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[4 points Suppose A, B, and Care 5 x 5 matrices with det(A) = -2, det(B) = 10 and the columns of C are linearly dependent. Fi

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\small \\det(A)=-2\\ det(B)=10\\ $Columns of C are linearly dependent ,So $det(C)=0\\ a)\\ det(10B^{-1})=10det(B^{-1})=\frac{10}{det(B)}=\frac{10}{-2}=-5\\ b)\\ det(A^5B^T)=det(A^5)det(B^T)=(det(A))^5det(B)=(-2)^510=-320\\ c)\\ det(CA+CB)\\ =det(C(A+B))=Det(C)Det(A+B)=(0)Det(A+B)=0

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