If omega is cube root of unity, then left| {begin{array}{*{20}{c}}1&omega &{{omega ^2}}omega

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3 and 4 .Determinants and Matrices

easy

If $\omega $ is the cube root of unity, then $\left| {\begin{array}{*{20}{c}}1&\omega &{{\omega ^2}}\\\omega &{{\omega ^2}}&1\\{{\omega ^2}}&1&\omega \end{array}} \right|$=

$1$

$0$

$\omega $

${\omega ^2}$

Solution

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

$ = \,\left| {\,\begin{array}{*{20}{c}}0&\omega &{{\omega ^2}}\\0&{{\omega ^2}}&1\\0&1&\omega \end{array}\,} \right| = 0$.

Standard 12

Mathematics

If $ab + bc + ca = 0$ and $\left| {\,\begin{array}{*{20}{c}}{a – x}&c&b\\c&{b – x}&a\\b&a&{c – x}\end{array}\,} \right| = 0$, then one of the value of $x$ is

hard

By using properties of determinants, show that:

$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

medium

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$

medium

Evaluate $\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|$

medium

Let $D_1 =$ $\left| {\,\begin{array}{{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a – b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is

normal

If $\omega $ is the cube root of unity, then $\left| {\begin{array}{*{20}{c}}1&\omega &{{\omega ^2}}\\\omega &{{\omega ^2}}&1\\{{\omega ^2}}&1&\omega \end{array}} \right|$=

Solution

Similar Questions

If $ab + bc + ca = 0$ and $\left| {\,\begin{array}{*{20}{c}}{a – x}&c&b\\c&{b – x}&a\\b&a&{c – x}\end{array}\,} \right| = 0$, then one of the value of $x$ is

By using properties of determinants, show that: $\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

Using the property of determinants and without expanding, prove that: $\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$

Evaluate $\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|$

Let $D_1 =$ $\left| {\,\begin{array}{*{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a – b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{*{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is

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By using properties of determinants, show that:

$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$

Let $D_1 =$ $\left| {\,\begin{array}{{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a – b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is