JEE Main 2025 — Matrices Question with Solution

$\begin{aligned} & X^{\top} A X=0 \\ & (x y z)\left[\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=0 \\ & (x y z)\left[\begin{array}{l} a_1 x+a_2 y+a_3 z \\ b_1 x+b_2 y+b_3 z \\ c_1 x+c_2 y+c_3 z \end{array}\right]=0 \\ & x\left(a_1 x+a_2 y+a_3 z\right)+y\left(b_1 x+b_2 y+b_3 z\right) \\ & +z\left(c_1 x+c_2 y+c_3 z\right) \\ & a_1=0, b_2=0, c_3=0 \\ & a_2+b_1=0, a_3+c_1=0, b_3+c_2=0 \\ & A=\text { skew symmetric matrix } \\ & A=\left[\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right] ; A\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]=\left[\begin{array}{c} 1 \\ 4 \\ -5 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]=\left[\begin{array}{c} 1 \\ 4 \\ -5 \end{array}\right] \\ & x+y=1 \\ & -x+z=4 \\ & y+z=5 \\ & {\left[\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right]=\left[\begin{array}{c} 1 \\ 4 \\ -8 \end{array}\right]} \\ & 2 x+y=0 \quad x=-1 \\ & -x+z=4 \quad y=2 \\ & -y-2 z=-8 \quad z=3 \end{aligned}$ $\begin{array}{rl}& A=\left[\begin{array}{ccc}0& -1& 2\\ 1& 0& 3\\ -2& -3& 0\end{array}\right]\\ & 2(A+l)=\left[\begin{array}{ccc}2& -2& 4\\ 2& 2& 6\\ -2& -6& 2\end{array}\right]\\ & 2(A+l)=120\\ & \Rightarrow \det (\mathrm{adj}(2A+l))\\ & =120^2=2^6\cdot 3^2\cdot 5^2\\ & \therefore \alpha =6,\beta =2,\gamma =2\\ & \text{ Hence }{\alpha }^2+{\beta }^2+{\gamma }^2=6^2+2^2+2^2=44\end{array}$