Explain solution for rd sharma class class 12 chapter Algebra of matrices exercise 4.3 question 21

$\left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

$w^3 = 1$ when w is a complex cube root of unity.

Given: $A= \left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

w is a complex cube root of unity

Prove: LHS=RHS

$\left(\left[\begin{array}{ccc}1 & w & w^{2} \\ w & w^{2} & 1 \\ w^{2} & 1 & w\end{array}\right]+\left[\begin{array}{ccc}w & w^{2} & 1 \\ w^{2} & 1 & w \\ w & w^{2} & 1\end{array}\right]\right)\left[\begin{array}{c}1 \\ w \\ w^{2}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

Consider the LHS\

$=\left[\begin{array}{ccc} 1+w & w+w^{2} & w^{2}+1 \\ w+w^{2} & w^{2}+1 & 1+w \\ w^{2}+w & 1+w^{2} & w+1 \end{array}\right]\left[\begin{array}{c} 1 \\ w \\ w^{2} \end{array}\right]$

We know that $1+w+w^{2}=0$ and $w^{3}=1$

$=\left[\begin{array}{ccc} -w^{2} & -1 & -w \\ -1 & -w & -w^{2} \\ -1 & -w & -w^{2} \end{array}\right]\left[\begin{array}{c} 1 \\ w \\ w^{2} \end{array}\right]$

Now by simplifying we get,

$\begin{array}{l} \\\\\\\\\\ =\left[\begin{array}{l} -w^{2}-w-w^{3} \\ -1-w^{2}-w^{4} \\ -1-w^{2}-w^{4} \end{array}\right] \\ =\left[\begin{array}{l} -w\left(w+1+w^{2}\right) \\ -1-w^{2}-w^{3} w \\ -1-w^{2}-w^{3} w \end{array}\right] \end{array}$

Again by substituting $1+w+w^{2}=0$ and $w^3 = 1$ in above matrix we get,

$=\left[\begin{array}{c} -w(0) \\ -1-w^{2}-w \\ -1-w^{2}-w \end{array}\right]=\left[\begin{array}{c} 0 \\ -\left(1+w+w^{2}\right) \\ -\left(1+w+w^{2}\right) \end{array}\right]\\\\ =\left[\begin{array}{c} 0 \\ -(0) \\ -(0) \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]$

Therefore LHS=RHS hence proved.