Which of the following is correct

Easy

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

_{$(a) w e h a v e | A B | = | A | | B |$}
Also for a square matrix of order3, _{$| K A | = K^{3} | A |$} because each element of the matrix A is multiplied by k and hence in this case we will have _$K^{3}$ common

Since A is invertible, therefore exists and

_{$∴ | 3 A B | = 3^{3} | A | | B | = 27 (- 1) (3) = - 81$}

_$(b)$ Since A is invertible, therefore_{$A^{- 1}$} for exists and

_{$A A^{- 1} = 1 \Rightarrow \det (A A^{- 1}) = \det (I)$}

_{$\Rightarrow \det (A) \det (A^{- 1}) = 1 \Rightarrow \det (A^{- 1}) = \frac{1}{\det (A)}$}

_$(c)$ _{${(A + B)}^{2} = (A + B) (A + B)$}

_{$= A^{2} + A B + B A + B^{2} = A^{2} + 2 A B + B^{2} i f A B = B A$}

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