$Which of the following is correct?$

Moderate

A

$If A and B are square matrices$ $of order 3 such that$ $| A | = - 1, | B | = 3, then the$ $determinant of 3 AB is equal$ $to 27 .$
B

$If A is an invertible matrix, then$ $det (A^{- 1}) is equal to det (A)$
C

$If A and B are matrices of the$ same order, then $(A + B)^{2} = A^{2} + 2 A B + B^{2} is$ $possible if A B = I$
D

None of these

Solution

$\begin{matrix} We have | AB | = | A | | B | \\ Also for a square matrix of order 3, \\ | kA | = k^{3} ∣ A| \\ {because each element of the matrix A is multiplied by k and hence in this case we will have k}^{3} common \\ | 3 AB | = 3^{3} | A ‖ B | = 27 (- 1) (3) = - 81 \\ {Since A is invertible, therefore A}^{- 1} exists and \\ A A^{- 1} = I \Rightarrow \det (A A^{- 1}) = \det (I) \\ \Rightarrow \det (A) \det (A^{- 1}) = 1 \\ \Rightarrow \det (A^{- 1}) = \frac{1}{\det (A)} \\ {(A + B)}^{2} = (A + B) (A + B) \\ = A^{2} + A B + B A + B^{2} \\ = A^{2} + 2 A B + B^{2} if A B = B A \end{matrix}$

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