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Supportive the vectors x_{1}, x_{2} and x_{3} solutions of the system of linear equations, Ax = b when the vector b on the right side is equal to b_{1},b_{2} and b_{3} respectively. If \\\mathrm{x}_{1}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right], \mathrm{x}_{2}=\left[\begin{array}{l} 0 \\ 2 \\ 1 \end{array}\right], \mathrm{x}_{3}=\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right], \mathrm{b}_{1}\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right], \mathrm{b}_{2}=\left[\begin{array}{l} 0 \\ 2 \\ 0 \end{array}\right] \&\; \mathrm{b}_{3}=\left[\begin{array}{l} 0 \\ 0 \\ 2 \end{array}\right] then the determinant A is equal to:
Option: 1 4
Option: 2 2
Option: 3 \frac{1}{2}
Option: 4 \frac{3}{2}

Answers (1)

best_answer

\\\mathrm{Ax}_{1}=\mathrm{b}_{1} \\ \mathrm{Ax}_{2}=\mathrm{b}_{2} \\ \mathrm{Ax}_{3}=\mathrm{b}_{3} \\

\Rightarrow |A|\left|\begin{array}{lll} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 1 \end{array}\right|=\left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right|

\Rightarrow|A|=\frac{4}{2}=2

Posted by

himanshu.meshram

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