Let $A$ be a $3\times 3$ matrix such that $X^{T}AX=O$ for all nonzero $3\times 1$ matrices $X=\begin{pmatrix}x\\y\\z\end{pmatrix}$. If $A\!\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}1\\4\\-5\end{pmatrix}$ and $A\!\begin{pmatrix}1\\2\\1\end{pmatrix}=\begin{pmatrix}0\\4\\-8\end{pmatrix}$, then $\det\bigl(\operatorname{adj}(2(A+I))\bigr)=2^{\alpha}\cdot 3^{\beta}\cdot 5^{\gamma}$ and $\alpha^{2}+\beta^{2}+\gamma^{2}$ is \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: $X^{T}AX=0$ for all $X$ forces the symmetric part of $A$ to be zero, i.e.\ $A$ is skew-symmetric — diagonal zero and $a_{ij}=-a_{ji}.$ Only $3$ entries to determine.
$X^{T}AX=0\,\forall X\Rightarrow A^{T}=-A.$ Write $A=\begin{pmatrix}0&x&y\\-x&0&z\\-y&-z&0\end{pmatrix}.$
From $A(1,1,1)^{T}=(1,4,-5)^{T}$: $x+y=1,\ -x+z=4,\ -y-z=-5.$
From $A(1,2,1)^{T}=(0,4,-8)^{T}$: $2x+y=0,\ -x+z=4,\ -y-2z=-8.$
Solve: from $x+y=1,\ 2x+y=0\Rightarrow x=-1,\,y=2;\ z=3.$
$A+I=\begin{pmatrix}1&-1&2\\1&1&3\\-2&-3&1\end{pmatrix},\ |A+I|=1(1+9)-(-1)(1+6)+2(-3+2)=10+7-2=15.$
$|2(A+I)|=2^{3}\cdot 15=120.$ Then $\det\bigl(\operatorname{adj}(2(A+I))\bigr)=120^{2}=14400=2^{6}\cdot 3^{2}\cdot 5^{2}.$
$\alpha=6,\beta=2,\gamma=2\Rightarrow \alpha^{2}+\beta^{2}+\gamma^{2}=36+4+4=44.$
Correct Answer: 44