First slide

Operations of matrix

Question

if $w$ is a complex cube root of unity, than $([\begin{array}{l} 1 {w w}^{2} \\ w w^{2} 1 \\ w^{2} 1 w \end{array}] + [\begin{array}{l} w w^{2} 1 \\ w^{2} 1 w \\ w w 1 \end{array}]) [\begin{array}{l} 1 \\ w \\ w^{2} \end{array}] =$

Easy

A

$[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}]$
B

$[\begin{array}{l} 2 \\ 4 \\ 8 \end{array}]$
C

$[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}]$
D

$[\begin{array}{l} 3 \\ 9 \\ 27 \end{array}]$

Solution

$W e h a v e ([\begin{array}{l} 1 {w w}^{2} \\ w w^{2} 1 \\ w^{2} 1 w \end{array}] + [\begin{array}{l} w w^{2} 1 \\ w^{2} 1 w \\ w w 1 \end{array}]) [\begin{array}{l} 1 \\ w \\ w^{2} \end{array}] = ([\begin{array}{l} 1 {w w}^{2} \\ w w^{2} 1 \\ w^{2} 1 w \end{array}] + w [\begin{array}{l} 1 {w w}^{2} \\ w w^{2} 1 \\ w^{2} 1 w \end{array}]) [\begin{array}{l} 1 \\ w \\ w^{2} \end{array}] = (1 + w) [\begin{array}{l} 1 {w w}^{2} \\ w w^{2} 1 \\ w^{2} 1 w \end{array}] [\begin{array}{l} 1 \\ w \\ w^{2} \end{array}] = (1 + w) [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}]$

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