Question

Investigate the values λ and μ for the system x + 2y + 3z = 6, x + 3y + 5z = 9, 2x + 5y + λz = μ
i) no soln.
ii) unique soln
iii) infinite no. of solns

Answer 1

Answer:

$\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: For\:no\:solution : \:\lambda=8,\:\mu\:\ne\:15\qquad \: \\ \\& \qquad \:\sf \: For\:unique\:solution : \:\lambda \: \ne \: 8,\:\mu\:\in\:R\\ \\& \qquad \:\sf \: For\:infinitely \:many\:solution : \:\lambda=8,\:\mu = 15\end{aligned}} \qquad \:$

Step-by-step explanation:

Given system of linear equations is

$\sf \: x + 2y + 3z = 6 \: \: \: \\ \\ \sf \: x + 3y + 5z = 9 \: \: \: \\ \\ \sf \: 2x + 5y + \lambda z = mu$

The above system of linear equations can be represented in matrix form as

$\sf \: \left[\begin{array}{ccc}1&2&3\\1&3&5\\2&5& \lambda\end{array}\right] \: \left[\begin{array}{c}x\\y\\z\end{array}\right] \: = \: \left[\begin{array}{c}6\\9\\ \mu\end{array}\right]$

where,

$\sf \: A = \left[\begin{array}{ccc}1&2&3\\1&3&5\\2&5& \lambda\end{array}\right]$

$\sf \: B = \left[\begin{array}{c}6\\9\\\mu\end{array}\right]$

$\sf \: X = \left[\begin{array}{c}x\\y\\z\end{array}\right]$

Now, Consider

$\sf \: \rho[A:B ]$

$\qquad\sf \: = \: \left[\begin{array}{ccc}1&2&3\: \: :&6\\1&3&5\: \: :&9\\2&5& \lambda\: \: :&\mu\end{array}\right]$

$\qquad\qquad\sf \: OP\:R_2\:\to\:R_2 - R_1$

$\qquad\sf \: = \: \left[\begin{array}{ccc}1&2&3\: \: :&6\\0&1&2\: \: :&3\\2&5& \lambda\: \: :&\mu\end{array}\right]$

$\qquad\qquad\sf \: OP\:R_3\:\to\:R_3 - 2R_1$

$\qquad\sf \: = \: \left[\begin{array}{ccc}1&2&3\: \: :&6\\0&1&2\: \: :&3\\0&1& \lambda - 6\: \: :&\mu - 12\end{array}\right]$

$\qquad\qquad\sf \: OP\:R_3\:\to\:R_3 - R_2$

$\qquad\sf \: = \: \left[\begin{array}{ccc}1&2&3\: \: :&6\\0&1&2\: \: :&3\\0&0& \lambda - 8\: \: :&\mu - 15\end{array}\right]$

(i) Now, for system has no solution

$\sf \: \rho(A) \: \ne \: \rho[ A:B]$

$\sf\implies \: \lambda - 8 = 0 \: \: and \: \: \mu - 15 \: \ne \: 0$

$\sf\implies \: \lambda = 8 \: \: and \: \: \mu \: \ne \: 15$

(ii) Now, for system has unique solution

$\sf \: \rho(A) = \rho[ A:B] = number \: of \: variables$

$\sf \: \rho(A) = \rho[ A:B] = 3$

$\sf\implies \sf \: \lambda - 8 \: \ne \: 0 \: \: and \: \: \mu \: \in \: R$

$\sf\implies \sf \: \lambda \: \ne \: 8\: \: and \: \: \mu \: \in \: R$

(iii) Now, for system of equations has infinitely many solutions.

$\sf \: \rho(A) = \rho[ A:B] < number \: of \: variables$

So, using this result, we get

$\sf \: \rho(A) = \rho[ A:B] < 3$

So, it means

$\sf \: \lambda - 8 = 0 \: \: and \: \: \mu - 15 = 0$

$\sf\implies \sf \: \lambda = 8 \: \: and \: \: \mu = 15$

Hence,

$\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: For\:no\:solution : \:\lambda=8,\:\mu\:\ne\:15\qquad \: \\ \\& \qquad \:\sf \: For\:unique\:solution : \:\lambda \: \ne \: 8,\:\mu\:\in\:R\\ \\& \qquad \:\sf \: For\:infinitely \:many\:solution : \:\lambda=8,\:\mu = 15\end{aligned}} \qquad \:$