Let $A=[a_{ij}]$ be $3 imes3$ 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 :

Question

Let $A=[a_{ij}]$ be $3	imes3$ 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 :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Let the columns be $c_1,c_2,c_3$. The first condition gives $c_2=(0,0,1)^T$. Then $4c_1+c_2+3c_3=(0,1,0)^T$ and $2c_1+c_2+2c_3=(1,0,0)^T$. Subtracting twice the second from the first: $-c_2-c_3=(-2,1,0)^T$, so $c_3=(2,-1,0)^T-c_2=(2,-1,-1)^T$. Hence $a_{23}=-1$. Answer (1).

Explanation

Reviewed by Dr. Abubacker

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