Question

If the system of linear equations \(x + y + z = 6\), \(x + 2y + 5z = 10\), \(2x + 3y + \lambda z = \mu\) has infinitely many solutions, then the value of \(\lambda + \mu\) equals:

• A. 12
• B. 16
• C. 22
• D. 28

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A system of three linear equations in three variables has infinitely many solutions if the determinant of the coefficient matrix is zero (\( \Delta = 0 \)), and all the \( \Delta_x, \Delta_y, \Delta_z \) determinants are also zero, meaning the planes intersect along a common line.
Step 2: Key Formula or Approach:
We can use row reduction to find the conditions for infinite solutions. By eliminating variables, we aim to obtain an identity of the form \( 0 = 0 \). Alternatively, we can check if the third equation is a linear combination of the first two.
Step 3: Detailed Explanation:
The given equations are:
\( E_1: x + y + z = 6 \)
\( E_2: x + 2y + 5z = 10 \)
\( E_3: 2x + 3y + \lambda z = \mu \)
Let's see if we can form \( E_3 \) by a linear combination of \( E_1 \) and \( E_2 \).
Notice the coefficients of \( x \) and \( y \) in \( E_3 \) are 2 and 3 respectively.
If we add \( E_1 \) and \( E_2 \):
\( (x + y + z) + (x + 2y + 5z) = 6 + 10 \)
\( 2x + 3y + 6z = 16 \)
For the system to have infinitely many solutions, this resulting equation must perfectly match the third equation \( E_3 \):
\( 2x + 3y + \lambda z = \mu \)
Comparing the coefficients of \( z \) and the constant term, we get:
\( \lambda = 6 \)
\( \mu = 16 \)
We are asked to find the value of \( \lambda + \mu \):
\( \lambda + \mu = 6 + 16 = 22 \).
Step 4: Final Answer:
The value of \( \lambda + \mu \) is 22.