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:
\( 3x + 5y + \lambda z = 3 \)
\( 7x + 11y - 9z = 2 \)
\( 97x + 155y - 189z = \mu \)
has infinitely many solutions, then \( \mu + 2\lambda \) is equal to:
- 25
- 24
- 27
- 22
The Correct Option is A
Approach Solution - 1
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:
\( \boxed{25} \)
Approach Solution -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).
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