JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Suppose the vectors \(x_{1}, x_{2}\) and \(x_{3}\) are the 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 \(x =\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right], x _{2}=\left[\begin{array}{l}0 \\ 2 \\ 1\end{array}\right], x _{3}=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], b _{1}=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\) \(b _{2}=\left[\begin{array}{l}0 \\ 2 \\ 0\end{array}\right]\) and \(b _{3}=\left[\begin{array}{l}0 \\ 0 \\ 2\end{array}\right],\) then the determinant of \(A\) is equal to
- A \(\frac{1}{2}\)
- B \(4\)
- C \(\frac{3}{2}\)
- D \(2 \)
Answer & Solution
Correct Answer
(D) \(2 \)
Step-by-step Solution
Detailed explanation
\(A x_{1}=b_{1}\) \(A x_{2}=b_{2}\) \(A x_{3}=b_{3}\) \(\Rightarrow\left|\begin{array}{lll}1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 1\end{array}\right|=\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right|\) \(\Rightarrow|A|=\frac{4}{2}=2\)
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