\left(B\right)}$$
4. det(xA) = $$x^n$$ x det(A)
Hence, now det$$(2AB^{-1}C^{3}B^{T})$$ = det(2A) x det($$B^{-1}$$) x $$[\det(C)]^3$$ x det(B)
= det(2A) x $$\dfrac{\ 1}{\det\left(B\right)}$$ x $$[\det(C)]^3$$ x det(B)
= $$2^4$$ x 5 x $$\dfrac{\ -1}{3}$$ x $$\dfrac{\ 1}{2^3}$$ x (-3)
= 16 x 5 x 1 x $$\dfrac{\ 1}{8}$$
= 10
Therefore, the answer is 10.
Create a FREE account and get: