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\[{{(A+B)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}\]is possible if\[AB=I\]
D) None of these
Correct Answer: D
Solution :
[a] 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 k3 common \[\therefore \]\[|3AB|\,\,={{3}^{3}}|A||B|\,\,=27(-1)(3)=-81\] [b] Since \[A\] is invertible, therefore \[{{A}^{-1}}\] exists and \[A{{A}^{-1}}=I=\det (A{{A}^{-1}})=\det (I)\] \[\Rightarrow \] \[\det (A)\det ({{A}^{-1}})=1\] \[\Rightarrow \] \[\det ({{A}^{-1}})=\frac{1}{\det (A)}\] [c]\[{{(A+B)}^{2}}=(A+B)(A+B)\] \[={{A}^{2}}+AB+BA+{{B}^{2}}\] \[={{A}^{2}}+2AB+{{B}^{2}}\]if\[AB=BA\].
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