Let A be a 33 matrix such that [A^Tpmatrix101pmatrix=pmatrix522pmatrix, A^Tpmatrix001pmatr…

Question

Let $A$ be a $3\times3$ matrix such that
$$A^T\begin{pmatrix}1\0\1\end{pmatrix}=\begin{pmatrix}5\2\2\end{pmatrix},\quad A^T\begin{pmatrix}0\0\1\end{pmatrix}=\begin{pmatrix}3\1\1\end{pmatrix},\quad A\begin{pmatrix}1\0\1\end{pmatrix}=\begin{pmatrix}3\4\4\end{pmatrix},\quad A\begin{pmatrix}0\0\1\end{pmatrix}=\begin{pmatrix}1\3\1\end{pmatrix}.$$
If $\det(A)=1$, then $\det(\operatorname{adj}(A^2+A))$ is equal to:

Accepted Answer

Let $u=\begin{pmatrix}1\0\1\end{pmatrix},\ v=\begin{pmatrix}0\0\1\end{pmatrix}$. Then $Au-Av=\begin{pmatrix}3\4\4\end{pmatrix}-\begin{pmatrix}1\3\1\end{pmatrix}=\begin{pmatrix}2\1\3\end{pmatrix}$, and since $u-v=\begin{pmatrix}1\0\0\end{pmatrix}$, the first column of $A$ is $\begin{pmatrix}2\1\3\end{pmatrix}$.

The third column of $A$ (from $Av$) is $\begin{pmatrix}1\3\1\end{pmatrix}$.

Similarly, from the transpose conditions, $A^Tu-A^Tv=\begin{pmatrix}2\1\1\end{pmatrix}$ gives the first row of $A^T$ (i.e. first column of $A$, consistent) and $A^Tv=\begin{pmatrix}3\1\1\end{pmatrix}$ gives the third row of $A$, so the third row of $A$ is $[3,1,1]$, fixing $c=1$ below.

So $A=\begin{pmatrix}2&a&1\1&b&3\3&1&1\end{pmatrix}$. Using $\det(A)=1$: expanding gives $8a-b-5c=1$ with $c=1$, i.e. $8a-b=6$.

Using $A^Tu=\begin{pmatrix}5\2\2\end{pmatrix}$ (first row of $A$ plus third row of $A$ equal this vector componentwise) gives $a=1$, and then $8(1)-b=6\Rightarrow b=2$.

So $A=\begin{pmatrix}2&1&1\1&2&3\3&1&1\end{pmatrix}$.

$A^2=\begin{pmatrix}8&5&6\13&8&10\10&6&7\end{pmatrix}$, so $A^2+A=\begin{pmatrix}10&6&7\14&10&13\13&7&8\end{pmatrix}$.

For a $3	imes3$ matrix $M$, $\det(\operatorname{adj}M)=\det(M)^2$. Also, since $A^2+A=A(A+I)$, $\det(A^2+A)=\det(A)\det(A+I)$.

$A+I=\begin{pmatrix}3&1&1\1&3&3\3&1&2\end{pmatrix}$, and expanding gives $\det(A+I)=8$.

So $\det(A^2+A)=1\cdot8=8$, and $\det(\operatorname{adj}(A^2+A))=8^2=64$.

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