A) \[2\]
B) \[4\]
C) \[6\]
D) \[8\]
Correct Answer: C
Solution :
[c] \[A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]\] and \[|A|=3\] According to the question, \[B=\left[ \begin{matrix} {{a}_{12}}+{{a}_{13}} & {{a}_{11}}+{{a}_{13}} & {{a}_{11}}+{{a}_{12}} \\ {{a}_{22}}+{{a}_{23}} & {{a}_{21}}+{{a}_{23}} & {{a}_{21}}+{{a}_{22}} \\ {{a}_{32}}+{{a}_{33}} & {{a}_{31}}+{{a}_{33}} & {{a}_{31}}+{{a}_{32}} \\ \end{matrix} \right]\] \[\Rightarrow \,\,|B|\,\,=\,\,\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|\,\,\left| \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{matrix} \right|=3\times 2=6\]You need to login to perform this action.
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