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Question 4 : Let A, B, C be three 4 X 4 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

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Let us recall a few properties of determinants.

det(kA) = k^{n} det(A), where A is a Square Matrix

n is the order of the Square Matrix

k is some scalar constant

det(AB) = det(A) * det(B), where A,B are two matrices compatible for multiplication.

det(A^{-1})=\\frac{1}{det(A)}\\) where A^{-1} is the inverse of matrix A.

det(A^{T})= det(A) where A^{T} is the transpose of matrix A.

Using the above properties,

det(2AB^{-1}C^{3}B^{T}) = det(2A) * det(B^{-1}) * det(C^{3}) * det(B^{T})

= 24 det(A) * \\frac{1}{det(B)}\\) * det(C) * det(C) * det(C) * det(B)

= 16 * 5 * \\frac{1}{-3}\\) * \\frac{1}{2}\\) * \\frac{1}{2}\\) * \\frac{1}{2}\\) * -3 = 10.

The question is **"Let A, B, C be three 4 X 4 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" **

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