Let $A = [a_{ij}]$ be a $3 \times 3$ matrix such that $A\begin{bmatrix}0\\1\\0\end{bmatrix} = \begin{bmatrix}0\\0\\1\end{bmatrix}$, $A\begin{bmatrix}4\\1\\3\end{bmatrix} = \begin{bmatrix}0\\1\\0\end{bmatrix}$, and $A\begin{bmatrix}2\\1\\2\end{bmatrix} = \begin{bmatrix}1\\0\\0\end{bmatrix}$. Then $a_{23}$ equals:
Step-by-Step Solution
Key Concept: From the first product, directly read off the second column of $A$. From the other two products, set up linear equations to determine the remaining entries.
From first product: column 2 of $A$ gives $b=0, e=0, h=1$. From second product: $4a+3c=0, 4d+3f=1, 4g+1+3i=0$. From third product: $2a+2c=1, 2d+2f=0, 2g+1+2i=0$. Solving: $d=1, f=-1$, so $a_{23} = f = -1$.
Correct Answer: -1