Question

If the following system of linear equations
\( 2x + y + z = 5 \)
\( x - y + z = 3 \)
\( x + y + az = b \)
has no solution, then :

• A. \( a = \frac{1}{3}, b \neq \frac{7}{3} \)
• B. \( a \neq \frac{1}{3}, b = \frac{7}{3} \)
• C. \( a = -\frac{1}{3}, b \neq \frac{7}{3} \)
• D. \( a \neq -\frac{1}{3}, b = \frac{7}{3} \)

Hint:

Cramer's Rule: \( \Delta = 0 \) is the threshold. If all \( \Delta_i = 0 \), there are usually infinite solutions; if any \( \Delta_i \neq 0 \), there are no solutions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a system of 3 linear equations to have no solution, the determinant of the coefficient matrix (\( \Delta \)) must be zero, but at least one of the determinants \( \Delta_x, \Delta_y, \) or \( \Delta_z \) must be non-zero.
Step 2: Detailed Explanation:
1. Calculate \( \Delta \):
\[ \Delta = \begin{vmatrix} 2 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & a \end{vmatrix} \]
\[ \Delta = 2(-a - 1) - 1(a - 1) + 1(1 - (-1)) = -2a - 2 - a + 1 + 2 = -3a + 1 \]
Set \( \Delta = 0 \implies -3a + 1 = 0 \implies a = \frac{1}{3} \).
2. Calculate \( \Delta_z \): (for no solution, \( \Delta_z \neq 0 \))
\[ \Delta_z = \begin{vmatrix} 2 & 1 & 5 \\ 1 & -1 & 3 \\ 1 & 1 & b \end{vmatrix} \]
\[ \Delta_z = 2(-b - 3) - 1(b - 3) + 5(1 - (-1)) = -2b - 6 - b + 3 + 10 = -3b + 7 \]
For no solution, \( \Delta_z \neq 0 \implies -3b + 7 \neq 0 \implies b \neq \frac{7}{3} \).
Step 3: Final Answer:
The condition is \( a = \frac{1}{3}, b \neq \frac{7}{3} \).