Mathematics - Matrices Question with Solution | TestHub

$\begin{array}{rl}& X^{\top }AX=0\\ & (xyz)\left[\begin{array}{lll}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=0\\ & (xyz)\left[\begin{array}{l}{a}_{1}x+a_{2}y+a_{3}z\\ b_{1}x+b_{2}y+b_{3}z\\ c_{1}x+c_{2}y+c_{3}z\end{array}\right]=0\\ & x\left(a_{1}x+a_{2}y+a_{3}z\right)+y\left(b_{1}x+b_{2}y+b_{3}z\right)\\ & +z\left(c_{1}x+c_{2}y+c_{3}z\right)\\ & a_1=0,b_2=0,c_3=0\\ & a_2+b_1=0,a_3+c_1=0,b_3+c_2=0\\ & A=\text{ skew symmetric matrix }\\ & A=\left[\begin{array}{ccc}0& x& y\\ -x& 0& z\\ -y& -z& 0\end{array}\right];A\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 4\\ -5\end{array}\right]\\ & \Rightarrow \left[\begin{array}{ccc}0& x& y\\ -x& 0& z\\ -y& -z& 0\end{array}\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 4\\ -5\end{array}\right]\\ & x+y=1\\ & -x+z=4\\ & y+z=5\\ & \left[\begin{array}{ccc}0& x& y\\ -x& 0& z\\ -y& -z& 0\end{array}\right]\left[\begin{array}{l}1\\ 2\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 4\\ -8\end{array}\right]\\ & 2x+y=0x=-1\\ & -x+z=4y=2\\ & -y-2z=-8z=3\end{array}$ $\begin{array}{rl}& A=\left[\begin{array}{ccc}0& -1& 2\\ 1& 0& 3\\ -2& -3& 0\end{array}\right]\\ & 2(A+l)=\left[\begin{array}{ccc}2& -2& 4\\ 2& 2& 6\\ -2& -6& 2\end{array}\right]\\ & 2(A+l)=120\\ & \Rightarrow \det (\mathrm{adj}(2A+l))\\ & =120^2=2^6\cdot 3^2\cdot 5^2\\ & \therefore \alpha =6,\beta =2,\gamma =2\\ & \text{ Hence }{\alpha }^2+{\beta }^2+{\gamma }^2=6^2+2^2+2^2=44\end{array}$