Let A be a 3 3 matrix such that A+A^T=O (i.e.

Question

Let $A$ be a $3	imes 3$ matrix such that $A+A^T=O$ (i.e. $A$ is skew-symmetric). Suppose $A\begin{pmatrix}1\-1\0\end{pmatrix}=\begin{pmatrix}3\3\2\end{pmatrix}$, and using this and a second condition the entries of $A$ can be determined. If $\det\big(\operatorname{adj}(2\,\operatorname{adj}(A+I))\big)=(2)^{\alpha}\cdot(3)^{\beta}\cdot(11)^{\gamma}$ where $\alpha,\beta,\gamma$ are non-negative integers, then $\alpha+\beta+\gamma$ is equal to \_\_\_\_.

Answer: 18

Accepted Answer

*Note: the source text for this question's stem is corrupted; the reconstruction below follows the surviving worked solution, which is internally consistent and yields the stated answer.*

Write the general skew-symmetric $3\times3$ matrix as $A=\begin{pmatrix}0 & p & q\ -p & 0 & r\ -q & -r & 0\end{pmatrix}$. Using the given image vector, $p=-3$ and (from a second given image) $-q+r=2$, and further data gives $q=3$, hence $r=5$. So
$$A=\begin{pmatrix}0&-3&3\3&0&5\-3&-5&0\end{pmatrix}.$$

Then $A+I=\begin{pmatrix}1&-3&3\3&1&5\-3&-5&1\end{pmatrix}$, and direct expansion gives $\det(A+I)=44=2^2\cdot 11$.

For an $n\times n$ matrix, $\det(\operatorname{adj}(kM))=k^{n(n-1)}\det(\operatorname{adj}(M))$ and $\det(\operatorname{adj}(\operatorname{adj}(M)))=\det(M)^{(n-1)^2}$. With $n=3$, $k=2$:
$$\det\big(\operatorname{adj}(2\operatorname{adj}(A+I))\big)=2^{6}\cdot\det(A+I)^{4}=2^{6}\cdot(2^2\cdot 11)^4=2^{6}\cdot 2^{8}\cdot 11^{4}=2^{14}\cdot 11^{4}.$$

So $\alpha=14,\ \beta=0,\ \gamma=4$, giving $\alpha+\beta+\gamma=18$.

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