A) \[1+{{\omega }^{2}}\]
B) \[{{\omega }^{2}}-1\]
C) \[1+\omega \]
D) \[{{(1+\omega )}^{2}}\]
E) \[{{\omega }^{2}}\]
Correct Answer: C
Solution :
Given, \[\left[ \begin{matrix} 1+\omega & 2\omega \\ -2\omega & -b \\ \end{matrix} \right]+\left[ \begin{matrix} a & -\omega \\ 3\omega & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & \omega \\ \omega & 1 \\ \end{matrix} \right]\] \[\Rightarrow \] \[\left[ \begin{matrix} 1+\omega +a & \omega \\ \omega & 2-b \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & \omega \\ \omega & 1 \\ \end{matrix} \right]\] \[\Rightarrow \] \[1+\omega +a=0,2-b=1\] \[\Rightarrow \] \[a=-1-\omega ,b=1\] \[\therefore \] \[{{a}^{2}}+{{b}^{2}}=(-1-{{\omega }^{2}})+{{1}^{2}}\] \[=1+{{\omega }^{2}}+2\omega +{{1}^{2}}\] \[=0+\omega +1\] \[(\because 1+\omega +{{\omega }^{2}}=0)\] \[=1+\omega \]
You need to login to perform this action.
You will be redirected in
3 sec
