Let α and β be roots of equation x^2 + x + 1 = 0. Then for y ne 0 in R, $left| {begin{array}

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3 and 4 .Determinants and Matrices

hard

Let $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ is equal to

$y\,({y^2} - \,3)$

${y^3} - \,1$

$y^3$

$y\,({y^2} - \,1)$

(JEE MAIN-2019)

Solution

Roots of the equation ${x^2} + x + 1 = 0$ are $\alpha = \omega $ and $\beta = {\omega ^2}$ where $\omega ,{\omega ^2}$are complex cube roots of unity

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

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

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

Expanding along ${R_1}$, we get

$\Delta = y.{y^2} \Rightarrow D = {y^3}$

If $\alpha = {\omega ^2},\beta = \omega $ we get same value or on expansion using $\alpha + \beta = – 1,\alpha \beta = 1$ we get value ${y^3}$.

Standard 12

Mathematics

The values of $\alpha$, for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$, lie in the interval

hard

(JEE MAIN-2024)

The number of distinct real roots of $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$ in the interval $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ is

hard

(JEE MAIN-2021)

The number of integers $x$ satisfying $-3 x^4+\operatorname{det}\left[\begin{array}{ccc}1 & x & x^2 \\ 1 & x^2 & x^4 \\ 1 & x^3 & x^6\end{array}\right]=0$ is equal to

hard

(KVPY-2019)

Let $\alpha, \beta$ and $\gamma$ be real numbers. consider the following system of linear equations

$x+2 y+z=7$

$x+\alpha z=11$

$2 x-3 y+\beta z=\gamma$

Match each entry in List – $I$ to the correct entries in List-$II$

List – $I$ List – $II$

($P$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has ($1$) a unique solution

($Q$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has ($2$) no solution

($R$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,

then the system has
($3$) infinitely many solutions

($S$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has ($4$) $x=11, y=-2$ and $z=0$ as a solution

($5$) $x=-15, y=4$ and $z=0$ as a solution

Then the system has

medium

(IIT-2023)

If $\left| {\begin{array}{*{20}{c}}
{\cos 2x}&{{{\sin }^2}x}&{\cos 4x} \\
{{{\sin }^2}x}&{\cos 2x}&{{{\cos }^2}x} \\
{\cos 4x}&{{{\cos }^2}x}&{\cos 2x}
\end{array}} \right| = {a_0} + {a_1}\sin x + {a_2}{\sin ^2}x + …..$ then $a_0$ is equal to

normal

List – $I$	List – $II$
($P$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has	($1$) a unique solution
($Q$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has	($2$) no solution
($R$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$, then the system has	($3$) infinitely many solutions
($S$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has	($4$) $x=11, y=-2$ and $z=0$ as a solution
	($5$) $x=-15, y=4$ and $z=0$ as a solution

Let $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}} {y\, + \,1}&\alpha &\beta \\ \alpha &{y\, + \,\beta }&1\\ \beta &1&{y\, + \,\alpha } \end{array}} \right|$ is equal to

Solution

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Let $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ is equal to

If $\left| {\begin{array}{*{20}{c}}
{\cos 2x}&{{{\sin }^2}x}&{\cos 4x} \\
{{{\sin }^2}x}&{\cos 2x}&{{{\cos }^2}x} \\
{\cos 4x}&{{{\cos }^2}x}&{\cos 2x}
\end{array}} \right| = {a_0} + {a_1}\sin x + {a_2}{\sin ^2}x + …..$ then $a_0$ is equal to