# Let $\lambda$ be a real number for which the system of linear equations$x+y+z=6$$4x+\lambda y-\lambda z=\lambda -2$$3x+2 y-4 z= -5$has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation : Option 1) $\lambda^{2}+3\lambda-4=0$ Option 2) $\lambda^{2}-3\lambda-4=0$ Option 3) $\lambda^{2}+\lambda-6=0$ Option 4) $\lambda^{2}-\lambda-6=0$

linear equations

$x+y+z=6$

$4x+\lambda y-\lambda z=\lambda -2$

$3x+2 y-4 z= -5$

$\Delta =\begin{vmatrix} 1 &1 &1 \\ 4& \lambda & -\lambda\\ 3& 2 &-4 \end{vmatrix}$

$=(-4\lambda+2\lambda)-1(-16+3\lambda)+1(8-3\lambda)$

$=-2\lambda+16-3\lambda+8-3\lambda$

$=-8\lambda+24$

Now, using cramers law for infinite solution

$\Delta ,\Delta _1,\Delta _1$ all will be zero

$-8\lambda+24=0$

$\lambda=3$

Now put $\lambda=3$ in options and check for the correct one

(1)$\lambda^{2}+3\lambda-4=(3)^{2}+3.3-4\neq 0$

(2)$\lambda^{2}-3\lambda-4=(3)^{2}-3.3-4\neq 0$

(3)$\lambda^{2}+\lambda-6=(3)^{2}+3-6\neq 0$

(4)$\lambda^{2}-\lambda-6=(3)^{2}-3-6=0$

So, option (4) is correct.

Option 1)

$\lambda^{2}+3\lambda-4=0$

Option 2)

$\lambda^{2}-3\lambda-4=0$

Option 3)

$\lambda^{2}+\lambda-6=0$

Option 4)

$\lambda^{2}-\lambda-6=0$

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