Question

Consider the system of linear equations
\(-x + y + 2z = 0\)
\(3x - ay + 5z = 1\)
\(2x - 2y - az = 7\)
Let \(S_1\) be the set of all \(a \in \mathbb{R}\) for which the system is inconsistent and \(S_2\) be the set of all \(a \in \mathbb{R}\) for which the system has infinitely many solutions. If \(n(S_1)\) and \(n(S_2)\) denote the number of elements in \(S_1\) and \(S_2\) respectively, then

• A. \(n(S_1) = 2,\ n(S_2) = 2\)
• B. \(n(S_1) = 1,\ n(S_2) = 0\)
• C. \(n(S_1) = 2,\ n(S_2) = 0\)
• D. \(n(S_1) = 0,\ n(S_2) = 2\)

Hint:

Find the values of \(a\) that make the coefficient determinant zero, then test the system for those values.

Answer 1

The Correct Option is C

Coefficient determinant \(\Delta=\begin{vmatrix}-1&1&2\\3&-a&5\\2&-2&-a\end{vmatrix} = -1(a^2+10) - 1(-3a-10) + 2(-6+2a) = -a^2+7a-12 = -(a-3)(a-4)\).
For \(a\ne3,4\), \(\Delta\ne0\): unique solution.
\(a=3\): from the first equation \(x-y=2z\); the second becomes \(3(x-y)+5z = 11z = 1\), so \(z=\frac{1}{11}\); the third becomes \(2(x-y)-3z = z = 7\). Contradiction, so inconsistent.
\(a=4\): the third equation becomes \(2(x-y)-4z = 4z-4z = 0 = 7\), impossible, so inconsistent.
Hence \(S_1=\{3,4\}\), \(n(S_1)=2\), and no value of \(a\) gives infinitely many solutions: \(S_2=\varnothing\), \(n(S_2)=0\).
Answer: (3) \(n(S_1)=2,\ n(S_2)=0\)