Let $A$ be a $3 imes3$ matrix such that $X^TAX=O$ for all nonzero $3 imes1$ matrices $X=\begin{bmatrix}x\y\z\end{bmatrix}$. If $A\begin{bmatrix}1\1\1\end{bmatrix}=\begin{bmatrix}1\4\-5\end{bmatrix}$, $A\begin{bmatrix}1\2\1\end{bmatrix}=\begin{bmatrix}0\4\-8\end{bmatrix}$, and $\det(\operatorname{adj}(2(A+I)))=2^\alpha3^\beta5^\gamma$, $\alpha,\beta,\gamma\in N$, then $\alpha^2+\beta^2+\gamma^2$ is ______.

Question

Let $A$ be a $3	imes3$ matrix such that $X^TAX=O$ for all nonzero $3	imes1$ matrices $X=\begin{bmatrix}x\y\z\end{bmatrix}$. If $A\begin{bmatrix}1\1\1\end{bmatrix}=\begin{bmatrix}1\4\-5\end{bmatrix}$, $A\begin{bmatrix}1\2\1\end{bmatrix}=\begin{bmatrix}0\4\-8\end{bmatrix}$, and $\det(\operatorname{adj}(2(A+I)))=2^\alpha3^\beta5^\gamma$, $\alpha,\beta,\gamma\in N$, then $\alpha^2+\beta^2+\gamma^2$ is ______.

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $X^TAX=0$ for all $X$ means $A$ is skew-symmetric: $A=\begin{bmatrix}0&a&b\-a&0&c\-b&-c&0\end{bmatrix}$. From the two given products: $a+b=1$, $2a+b=0$, $c-a=4$, giving $a=-1,\ b=2,\ c=3$. $\det(A+I)=\begin{vmatrix}1&-1&2\1&1&3\-2&-3&1\end{vmatrix}=15$. For a $3	imes3$ matrix $\det(\operatorname{adj}M)=(\det M)^2$, so $\det(\operatorname{adj}(2(A+I)))=(8\cdot15)^2=120^2=2^6\cdot3^2\cdot5^2$. $\alpha^2+\beta^2+\gamma^2=36+4+4=44$. Answer: 44.

Explanation

Reviewed by Dr. Abubacker

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