A) 4
B) 3
C) 2
D) 1
Correct Answer: D
Solution :
We have \[\det B=\det .\,C=\det .\,A=2,\] because \[\det .B=\left| \begin{matrix} {{a}_{11}} & \frac{{{a}_{12}}}{3} & \frac{{{a}_{13}}}{{{3}^{2}}} \\ 3{{a}_{21}} & {{a}_{22}} & \frac{1}{3}{{a}_{23}} \\ 9{{a}_{31}} & 3{{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|\,=|A|=2\] Similarly, \[\det .C=\det .B=|A|=2\] Hence \[\det .B+\det .C=2+2=4\].You need to login to perform this action.
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