If the system of linear equations $x + y + z = 5$,$x + 2y + 2z = 6$,and $x + 3y + \lambda z = \mu$ (where $\lambda, \mu \in \mathbb{R}$) has infinitely many solutions,then the value of $\lambda + \mu$ is:

  • A
    $12$
  • B
    $7$
  • C
    $10$
  • D
    $9$

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Similar Questions

Consider the following system of equations: $\alpha x + 2y + z = 1$; $2\alpha x + 3y + z = 1$; $3x + \alpha y + 2z = \beta$. For some $\alpha, \beta \in \mathbb{R}$. Which of the following is $NOT$ correct?

If the system of equations $(\lambda-1) x+(\lambda-4) y+\lambda z=5$,$\lambda x+(\lambda-1) y+(\lambda-4) z=7$,and $(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$ has infinitely many solutions,then $\lambda^2+\lambda$ is equal to

Let $\alpha, \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations:
$x+2y+z=7$
$x+\alpha z=11$
$2x-3y+\beta z=\gamma$
Match each entry in List-$I$ to the correct entries in List-$II$:
List-$I$ List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has $(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has $(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has $(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has $(4)$ $x=11, y=-2$ and $z=0$ as a solution
$(5)$ $x=-15, y=4$ and $z=0$ as a solution

The number of non-trivial solutions of the system $x-y+z=0, x+2y-z=0, 2x+y+3z=0$ is

The value of $\lambda$ such that the system of equations $2x-y-2z=2$, $x-2y+z=-4$, and $x+y+\lambda z=4$ has no solution, is:

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