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

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

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Answer: Hence proved,

\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.

Posted by

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