CGPET 2015 Solved Paper

Given, $(\hat{i}+\hat{j}+3\hat{k})x+(3\hat{i}-3\hat{j}+\hat{k})y+(-4\hat{i}+5\hat{j})z$ $=\lambda(\hat{i}x+\hat{j}y+\hat{k}z)$ On equating the coefficients of $\hat{i}, \hat{j} \text{ and } \hat{k}$ both sides, we have $x+3y-4z = \lambda x,$ $x-3y+5z=\lambda y$ and $3x+y+0=\lambda z$ Above three equations can b e rewritten as $(1-\lambda)x+3y-4z=0$ $x-(3+\lambda)y+5z=0$ $3x+y-\lambda z=0$ This is homogeneous systemn of equations in three variables x, y and z. It is consistent and have non - zero solution . i.e (x, y, z) ≠ (0, 0, 0), if determinant of coefficient matrix is zero. ⇒$\begin{bmatrix} 1-\lambda & 3 & -4 \\ 1 & -(3+\lambda) & 5 \\ 3 & 1 & -\lambda \end{bmatrix}$ On expanding along first row, we have $(1-\lambda)[\lambda(3+\lambda)-5]-3(-\lambda-15)-4(1+9+3\lambda)=0$ ⇒$(1-\lambda)(\lambda^2+3\lambda-5)+3\lambda+45-40-12\lambda=0$ ⇒$\lambda^2+3\lambda-5-\lambda^3-3\lambda^2+5\lambda-9\lambda+5=0$ ⇒$-\lambda^3-2\lambda^2-\lambda=0$ ⇒$\lambda(\lambda^2+2\lambda+1)=0$ ⇒$\lambda(\lambda+1)^2=0$ ⇒$\lambda=0, -1$