At what value of x, will left| {,begin{array}{*{20}{c}}{x + {omega ^2}}&omega &1omega &{

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3 and 4 .Determinants and Matrices

easy

At what value of $x,$ will $\left| {\,\begin{array}{*{20}{c}}{x + {\omega ^2}}&\omega &1\\\omega &{{\omega ^2}}&{1 + x}\\1&{x + \omega }&{{\omega ^2}}\end{array}\,} \right| = 0$

$x = 0$

$x = 1$

$x = - 1$

None of these

Solution

(a) $\left| {\,\begin{array}{*{20}{c}}{x + {\omega ^2}}&\omega &1\\\omega &{{\omega ^2}}&{1 + x}\\1&{x + \omega }&{{\omega ^2}}\end{array}\,} \right| = 0$

Check at $x = 0,$ we get $\left| {\,\begin{array}{*{20}{c}}{{\omega ^2}}&\omega &1\\\omega &{{\omega ^2}}&1\\1&\omega &{{\omega ^2}}\end{array}\,} \right|$

= ${\omega ^2}({\omega ^4} – \omega ) – \omega ({\omega ^3} – 1) + 1({\omega ^2} – {\omega ^2})$

= ${\omega ^2}(\omega – \omega ) – \omega \,(1 – 1) + 0 = 0$ Or

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

by ${C_1} \to {C_1} + {C_2} + {C_3}$

$ = \,\left| {\,\begin{array}{*{20}{c}}x&\omega &1\\x&{{\omega ^2}}&{1 + x}\\x&{x + \omega }&{{\omega ^2}}\end{array}\,} \right|$, ($\because \,1 + \omega + {\omega ^2} = 0$)

$= 0$, if $x = 0$.

Standard 12

Mathematics

The value of $\left| {\,\begin{array}{*{20}{c}}{441}&{442}&{443}\\{445}&{446}&{447}\\{449}&{450}&{451}\end{array}\,} \right|$ is

easy

Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a b c+b c+c a+a b$

hard

A value of $\theta \in (0, \pi /3)$, for which $\left| {\begin{array}{*{20}{c}}
{1 + {{\cos }^2}\,\theta }&{{{\sin }^2}\,\theta }&{4\,\cos \,6\theta } \\
{{{\cos }^2}\,\theta }&{1 + {{\sin }^2}\,\theta }&{4\,\cos \,6\theta } \\
{{{\cos }^2}\,\theta }&{{{\sin }^2}\,\theta }&{1 + 4\,\cos \,6\theta }
\end{array}} \right| = 0$, is

hard

(JEE MAIN-2019)

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 $\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&x&{{x^2}}\\{{b^2}}&{ab}&{{a^2}} \end{array}} \right|$ $= 0$ , then :