If omega is one of imaginary cube roots of unity, then value of determinant left| {begin{array}{*{20

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3 and 4 .Determinants and Matrices

medium

If $\omega$ is one of the imaginary cube roots of unity, then the value of the determinant $\left| {\begin{array}{*{20}{c}}1&{{\omega ^3}}&{{\omega ^2}}\\ {{\omega ^3}}&1&\omega \\{{\omega ^2}}&\omega &1\end{array}} \right|$ $=$

$1$

$2$

$3$

none

Solution

Put $\omega ^3 = 1$ $\left| {\,\begin{array}{*{20}{c}}1&1&{{\omega ^2}}\\ 1&1&\omega \\{{\omega ^2}}&\omega &1\end{array}\,} \right|$ and open by $R_1$ to get $(1 – \omega ^2) + (1 – \omega ) = 3$

Standard 12

Mathematics

Find the equation of the line joining $\mathrm{A}(1,3)$ and $\mathrm{B}(0,0)$ using determinants and find $\mathrm{k}$ if $\mathrm{D}(\mathrm{k}, 0)$ is a point such that area of triangle $\mathrm{ABD}$ is $3 \,\mathrm{sq}$ $\mathrm{units}$.

medium

Let $\alpha, \beta(\alpha \neq \beta)$ be the values of $m$, for which the equations $x+y+z=1 ; x+2 y+4 z=m$ and $x+4 y+10 z=m^2$ have infinitely many solutions. Then the value of $\sum_{n=1}^{10}\left(n^\alpha+n^\beta\right)$ is equal to :

hard

(JEE MAIN-2025)

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$x+y+\sqrt{3} z=0$

$-x+(\tan \theta) y+\sqrt{7} z=0$

$x+y+(\tan \theta) z=0$

has non-trivial solution. Then $\frac{120}{\pi} \sum_{\theta \in s} \theta$ is equal to

hard

(JEE MAIN-2023)

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normal

$\left| {\,\begin{array}{*{20}{c}}{a – 1}&a&{bc}\\{b – 1}&b&{ca}\\{c – 1}&c&{ab}\end{array}\,} \right| = $

easy