JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
If the following system of linear equations \(2 x+y+z=5\) \(x-y+z=3\) \(x+y+a z=b\) has no solution, then :
- A \(\mathrm{a}=-\frac{1}{3}, \mathrm{~b} \neq \frac{7}{3}\)
- B \(a \neq \frac{1}{3}, b=\frac{7}{3}\)
- C \(\mathrm{a} \neq-\frac{1}{3}, \mathrm{~b}=\frac{7}{3}\)
- D \(\mathrm{a}=\frac{1}{3}, \mathrm{~b} \neq \frac{7}{3}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{a}=\frac{1}{3}, \mathrm{~b} \neq \frac{7}{3}\)
Step-by-step Solution
Detailed explanation
Here \(\mathrm{D}=\left|\begin{array}{ccc}2 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & \mathrm{a}\end{array}\right|\) \(=2(-\mathrm{a}-1)-1(\mathrm{a}-1)+1+1\) \(=1-3 \mathrm{a}\) \(\mathrm{D}_{3}=\left|\begin{array}{ccc}2 & 1 & 5 \\ 1 & -1 & 3 \\ 1 & 1 & \mathrm{~b}\end{array}\right|\)…
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