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(d) Given, $\Delta = \left| {\,\begin{array}{*{20}{c}}a&b&0\0&a&b\b&0&a\end{array}\,} ight|\, = \,0.$
Expanding the given

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$\left( {\frac{a}{b}} \right)$is one of the cube roots of $ -1$

Vedclass · Answer

(d) Given, $\Delta = \left| {\,\begin{array}{*{20}{c}}a&b&0\0&a&b\b&0&a\end{array}\,} ight|\, = \,0.$
Expanding the given determinant, we get $a({a^2} - 0) - b(0 - {b^2}) = 0$ or ${a^3} + {b^3} = 0.$
This equation may be written as ${\left( {\frac{a}{b}} ight)^3} = - 1.$
Therefore, $\left( {\frac{a}{b}} ight)$ is one of the cube roots of $ -1$.

If $\left| {\,\begin{array}{*{20}{c}}a&b&0\\0&a&b\\b&0&a\end{array}\,} \right| = 0$, then

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