Let A be a 33 matrix such that X^TAX=O for all nonzero 31 matrices X=pmatrixxpmatrix.

Question

Let $A$ be a $3	imes3$ matrix such that $X^TAX=O$ for all nonzero $3	imes1$ matrices $X=\begin{pmatrix}x\y\z\end{pmatrix}$. Given $A\begin{pmatrix}1\1\1\end{pmatrix}=\begin{pmatrix}1\4\-5\end{pmatrix}$ and $A\begin{pmatrix}1\2\1\end{pmatrix}=\begin{pmatrix}0\4\-8\end{pmatrix}$, and $\det(\operatorname{adj}(2(A+I)))=2^{\alpha}3^{\beta}5^{\gamma}$ for $\alpha,\beta,\gamma\in\mathbb{N}$, find $\alpha^2+\beta^2+\gamma^2$.

Answer: 44

Accepted Answer

$X^TAX=0$ for all $X$ forces $A$ to be skew-symmetric: $A=\begin{pmatrix}0&x&y\-x&0&z\-y&-z&0\end{pmatrix}$.

Using $A\begin{pmatrix}1\1\1\end{pmatrix}=\begin{pmatrix}1\4\-5\end{pmatrix}$: $x+y=1,\ -x+z=4,\ -y-z=-5$.

Using $A\begin{pmatrix}1\2\1\end{pmatrix}=\begin{pmatrix}0\4\-8\end{pmatrix}$: $2x+y=0,\ -x+z=4,\ -y-2z=-8$.

Solving: from $x+y=1$ and $2x+y=0$: $x=-1,\ y=2$. Then $-x+z=4\Rightarrow z=3$.

So $A=\begin{pmatrix}0&-1&2\1&0&3\-2&-3&0\end{pmatrix}$, and $A+I=\begin{pmatrix}1&-1&2\1&1&3\-2&-3&1\end{pmatrix}$.

Direct computation gives $\det(2(A+I))=120=2^3\cdot3\cdot5$, so $\det(\operatorname{adj}(2(A+I)))=120^2=2^6\cdot3^2\cdot5^2$.

So $\alpha=6,\ \beta=2,\ \gamma=2$, and $\alpha^2+\beta^2+\gamma^2=36+4+4=44$.

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