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

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 $k \in R$