${Suppose the vectors X}_{1}, X_{2} {and X}_{3} are the solutions of the system of linear equations, A X = b$
${when the vector B on the right side is equal to B}_{1}, B_{2} {and B}_{3} respectively. If$ $X_{1} = [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}], X_{2} = [\begin{matrix} 0 \\ 2 \\ 1 \end{matrix}], X_{3} = [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}], B_{1} = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}]$
$B_{2} = [\begin{array}{l} 0 \\ 2 \\ 0 \end{array}] {and B}_{3} = [\begin{array}{l} 0 \\ 0 \\ 2 \end{array}], then the determinant of A is equal to$

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$\frac{3}{2}$

$\frac{1}{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} \begin{array}{l} A X = B \\ A X_{1} = B_{1} {Ax}_{2} = b_{2} {Ax}_{3} = b_{3} \end{array} \\ Let A = [\begin{matrix} ℓ_{1} & ℓ_{2} & ℓ_{3} \\ m_{1} & m_{2} & m_{3} \\ n_{1} & n_{2} & n_{3} \end{matrix}] \\ A X_{3} = B_{3} \Rightarrow [\begin{matrix} ℓ_{1} & ℓ_{2} & ℓ_{3} \\ m_{1} & m_{2} & m_{3} \\ n_{1} & n_{2} & n_{3} \end{matrix}] [\begin{array}{l} 0 \\ 0 \\ 1 \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 2 \end{array}] [\begin{array}{l} ℓ_{3} \\ m_{3} \\ n_{3} \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 2 \end{array}] \\ ℓ_{3} = 0, m_{3} = 0, n_{3} = 2 \\ A X_{2} = B_{2} \Rightarrow [\begin{matrix} ℓ_{1} & ℓ_{2} & 0 \\ m_{1} & m_{2} & 0 \\ n_{1} & n_{2} & 2 \end{matrix}] [\begin{array}{l} 0 \\ 2 \\ 1 \end{array}] = [\begin{array}{l} 0 \\ 2 \\ 0 \end{array}] \\ [\begin{matrix} 2 ℓ_{2} \\ 2 m_{2} \\ 2 n_{2} + 2 \end{matrix}] = [\begin{array}{l} 0 \\ 2 \\ 0 \end{array}] \\ now A X_{3} = B_{3} \\ [\begin{array}{l} ℓ_{1} & 0 & 0 \\ m_{1} & 1 & 0 \\ n_{1} & - 1 & 2 \end{array}] [\begin{array}{l} 1 \\ 1 \\ 1 \end{array}] = [\begin{array}{l} 1 \\ 0 \\ 0 \end{array}] ℓ_{1} = 1 m_{1} = - 1 n_{1} = - 1 \end{array}$