Let $S = \left\{ {\sqrt n :1 \leqslant n \leqslant 50\,\,{\text{and}}\,\,n\,\,{\text{is}}\,\,{\text{odd}}} \right\}.$ Let $a \in S\,\,{\text{and}}\,\,A = \left[ {\begin{array}{*{20}{c}} 1&0&a \\ { - 1}&1&0 \\ { - a}&0&1 \end{array}} \right]$ If $\sum\limits_{a \in S} {\det \left( {adj\,A} \right)} = 100\lambda ,$ then $\lambda $ is equal to:

JEE MATHEMATICS MCQ: MATRICES

1 Easy

1 of 1

Application +4 | -1

Let

S = \left\{ {\sqrt n :1 \leqslant n \leqslant 50\,\,{\text{and}}\,\,n\,\,{\text{is}}\,\,{\text{odd}}} \right\}.

Let

a \in S\,\,{\text{and}}\,\,A = \left[ {\begin{array}{cccccccccccccccccccc} 1&0&a \\ { - 1}&1&0 \\ { - a}&0&1 \end{array}} \right]

\sum\limits_{a \in S} {\det \left( {adj\,A} \right)} = 100\lambda ,

then

\lambda

is equal to:

View answer and explanation

Correct option: B

221

Option B (221) is correct because the sum of the determinants of the adjoint matrices is calculated and simplified to find the value of lambda.

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|