$x + y + z = 6$,
$x + 2y + 5z = 10$,
$2x + 3y + \lambda z = \mu$
has infinitely many solutions, then the value of $\lambda + \mu$ equals :
Analyzing Linear Equations for Infinitely Many Solutions
We are given a system of three linear equations:
- $x + y + z = 6$
- $x + 2y + 5z = 10$
- $2x + 3y + \lambda z = \mu$
For this system to have infinitely many solutions, two conditions must be met:
- The determinant of the coefficient matrix must be zero.
- The system must be consistent (the ranks of the coefficient matrix and the augmented matrix must be equal).
Calculating the Determinant
The coefficient matrix is:
$ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda \end{bmatrix} $The determinant is calculated as:
$ \det(A) = 1 \begin{vmatrix} 2 & 5 \\ 3 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 5 \\ 2 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} $ $ \det(A) = 1(2\lambda - 15) - 1(\lambda - 10) + 1(3 - 4) $ $ \det(A) = 2\lambda - 15 - \lambda + 10 - 1 $ $ \det(A) = \lambda - 6 $For infinitely many solutions, $det(A) = 0$.
$ \lambda - 6 = 0 \implies \lambda = 6 $Checking Consistency with Gaussian Elimination
Substitute $\lambda = 6$ into the augmented matrix and perform row operations:
$ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 5 & | & 10 \\ 2 & 3 & 6 & | & \mu \end{bmatrix} $Apply $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - 2R_1$:
$ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 4 & | & 4 \\ 0 & 1 & 4 & | & \mu - 12 \end{bmatrix} $Apply $R_3 \rightarrow R_3 - R_2$:
$ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 4 & | & 4 \\ 0 & 0 & 0 & | & (\mu - 12) - 4 \end{bmatrix} $ $ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 4 & | & 4 \\ 0 & 0 & 0 & | & \mu - 16 \end{bmatrix} $For infinitely many solutions, the last row must be entirely zeros:
$ \mu - 16 = 0 \implies \mu = 16 $Final Calculation
We found $\lambda = 6$ and $\mu = 16$.
Therefore, the value of $\lambda + \mu$ is:
$ \lambda + \mu = 6 + 16 = 22 $Similar Questions
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