Question

Let $ a,\lambda ,\mu \in R $ consider the system of linear equation
 $ \begin{gathered}
  ax\, + \,2y\, = \lambda \\
  3x\, - 2y\, = \mu \\
\end{gathered} $
Which of the following statement(s) is (are) correct?
 THIS QUESTION HAS MULTIPLE CORRECT OPTIONS
(A) If $ a\, = - 3 $ , then the system has infinitely many solutions for all values of $ \lambda $ and $ \mu $
(B) If $ a\, \ne - 3 $ , then the system has a unique solution for all values of $ \lambda $ and $ \mu $
(c) If $ \lambda + \mu = 0 $ then the system has infinitely many solutions for a =-3
(D) If $ \lambda + \mu \ne 0 $ , the system has no solution for a = -3

Answer 1

Hint: The following steps provide a good method to use when solving linear equations.
Simplify each side of the equation by removing parentheses and combining like terms.
Use addition or subtraction to isolate the variable term on one side of the equation.
Use multiplication or division to solve for the variable.

Complete step-by-step answer:
Here $ a,\lambda ,\mu \in R $ which mean $ a,\lambda ,\mu $ all belongs to real number.
If $ \begin{gathered}
  ax\, + \,2y\, = \lambda \\
  3x\, - 2y\, = \mu \\
\end{gathered} $
We have 4 options if we consider option number 1 that is:-
It is $ a\, = - 3 $
So the system would be
 $ - 3x + 2y = \lambda $ ….1
And the second equation will be
 $ 3x - 2y = \mu $ …..2
Then equation 1 and 2 are equal
 $ \lambda = - \mu $ Or $ \lambda + \mu = 0 $
But not all the values of $ \lambda $ and $ \mu $ so this option is not correct.
(B) Now we will consider the option B that is

 $ \begin{gathered}
  \Delta = \left( {\begin{array}{*{20}{c}}
  a&2 \\
  3&{ - 2}
\end{array}} \right)\, \\
  \,\,\,\,\, = \, - 2a - 6 \\
  \,\,\,\, \ne 0 \\
    \\
\end{gathered} $
(Unique solution)
So option B is correct.
(C) Now we will consider option the C that is
Already checked in option A so it is correct
(D) Now we can consider Option D If $ \lambda + \mu \ne 0 $
That we system not solution for $ a = - 3 $
So $ \begin{gathered}
  \left( {a + 3} \right)x + 0 = \lambda + \mu \\
  ,\,\lambda + \mu \ne 0 \\
\end{gathered} $
That means if a=-3 then the system would be so it can have many solutions so the option D would also be correct.

Option B, C, D are the correct options.

Note: Any term of an equation may be shifted to the other side with a change in its sign without affecting the equality. This process is called transposition.
So, by transposing a term —
* We simply change its sign and carry it to the other side of the equation.
* ‘+‘ sign of the term changes to ‘—‘ sign to the other side and vice-versa.
* ‘×’ sign of the factor changes to ‘÷‘ sign to the other side and vice-versa.
* Now, simplify L.H.S. such that each side contains just one term.
* Finally, simplify the equation to get the value of the variable.
Fractions may be removed by multiplying each side of the equation by the common denominator.