JEE Main 2022 · 24 Jun · Shift 1 · Mathematics Q9

← JEE Main 2022 · 24 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 9 of 90

Let $S=\{\sqrt n:1\le n\le50$ and $n$ is odd $\}$ .

Let $a\in S$ and $A=\begin{bmatrix}1&0&a\\-1&1&0\\-a&0&1\end{bmatrix}$ .

If $\displaystyle\sum_{a\in S}\det(\operatorname{adj}A)=100\lambda$ , then $\lambda$ is equal to :

218

221

663

1717

Answer and worked solution

Official answer: B

Determinant. Expanding along the first row, $\det A=1+a\cdot a=1+a^2$ .

Adjugate. For a $3\times3$ matrix $\det(\operatorname{adj}A)=(\det A)^2=(1+a^2)^2$ .

Sum. With $a=\sqrt n$ , $n=2k-1$ ( $k=1,\dots,25$ ): $(1+n)^2=4k^2$ , so
$\sum=4\cdot\frac{25\cdot26\cdot51}6=22100=100\lambda\implies\lambda=221.$

(B)

Topics:Adjoint and inverse Arithmetic progression

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JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 9