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

The Correct Option is A

We start with the given equations:

\( 3x + 5y + \lambda z = 3 \)

\( 7x + 11y - 9z = 2 \)

\( 97x + 155y - 189z = \mu \)

By Cramer's Rule:

\( \Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0 \)

So,

\( \begin{vmatrix} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{vmatrix} = 0 \)

Perform \( R_3 \to R_3 - 14R_2 \):

\( \begin{vmatrix} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ -1 & -1 & -63 \end{vmatrix} = 0 \)

Next, perform \( C_1 \to C_1 + C_2 \):

\( \begin{vmatrix} 4 & 5 & \lambda \\ 9 & 11 & -9 \\ 0 & -1 & -63 \end{vmatrix} = 0 \)

Expanding along the third row:

\( -1( -36 - 9\lambda ) - 63(44 - 45) = 0 \)

\( 36 + 9\lambda + 63 = 0 \)

\( 9\lambda = -99 \Rightarrow \lambda = -11 \)

Now, \(\Delta_3 = 0\):

\( \begin{vmatrix} 3 & 5 & 3 \\ 7 & 11 & 2 \\ 97 & 155 & \mu \end{vmatrix} = 0 \)

Perform \( C_2 \to C_2 - C_1 \):

\( \begin{vmatrix} 3 & 2 & 3 \\ 7 & 4 & 2 \\ 97 & 58 & \mu \end{vmatrix} = 0 \)

Next, \( C_1 \to C_1 - C_3 \):

\( \begin{vmatrix} 0 & 2 & 3 \\ 5 & 4 & 2 \\ 97 & 58 & \mu \end{vmatrix} = 0 \)

Expanding along the first column:

\( -5(2\mu - 174) + (97 - \mu)(4 - 12) = 0 \)

\( 10\mu - 870 + 776 - 8\mu = 0 \)

\( 2\mu = 94 \Rightarrow \mu = 47 \)

Finally, substitute the values:

\( \mu + 2\lambda = 47 - 22 = 25 \)

Hence, the final result is:

\( \boxed{25} \)

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).