$x + y + z = 5$
$x + 2y + 3z = 9$
$x + 3y + \lambda z = \mu$
has infinitely many solutions, then the value of $\lambda + \mu$ is:
System Equations Analysis for Infinite Solutions
We are given a system of three linear equations:
- $x + y + z = 5$
- $x + 2y + 3z = 9$
- $x + 3y + \lambda z = \mu$
For this system to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the rank of the coefficient matrix must be equal to the rank of the augmented matrix.
Coefficient Matrix Determinant
The coefficient matrix $A$ is:
$ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{pmatrix} $Calculate the determinant of $A$ ($\det(A)$):
$\det(A) = 1 \begin{vmatrix} 2 & 3 \\ 3 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 3 \\ 1 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix}$
$\det(A) = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2)$
$\det(A) = 2\lambda - 9 - \lambda + 3 + 1 = \lambda - 5$
For infinitely many solutions, $\det(A) = 0$. Therefore:
$\lambda - 5 = 0 \implies \lambda = 5$
Augmented Matrix Row Reduction
Now, substitute $\lambda = 5$ into the system and form the augmented matrix:
$ \begin{pmatrix} 1 & 1 & 1 & | & 5 \\ 1 & 2 & 3 & | & 9 \\ 1 & 3 & 5 & | & \mu \end{pmatrix} $Apply row operations to simplify:
$R_2 \rightarrow R_2 - R_1$
$R_3 \rightarrow R_3 - R_1$
$ \begin{pmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 1 & 2 & | & 4 \\ 0 & 2 & 4 & | & \mu - 5 \end{pmatrix} $$R_3 \rightarrow R_3 - 2R_2$
$ \begin{pmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 1 & 2 & | & 4 \\ 0 & 0 & 0 & | & (\mu - 5) - 2(4) \end{pmatrix} $ $ \begin{pmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 1 & 2 & | & 4 \\ 0 & 0 & 0 & | & \mu - 13 \end{pmatrix} $For the system to have infinitely many solutions, the last row must represent the identity $0 = 0$. This requires:
$\mu - 13 = 0 \implies \mu = 13$
Calculate $\lambda + \mu$
We found $\lambda = 5$ and $\mu = 13$. The required value is:
$\lambda + \mu = 5 + 13 = 18$
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