• # question_answer $\text{Let}\,\,\text{M}\,=\,\left[ \begin{matrix} 0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1 \\ \end{matrix} \right]\,\,\text{and}\,\,\,\text{adj}\,\,\text{M}\,=\,\left[ \begin{matrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \\ \end{matrix} \right]$ Where a and b are real numbers. Which of the following .options is/are correct? A) $\text{det}\,(adj\,\,{{\text{M}}^{2}})\,=\,81$ B) $a+b=3$ C) $(adj\,{{\text{M}}^{-1}})\,+\,adj\,\,{{\text{M}}^{-1}}=\,-\text{M}$ D) $\text{M}\left[ \begin{matrix} \alpha \\ \beta \\ \gamma \\ \end{matrix} \right]\,=\,\left[ \begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} \right]\,,\,\text{the}\,\,\alpha \,-\,\beta +\,\gamma \,=\,3$

Correct Answer: B , C , D

Solution :

 $\text{M}=\left[ \begin{matrix} 0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1 \\ \end{matrix} \right]$ and adj $\text{M}=\left[ \begin{matrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \\ \end{matrix} \right]$$\Rightarrow$adj$\text{M}=\left[ \begin{matrix} 2-3b & ab-1 & -1 \\ 8 & -6 & 2 \\ b-6 & 3 & -1 \\ \end{matrix} \right]$$=\left[ \begin{matrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \\ \end{matrix} \right]$ $2-3b=-1;ab-1=1$ $b-6=-\,5;a=2$ $b=1$ Now,     $\text{M}=\left[ \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \\ \end{matrix} \right]$ $\left| \text{M} \right|=8-10=-2$ $\Rightarrow$   $a+b=3$option [b] $\left| \text{adj}\,({{\text{M}}^{2}}) \right|={{\left| {{\text{M}}^{2}} \right|}^{2}}$ $={{\left| \text{M} \right|}^{4}}=16$ [c] ${{\text{(adj}\,\text{M})}^{-1}}+\text{adj}\,({{\text{M}}^{-1}})\,\text{option}\,(c)$$=\,\text{adj}\,({{\text{M}}^{-1}})+\text{adj}\,({{\text{M}}^{-1}})$$=2\text{adj}\,({{\text{M}}^{-\,\,1}})$$=2(\left| {{\text{M}}^{-\,\,1}} \right|\text{M})=2\left( \frac{1}{-2}\text{M} \right)=\,-\,\text{M}$ [d] $\text{M}\left[ \begin{matrix} \alpha \\ \beta \\ \gamma \\ \end{matrix} \right]\,=\,\left[ \begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} \right]$ $\left[ \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} \alpha \\ \beta \\ \gamma \\ \end{matrix} \right]\,=\,\begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix}$ $\beta +2\gamma =1$ $\alpha +2\beta +3\gamma =2$ $3\alpha +\beta +\gamma =1$ $\alpha =1,$$\beta$= $-1,$$\gamma =1$ $\alpha -\beta +\gamma =3$Option [d]

Solution :

Same as Above

Solution :

Same as Above

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