1. Tardigrade
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  4. If 1,ω ,ω 2 are the cube roots of unity and if [ beginmatrix 1+ω 2ω -2ω -b endmatrix ]+[ beginmatrix a -ω 3ω 2 endmatrix ]=[ beginmatrix 0 ω ω 1 endmatrix ], then a2+b2 is equal to

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Q. If $ 1,\omega ,{{\omega }^{2}} $ are the cube roots of unity and if $ \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], $ then $ {{a}^{2}}+{{b}^{2}} $ is equal to

KEAMKEAM 2009Matrices

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 $