Let $\alpha \,\,\,{\rm{and}}\,\,\,\,\,\beta $ be the roots of equation ${x^2} - 6x - 2 = 0.$ If ${a_n} = {\alpha ^n} - {\beta ^n},$ for $n \ge 1$ then the value of $\frac{{{a_{10}} - 2{a_8}}}{{2{a_9}}}$ is equal to

JEE MATHEMATICS MCQ: MATRICES

1 Hard

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Application +4 | -1

Let

\alpha \,\,\,\mathrm{and}\,\,\,\,\,\beta

be the roots of equation

{x^2} - 6x - 2 = 0.

{a_n} = {\alpha ^n} - {\beta ^n},

for

n \ge 1

then the value of

\frac{{{a_{10}} - 2{a_8}}}{{2{a_9}}}

is equal to

View answer and explanation

Correct option: C

Option C (3) is correct because, after substituting the roots into the given expression and simplifying using the quadratic equation, the value reduces to 3.

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\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

and

a, b, c

are real numbers then the line

a x+b y+c=0

passes through the fixed point is The number of solutions of the system of equations

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

is If the equations

2x + 3y + 1 = 0,3x + y - 2 = 0

and

ax + 2y - b = 0

are consistent, then The value of the determinant

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

, is If

\omega

is the cube root of unity, then

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

is equal to For any matrix

A

of order

2 \times 2

, if

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

, then

|A|

is equal to