A) \[\left[ \begin{matrix} 9 & 19 & -4 \\ -4 & 8 & -1 \\ 4 & -1 & 2 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 9 & 19 & 4 \\ -2 & 14 & 1 \\ -4 & 1 & x \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 9 & 19 & -4 \\ 2 & 8 & 1 \\ 4 & -1 & 2 \\ \end{matrix} \right]\]
D) None of these
Correct Answer: C
Solution :
\[A=\left[ \begin{matrix} 1 & -2 & 3 \\ 0 & 2 & 1 \\ -2 & 5 & 2 \\ \end{matrix} \right]\] \[{{a}_{11}}=\left| \begin{matrix} 2 & -1 \\ 5 & 2 \\ \end{matrix} \right|=4+5=9\] \[{{a}_{12}}=-\left| \begin{matrix} 0 & -1 \\ -2 & 2 \\ \end{matrix} \right|=-(0-2)=2\] \[{{a}_{13}}=\left| \begin{matrix} 0 & 2 \\ -2 & 5 \\ \end{matrix} \right|=0+4=4\] \[{{a}_{21}}=-\left| \begin{matrix} -2 & 3 \\ 5 & 2 \\ \end{matrix} \right|=-(-4-15)=19\] \[{{a}_{22}}=\left| \begin{matrix} 1 & 3 \\ -2 & 2 \\ \end{matrix} \right|=2+6=8\] \[{{a}_{23}}=-\left| \begin{matrix} 1 & -2 \\ -2 & 5 \\ \end{matrix} \right|=-(5-4)=-1\] \[{{a}_{31}}=\left| \begin{matrix} -2 & 3 \\ 2 & -1 \\ \end{matrix} \right|=2-6=-4\] \[{{a}_{32}}=-\left| \begin{matrix} 1 & 3 \\ 0 & -1 \\ \end{matrix} \right|=-(-1-0)=1\] \[{{a}_{33}}=\left| \begin{matrix} 1 & -2 \\ 0 & 2 \\ \end{matrix} \right|=2\] \[\therefore \]Adjoint of matrix \[A={{\left[ \begin{matrix} 9 & 2 & 4 \\ 19 & 8 & -1 \\ -4 & 1 & 2 \\ \end{matrix} \right]}^{T}}\]
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