Let $A=[a_{ij}]_{3\times 3}$ be a matrix such that $A\!\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\0\\1\end{pmatrix},\ A\!\begin{pmatrix}4\\1\\3\end{pmatrix}=\begin{pmatrix}0\\1\\0\end{pmatrix}$ and $A\!\begin{pmatrix}2\\1\\2\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}$. Then $a_{23}$ equals:
Step-by-Step Solution
Key Concept: $A\,e_{j}$ is the $j^{\text{th}}$ column of $A$. From $A\!\begin{pmatrix}0\\1\\0\end{pmatrix}$ read column $2$ directly; then write the other two products as linear systems for the remaining columns.
Column $2$ of $A$ is $\begin{pmatrix}0\\0\\1\end{pmatrix}$ from $A\,e_{2}$.
Let columns $1,3$ be $(a,d,g)^{T},(c,f,i)^{T}.$
From $A(4,1,3)^{T}=(0,1,0)^{T}$ (second component): $4d+0+3f=1.$
From $A(2,1,2)^{T}=(1,0,0)^{T}$ (second component): $2d+0+2f=0\Rightarrow d=-f.$
Substitute: $-4f+3f=1\Rightarrow f=-1\Rightarrow d=1.$ Therefore $a_{23}=f=-1.$
Correct Answer: 1