Let $A$ be a $3 imes 3$ matrix such that $X^T A X=O$ for all nonzero $3 imes 1$ matrices $X=\left[\begin{array}{l}x \ y \ z\end{array} ight]$. If $\mathbf{A}\left[\begin{array}{l}1 \ 1 \ 1\end{array} ight]=\left[\begin{array}{c}1 \ 4 \ -5\end{array} ight],…

Question

Let $A$ be a $3 	imes 3$ matrix such that $X^T A X=O$ for all nonzero $3 	imes 1$ matrices $X=\left[\begin{array}{l}x \ y \ z\end{array}ight]$. If $\mathbf{A}\left[\begin{array}{l}1 \ 1 \ 1\end{array}ight]=\left[\begin{array}{c}1 \ 4 \ -5\end{array}ight], \mathbf{A}\left[\begin{array}{l}1 \ 2 \ 1\end{array}ight]=\left[\begin{array}{c}0 \ 4 \ -8\end{array}ight]$, 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_____.

Aicharya · Accepted Answer

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

Let $A$ be a $3 \times 3$ matrix such that $X^T A X=O$ for all nonzero $3 \times 1$ matrices…

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