A) \[11.2\,L\,{{H}_{2}}(g)\]
B) 1
C) \[HCl(aq)\]
D) 5
Correct Answer: C
Solution :
Firstly, solve\[{{A}^{2}}=A.\text{ }A\]and then \[|{{A}^{2}}|=25.\] \[\because \] \[A=\left[ \begin{matrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{2}}=\left[ \begin{matrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \\ \end{matrix} \right]\left[ \begin{matrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 25 & 25\alpha +5{{\alpha }^{2}} & 10\alpha +25{{\alpha }^{2}} \\ 0 & {{\alpha }^{2}} & 5{{\alpha }^{2}}+25\alpha \\ 0 & 0 & 25 \\ \end{matrix} \right]\] \[=25\left| \begin{matrix} 25 & 25\alpha +5{{\alpha }^{2}} \\ 0 & {{\alpha }^{2}} \\ \end{matrix} \right|=625{{\alpha }^{2}}\] But \[|{{A}^{2}}|=25\] \[\therefore \]\[625{{\alpha }^{2}}=25\Rightarrow |\alpha |=\frac{1}{5}\]
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