The values of λ and μ such that the system of equations x + y + z =...

For the system of equations ${a}_{1}x + {b}_{1}y + {c}_{1}z = {d}_{1}, {a}_{2}x + {b}_{2}y + {c}_{2}z = {d}_{2}$ and ${a}_{3}x + {b}_{3}y + {c}_{3}z = {d}_{3},$

We have D = \left|\begin{cases}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{cases}\right|, {D}_{x} = \left|\begin{cases}{d}_{1} & {b}_{1} & {c}_{1} \\ {d}_{2} & {b}_{2} & {c}_{2} \\ {d}_{3} & {b}_{3} & {c}_{3}\end{cases}\right|, {D}_{y} = \left|\begin{cases}{a}_{1} & {d}_{1} & {c}_{1} \\ {a}_{2} & {d}_{2} & {c}_{2} \\ {a}_{3} & {d}_{3} & {c}_{3}\end{cases}\right| and {D}_{z} = \left|\begin{cases}{a}_{1} & {b}_{1} & {d}_{1} \\ {a}_{2} & {b}_{2} & {d}_{2} \\ {a}_{3} & {b}_{3} & {d}_{3}\end{cases}\right| .

And, the system has no solution, if $D = 0$ and atleast one of ${D}_{x}, {D}_{y} \text{ and } {D}_{z}$ is non-zero.

Thus, for the given system of equations,

$x + y + z = 6,$ $3x + 5y + 5z = 26$ and $x + 2y + \lambdaz = \mu$

We have D = \left|\begin{cases}1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \lambda\end{cases}\right| = 0

$⇒ 1\left(5\lambda - 10\right) - 1\left(3\lambda - 5\right) + 1\left(6 - 5\right) = 0$

$⇒ 5\lambda - 10 - 3\lambda + 5 + 1 = 0$

$⇒ 2\lambda = 4$

$⇒ \lambda = 2 .$

And, {D}_{z} = \left|\begin{cases}1 & 1 & 6 \\ 3 & 5 & 26 \\ 1 & 2 & \mu\end{cases}\right| \ne 0

$⇒ 1\left(5\mu - 52\right) - 1\left(3\mu - 26\right) + 6\left(6 - 5\right) \ne 0$

$⇒ 5\mu - 52 - 3\mu + 26 + 6 \ne 0$

$⇒ 2\mu \ne 20$

$⇒ \mu \ne 10 .$

$∴$ For no solution $\lambda = 2$ and $\mu \ne 10 .$