A) 80
B) 100
C) - 110
D) 92
Correct Answer: B
Solution :
\[A=\left[ \begin{matrix} 3 & 5 \\ 2 & 0 \\ \end{matrix} \right]\] and \[B=\left[ \begin{matrix} 1 & 17 \\ 0 & -10 \\ \end{matrix} \right]\] \[\therefore \] \[AB=\left[ \begin{matrix} 3 & 5 \\ 2 & 0 \\ \end{matrix} \right]\,\,\left[ \begin{matrix} 1 & 17 \\ 0 & -10 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 3+0 & 51-50 \\ 2+0 & 34-0 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & 1 \\ 2 & 34 \\ \end{matrix} \right]\] \[\Rightarrow \] \[|AB|=\left| \begin{matrix} 1 & 1 \\ 2 & 34 \\ \end{matrix} \right|\] = 102 - 2 = 100You need to login to perform this action.
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