JEE MATHEMATICS MCQ: MATRICES

1 Medium

1 of 1

Analyzing +4 | -1

Let

\quad \Delta=\left|\begin{array}{ccc}-b c & b^2+b c & c^2+b c \\ a^2+a c & -a c & c^2+a c \\ a^2+a b & b^2+a b & -a b\end{array}\right|

and the equation

p x^3+q x^2+r x+s=0

has roots

a, b, c

, where

a, b, c \in R^{+}

, then if

\Delta=27

and

a^2+b^2+c^2=2

find

\frac{\sqrt{2} p}{q}+1

View answer and explanation

Correct option: C

01-Feb

The correct answer is 0.5 because after simplifying the determinant and using the given conditions, we can find the value of

\frac{\sqrt{2} p}{q}+1

More Questions from MATRICES

Practice All Chapter MCQs →

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

a, b, c

a x+b y+c=0

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

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

ax + 2y - b = 0

\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|

\omega

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

A

2 \times 2

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

|A|