The system of equations \(x + y - z = 1, 3x + y - 2z = 3, x - y + \lambda z = 1\) has infinite number of solutions if \(\lambda\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
If the determinant of the coefficients is not zero, the system has a unique solution. So infinitely many solutions need
\(\displaystyle |A| = \begin{vmatrix}1&1&-1\\3&1&-2\\1&-1&\lambda\end{vmatrix} = 0\)
Expanding along the first row:
\(|A| = 1(\lambda - 2) - 1(3\lambda + 2) - 1(-3 - 1)\)
\(= \lambda - 2 - 3\lambda - 2 + 4 = -2\lambda\)
So \(|A| = 0\) only when \(\lambda = 0\). The system must then also be consistent. With \(\lambda = 0\) the third equation is \(x - y = 1\), so
\(y = x - 1\)
From the first equation:
\(z = x + y - 1 = 2x - 2\)
Putting these in the second equation:
\(3x + (x - 1) - 2(2x - 2) = 3\)
\(3 = 3\)
This holds for every x, so x = t, y = t − 1, z = 2t − 2 is a solution for every real t: infinitely many solutions.