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Algebra of Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices4 Marks
Question
If w is a complex cube root of unity, show that.
$\begin{pmatrix}\begin{bmatrix}1&w&w^2\\w&w^2&1\\w^2&1&w\end{bmatrix} +\begin{bmatrix}w&w^2&1\\w^2&1&w\\w&w^2&1\end{bmatrix}\end{pmatrix}\begin{bmatrix}1\\w\\w^2\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
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Answer

Here,
$\text{LHS}=\begin{pmatrix}\begin{bmatrix}1&w&w^2\\w&w^2&1\\w^2&1&w\end{bmatrix} +\begin{bmatrix}w&w^2&1\\w^2&1&w\\w&w^2&1\end{bmatrix}\end{pmatrix}\begin{bmatrix}1\\w\\w^2\end{bmatrix}$
$=\begin{bmatrix}1+w&w+w^2&w^2+1\\w+w^2&w^2+1&1+w\\w^2+w&1+w^2&w+1\end{bmatrix}\begin{bmatrix}1\\w\\w^2\end{bmatrix}$
$=\begin{bmatrix}-w^2&-1&-w\\-1&-w&-w^2\\-1&-w&-w^2\end{bmatrix}\begin{bmatrix}1\\w\\w^2\end{bmatrix}$ $\big(\because1+\text{w}+\text{w}^2=0\text{ and w}^3=1\big)$
$=\begin{bmatrix}-w^2-w-w^3\\-1-w^2-w^4\\-1-w^2-w^4\end{bmatrix}$
$=\begin{bmatrix}-w(1+w+w^2)\\-1-w^2-w^3w\\-1-w^2-w^3w\end{bmatrix}$
$=\begin{bmatrix}-w\times0\\-1-w^2-w\\-1-w^2-w\end{bmatrix}$ $\big(\because1+\text{w}+\text{w}^2=0\text{ and w}^3=1\big)$
$=\begin{bmatrix}0\\-0\\-0\end{bmatrix}$
$=\begin{bmatrix}0\\0\\0\end{bmatrix}$
$\therefore\ \begin{pmatrix}\begin{bmatrix}1&w&w^2\\w&w^2&1\\w^2&1&w\end{bmatrix} +\begin{bmatrix}w&w^2&1\\w^2&1&w\\w&w^2&1\end{bmatrix}\end{pmatrix}\begin{bmatrix}1\\w\\w^2\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$

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