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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 :

  1. A \(\mathrm{a}=-\frac{1}{3}, \mathrm{~b} \neq \frac{7}{3}\)
  2. B \(a \neq \frac{1}{3}, b=\frac{7}{3}\)
  3. C \(\mathrm{a} \neq-\frac{1}{3}, \mathrm{~b}=\frac{7}{3}\)
  4. D \(\mathrm{a}=\frac{1}{3}, \mathrm{~b} \neq \frac{7}{3}\)
Verified Solution

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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