If the system of equations x + 2y + 3z = 3 , 4x + 3y - 4z = 4 and...

Given,

The system of equations

$x + 2y + 3z = 3 . . . . . . \left(1\right)</p><p>$4x + 3y - 4z = 4 . . . . . . . . . \left(2\right)$</p><p>$8x + 4y - \lambdaz = 9 + \mu . . . . . \left(3\right)

Now using$the$operation, equation \left(1\right) \times 4 - \left(2\right) we get,

$5y + 16z = 8 . . . . . . . . \left(4\right)$

Now using the operation, equation \left(2\right)$\times 2 - \left(3\right)$ we get,

$2y + \left(\lambda - 8\right)$z = - 1 - \mu . . . . . . . . \left(5\right)$</p><p>Now using the operation, equation \left(4\right)$ \times 2 - \left(3\right)$\times 5$ we get,

\left(72 - 5\lambda\right)z = 21 + 5\mu$</p><p>Now given system of equation has infinitely many solutions, so making coefficient of$z$to be zero and equating$21 + 5\mu$with zero we get,</p><p>$72 - 5\lambda = 0 ⇒ \lambda = \dfrac{72}{5} a$nd$21 + 5\mu = 0 ⇒ \mu =\dfrac{- 21}{5}$</p><p>Hence, the ordered pair \left(\lambda, \mu\right)$ is equal to \left(\dfrac{72}{5}, \dfrac{- 21}{5}\right)$.