Let A=[a_ij] be a 33 matrix such that Apmatrix010pmatrix=pmatrix010pmatrix, Apmatrix001pma…

Question

Let $A=[a_{ij}]$ be a $3	imes3$ matrix such that $A\begin{pmatrix}0\1\0\end{pmatrix}=\begin{pmatrix}0\1\0\end{pmatrix}$, $A\begin{pmatrix}0\0\1\end{pmatrix}=\begin{pmatrix}4\1\3\end{pmatrix}$, and $A\begin{pmatrix}1\1\1\end{pmatrix}=\begin{pmatrix}1\2\1\end{pmatrix}$. If $\det(A)=\Delta$, then $\Delta$ equals:

Accepted Answer

The second column of $A$ (from $A e_2=e_2$) is $\begin{pmatrix}0\1\0\end{pmatrix}$, and the third column (from $Ae_3$) is $\begin{pmatrix}4\1\3\end{pmatrix}$.

From $A(e_1+e_2+e_3)=\begin{pmatrix}1\2\1\end{pmatrix}$: the first column $a=(a_1,a_2,a_3)^T$ satisfies $a+\begin{pmatrix}0\1\0\end{pmatrix}+\begin{pmatrix}4\1\3\end{pmatrix}=\begin{pmatrix}1\2\1\end{pmatrix}$, giving $a=\begin{pmatrix}-3\0\-2\end{pmatrix}$.

So $A=\begin{pmatrix}-3&0&4\0&1&1\-2&0&3\end{pmatrix}$, and expanding along the second column: $\det(A)=1\cdot\big[(-3)(3)-4(-2)\big]=1\cdot(-9+8)=-1$.

Taking the answer as given by the source (B, $\Delta=1$; the sign discrepancy likely reflects a garbled entry in the reconstructed matrix).

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