Question

If the system of linear equations \( 2x + y - z = 7 \), \( x - 3y + 2z = 1 \), \( x + 4y + \delta z = k \) (where \(\delta, k \in \mathbb{R}\)) has infinitely many solutions, then \(\delta + k\) is equal to:

• A. -3
• B. 3
• C. 6
• D. 9

Hint:

Before calculating bulky determinants, check if one equation is simply a sum or difference of the others. It can save you minutes of calculation!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept: 
For a system of linear equations \(AX = B\) to have infinitely many solutions, the determinant of the coefficient matrix (\(\Delta\)) must be zero, and all specific determinants (\(\Delta_x, \Delta_y, \Delta_z\)) must also be zero. 

Step 2: Key Formula or Approach: 
1. Set \(\Delta = 0\) to find \(\delta\).
2. Set \(\Delta_z = 0\) (or use row operations) to find \(k\). 

Step 3: Detailed Explanation: 
1. Find \(\delta\):
\[ \Delta = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta \end{vmatrix} = 0 \] \[ 2(-3\delta - 8) - 1(\delta - 2) - 1(4 - (-3)) = 0 \] \[ -6\delta - 16 - \delta + 2 - 7 = 0 \] \[ -7\delta - 21 = 0 \implies \delta = -3 \] 2. Find \(k\):
For infinitely many solutions, the third equation must be a linear combination of the first two. Observe the coefficients of \(x\) and \(y\):
Eq(1) \( - \) Eq(2): \[ (2-1)x + (1 - (-3))y + (-1 - 2)z = 7 - 1 \] \[ \implies x + 4y - 3z = 6 \] Comparing this with Eq(3) \(x + 4y + \delta z = k\):
We see \(\delta = -3\) (which matches) and \(k = 6\). 3. Calculate \(\delta + k\):
\[ \delta + k = -3 + 6 = 3 \] 

Step 4: Final Answer 
The value of \(\delta + k\) is \(3\).