Let A be a 3 × 3 matrix such that X^T A X=O for all nonzero 3 × 1 matrices X=[ array l x y z array ] . If A[ array l 1 1 1 array ]=[ array c 1 4 -5 array ], A[ array l 1 2 1 array ]=[ array c 0 4 -8 array ] , and

Correct answer: 0 aligned & X^ A X=0 & (x y z)[ array lll a_1 & a_2 & a_3 b_1 & b_2 & b_3 c_1 & c_2 & c_3 array ][ array l x y z array ]=0 & (x y z)[ array l a_1 x+a_2 y+a_3 z b_1 x+b_2 y+b_3 z c_1 x+c_2 y+c_3 z array ]=0 & x(a_1 x+a_2 y+a_3 z)+y(b_1 x+b_2 y+b_3 z) & +z(c_1 x+c_2 y+c_3 z) & a_1=0, b_2=0, c_3=0 & a_2+b_1=0, a_3+c_1=0, b_3+c_2=0 & A= skew symmetric matrix & A=[ array ccc 0 & x & y -x & 0 & z -y & -z & 0 array ] ; A[ array l 1 1 1 array ]=[ array c 1 4 -5 array ] & [ array ccc 0 & x & y -x & 0 & z -y & -z & 0 array ][ array l 1 1 1 array ]=[ array c 1 4 -5 array ] & x+y=1 & -x+z=4 & y+z=5 & [ array ccc 0 & x & y -x & 0 & z -y & -z & 0 array ][ array l 1 2 1 array ]=[ array c 1 4 -8 array ] & 2 x+y=0 x=-1 & -x+z=4 y=2 & -y-2 z=-8 z=3 aligned aligned & A=[ array ccc 0 & -1 & 2 1 & 0 & 3 -2 & -3 & 0 array ] & 2(A+l)=[ array ccc 2 & -2 & 4 2 & 2 & 6 -2 & -6 & 2 array ] & 2(A+l)=120 & det(adj(2 A+l)) & =120^2=2^6 · 3^2 · 5^2 & α=6, β=2, γ=2 & Hence α^2+β^2+γ^2=6^2+2^2+2^2=44 aligned

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Matrices

Let A be a 3 × 3 matrix such that X^T A X=O for all nonzero 3 × 1...

Let

A

be a

3 \times 3

matrix such that

X^T A X=O

for all nonzero

3 \times 1

matrices

X=\left[

\begin{array}{l}x \\ y \\ z\end{array}

\right]

. If

\mathbf{A}\left[

\begin{array}{l}1 \\ 1 \\ 1\end{array}

\right]=\left[

\begin{array}{c}1 \\ 4 \\ -5\end{array}

\right], \mathbf{A}\left[

\begin{array}{l}1 \\ 2 \\ 1\end{array}

\right]=\left[

\begin{array}{c}0 \\ 4 \\ -8\end{array}

\right]

, and

\operatorname{det}(\operatorname{adj}(2(\mathbf{A}+\mathbf{1})))-2^\alpha 3^\beta 5^\gamma, \alpha, \beta, \gamma \in N

, then

\alpha^2+\beta^2+\gamma^2

is_____.

Correct answer

Step-by-step solution

\begin{aligned} & X^{\top} A X=0 \\ & (x y z)\left[\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=0 \\ & (x y z)\left[\begin{array}{l} a_1 x+a_2 y+a_3 z \\ b_1 x+b_2 y+b_3 z \\ c_1 x+c_2 y+c_3 z \end{array}\right]=0 \\ & x\left(a_1 x+a_2 y+a_3 z\right)+y\left(b_1 x+b_2 y+b_3 z\right) \\ & +z\left(c_1 x+c_2 y+c_3 z\right) \\ & a_1=0, b_2=0, c_3=0 \\ & a_2+b_1=0, a_3+c_1=0, b_3+c_2=0 \\ & A=\text{ skew symmetric matrix } \\ & A=\left[\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right] ; A\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]=\left[\begin{array}{c} 1 \\ 4 \\ -5 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]=\left[\begin{array}{c} 1 \\ 4 \\ -5 \end{array}\right] \\ & x+y=1 \\ & -x+z=4 \\ & y+z=5 \\ & {\left[\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right]=\left[\begin{array}{c} 1 \\ 4 \\ -8 \end{array}\right]} \\ & 2 x+y=0 \quad x=-1 \\ & -x+z=4 \quad y=2 \\ & -y-2 z=-8 \quad z=3 \end{aligned}

\begin{aligned} & A=\left[\begin{array}{ccc}0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0\end{array}\right] \\ & 2(A+l)=\left[\begin{array}{ccc}2 & -2 & 4 \\ 2 & 2 & 6 \\ -2 & -6 & 2\end{array}\right] \\ & 2(A+l)=120 \\ & \Rightarrow \operatorname{det}(\operatorname{adj}(2 A+l)) \\ & \quad=120^2=2^6 \cdot 3^2 \cdot 5^2 \\ & \therefore \quad \alpha=6, \beta=2, \gamma=2 \\ & \text{ Hence } \alpha^2+\beta^2+\gamma^2=6^2+2^2+2^2=44\end{aligned}

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