Question

Let $\lambda, \mu \in \mathbb{R}$. If the system of equations
$3x + 5y + \lambda z = 3$
$7x + 11y - 9z = 2$
$97x + 155y - 189z = \mu$
has infinitely many solutions, then $\mu + 2\lambda$ is equal to:

• A. 25
• B. 24
• C. 27
• D. 22

Answer 1

Answer: (A)

25

Answer 2

Step 1: Condition for infinitely many solutions For the system of equations to have infinitely many solutions, the three equations must be linearly dependent.

Step 2: Manipulate the equations The given equations are:

$3x + 5y + \lambda z = 3, \cdots \cdots(1)$ $7x + 11y - 9z = 2, \cdots \cdots(2)$ $97x + 155y - 189z = \mu. \cdots \cdots(3)$

Multiply equation (1) by 31:

$93x + 155y + 31\lambda z = 93. \cdots \cdots(4)$

Subtract equation (4) from equation (3):

$(97x + 155y - 189z) - (93x + 155y + 31\lambda z) = \mu - 93.$ $4x - (31\lambda + 189)z = \mu - 93. \cdots \cdots(5)$

Step 3: Express further conditions Now consider equations (2) and (5). Multiply equation (2) by 9:

$63x + 99y - 81z = 18. \cdots \cdots({6})$

Multiply equation (5) by 9 and subtract from equation (6):

$(63x + 99y - 81z) - 9(4x - (31\lambda + 189)z) = 18 - 9(\mu - 93).$ $63x + 99y - 81z - 36x + 9(31\lambda + 189)z = 18 - 9\mu + 837.$ $36x + 1368z = 2(310 - 11\mu). \cdots \cdots({7})$

Step 4: Solve for λ and μ Expand equation (7):

$279\lambda + 3069z = 1457 - 31\mu. \cdots \cdots({8})$

For infinitely many solutions:

$279\lambda + 3069 = 0 \quad \Rightarrow \quad \lambda = -\frac{3069}{279} = -\frac{341}{31}.$

Substitute $\lambda = -\frac{341}{31}$ into the original equations to find $\mu$:

$\mu = \frac{1457}{31}.$

Step 5: Calculate $\mu + 2\lambda$

$\mu + 2\lambda = \frac{1457}{31} + 2\left(-\frac{341}{31}\right).$ $\mu + 2\lambda = \frac{1457 - 682}{31} = \frac{775}{31} = 25.$

Final Answer: Option (1).