Let be matrix such that , then equals: [2025]

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Solution

Q.
Let $A = [a_{i j}]$ be $3 \times 3$ matrix such that $A [\begin{matrix} \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \end{matrix}], A [\begin{matrix} \begin{matrix} 4 \\ 1 \\ 3 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \end{matrix}] and A [\begin{matrix} \begin{matrix} 2 \\ 1 \\ 2 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \end{matrix}]$ , then $a_{23}$ equals: [2025]

1 –1

2 0

3 2

4 1

Ans.
(1)

Let $A = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}]$

$A [\begin{matrix} \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \end{matrix}] \Rightarrow [\begin{matrix} \begin{matrix} a_{12} \\ a_{22} \\ a_{33} \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \end{matrix}] \Rightarrow a_{12} = 0; a_{22} = 0; a_{32} = 1$

$A [\begin{matrix} \begin{matrix} 4 \\ 1 \\ 3 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] \end{matrix} \Rightarrow \begin{matrix} 4 a_{11} + a_{12} + 3 a_{13} = 0 & . . . (i) \\ 4 a_{21} + a_{22} + 3 a_{23} \Rightarrow 4 a_{21} + 3 a_{23} = 1 & . . . (ii) \\ 4 a_{31} + a_{32} + 3 a_{33} = 0 & . . . (iii) \end{matrix}$

$A [\begin{matrix} \begin{matrix} 2 \\ 1 \\ 2 \end{matrix} \end{matrix}] = [\begin{matrix} \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}] \end{matrix} \Rightarrow \begin{matrix} 2 a_{11} + a_{12} + 2 a_{13} = 1 & . . . (iv) \\ 2 a_{21} + a_{22} + 2 a_{23} = 0 \Rightarrow a_{21} + a_{23} = 0 & . . . (v) \\ 2 a_{31} + a_{32} + 2 a_{33} = 0 & . . . (vi) \end{matrix}$

On solvig equations (ii) and (v), we get

$a_{21} = 1 and a_{23} = - 1$ .

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