Let omega be a complex number such that 2omega + 1 = z where z = √ {

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3 and 4 .Determinants and Matrices

hard

Let $\omega $ be a complex number such that $2\omega + 1 = z$ where $z = \sqrt { - 3} $ . If $\left| {\begin{array}{*{20}{c}}1&1&1\\1&{ - {\omega ^2} - 1}&{{\omega ^2}}\\1&{{\omega ^2}}&{{\omega ^7}}\end{array}} \right| = 3k$ then $k$ is equal to :

$1$

$-z$

$z$

$-1$

(JEE MAIN-2017)

Solution

Given $2\omega + 1 = z;$

$z = \sqrt {3i} $

$ \Rightarrow \omega = \frac{{\sqrt {3i} – 1}}{2}$

$ \Rightarrow \omega $ is complex cube root of unity

Applying ${R_1} \to {R_1} + {R_2} + {R_3}$

$ = \left| \begin{array}{l}
3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\
1\,\,\,\,\, – {\omega ^2} – 1\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^2}\\
1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\omega
\end{array} \right|\,$

$ = 3\left( { – 1 – \omega – \omega } \right) = – 3\left( {1 + 2\omega } \right)\, = – 3z$

$ \Rightarrow k = – z$

Standard 12

Mathematics

$\left| {\begin{array}{*{20}{c}}
{4 + {x^2}}&{ – 6}&{ – 2}\\
{ – 6}&{9 + {x^2}}&3\\
{ – 2}&3&{1 + {x^2}}
\end{array}} \right|$ $;(x\neq0)$ is not divisible by

normal

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medium

Let $A=\left(\begin{array}{cc}4 & -2 \\ \alpha & \beta\end{array}\right)$ . If $A ^{2}+\gamma A +18 I = O$, then $\operatorname{det}( A )$ is equal to

easy

(JEE MAIN-2022)

$\left| {\,\begin{array}{*{20}{c}}{{{({a^x} + {a^{ – x}})}^2}}&{{{({a^x} – {a^{ – x}})}^2}}&1\\{{{({b^x} + {b^{ – x}})}^2}}&{{{({b^x} – {b^{ – x}})}^2}}&1\\{{{({c^x} + {c^{ – x}})}^2}}&{{{({c^x} – {c^{ – x}})}^2}}&1\end{array}\,} \right| = $

easy

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easy

Let $\omega $ be a complex number such that $2\omega + 1 = z$ where $z = \sqrt { - 3} $ . If $\left| {\begin{array}{*{20}{c}}1&1&1\\1&{ - {\omega ^2} - 1}&{{\omega ^2}}\\1&{{\omega ^2}}&{{\omega ^7}}\end{array}} \right| = 3k$ then $k$ is equal to :

Solution

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$\left| {\begin{array}{*{20}{c}}
{4 + {x^2}}&{ – 6}&{ – 2}\\
{ – 6}&{9 + {x^2}}&3\\
{ – 2}&3&{1 + {x^2}}
\end{array}} \right|$ $;(x\neq0)$ is not divisible by