A) 1, 1
B) \[\pm \,1,1\]
C) \[1,0\]
D) none of these
Correct Answer: A
Solution :
\[\because \] \[A=\left[ \begin{matrix} 1 & x \\ {{x}^{2}} & 4y \\ \end{matrix} \right]\]and \[B=\left[ \begin{matrix} -3 & 1 \\ 1 & 0 \\ \end{matrix} \right]\] \[\text{Adj}\,\text{A}\,\text{=}\,\left[ \begin{matrix} 4y & -x \\ -{{x}^{2}} & 1 \\ \end{matrix} \right]\] \[\therefore \] \[\text{adj}\,\text{+}\,A+B=\left[ \begin{matrix} 4y & -x \\ -{{x}^{2}} & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -3 & 1 \\ 1 & 0 \\ \end{matrix} \right]\] \[\Rightarrow \] \[\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 4y-3 & -x+1 \\ -{{x}^{2}}+1 & 1+0 \\ \end{matrix} \right]\] \[\Rightarrow \] \[4y-3=1\Rightarrow y=1\] and \[-x+1=0\Rightarrow x=1\]
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