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explain solution rd sharma class 12 chapter determinants exercise 5.5 question 4 maths

Answers (1)

Answer: \lambda = 2, x = k, y =k/2, z = -k,  where k \in R

Hint: The system has non-trivial solution implies that the determinant is zero.

Given: 2\lambda x - 2y + 3z =0

             x x + \lambda y + 2z = 0

            2 x + \lambda z = 0 \lambda

Solution:

Firstly, form 3x3 matrix from the given equations:

                Let  A = [2 \lambda, 2,   3   ]

                                x   \lambda   2  

                             [   2  0     \lambda   ]

We know that, |A| = 0

Expanding through first row,

                                2\lambda (\lambda ^2 - 0) + 2 (\lambda - 4) + 3 (0 - 2 \lambda ) = 0

                                2\lambda ^3 + 2\lambda - 8 - 6\lambda = 0

                                2\lambda ^3 - 4\lambda - 8 = 0

                                \lambda ^3 - 2\lambda - 4= 0

Put \lambda = 2:

          = (2)^{3} - 2 (2) - 4

          = 8 - 4 - 4

          = 0

As it satisfies the given equation, \therefore \lambda = 2.

So, the given linear equation transforms into:

                               4x-2y+3z=0

                                 x+2y+2z=0

                                         2 x+2z=0

Now, the above equations can be represented in matrix form i.e.

        A = \begin{bmatrix} 4 & -2 & 3\\ 1& 2& 2\\ 2& 0& 2\end{bmatrix}, X = \begin{bmatrix} x\\ y\\ z\end{bmatrix}, B =\begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}

        or AX=B

Now, Put z = k from first two equations, we get:

    4 x - 2y = -3k

    x + 2y = -2k

Now, By Crammer’s rule:

x = \frac{D_{1}}{D} = \frac{\begin{vmatrix} -3k &-2 \\ -2k& 2\end{vmatrix}}{\begin{vmatrix} 4 & -2\\ 1 & 2\end{vmatrix}} =\frac{10k}{10} = k

y = \frac{D_{2}}{D} = \frac{\begin{vmatrix} 4-3k & 1-2k \end{vmatrix}}{\begin{vmatrix} 4-2 & 12 \end{vmatrix}}=\frac{-5k}{10} = \frac{-k}{2}

 

It also satisfies the third equation.

Hence,x = k ,y = \frac{-k}{2} and z = -k where k \in R

 

 

 

 

 

 

Posted by

infoexpert26

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