Question

Evaluate :

$\left|\begin{array}{ccc}x+\lambda & x& x\\ x& x+\lambda & x\\ x& x& x+\lambda \end{array}\right|$

Answer 1

$$\begin{gathered} ∆=\left|\begin{array}{ccc}x+\lambda & x& x\\ x& x+\lambda & x\\ x& x& x+\lambda \end{array}\right| \\ =\left|\begin{array}{ccc}\lambda & 0& x\\ -\lambda & \lambda & x\\ 0& -\lambda & x+\lambda \end{array}\right|\left[\mathrm{Applying}{C}_1\to C_1-C_2,C_2\to C_2-C_3\right] \\ =\left|\begin{array}{ccc}\lambda & 0& x\\ -\lambda & 0& 2x+\lambda \\ 0& -\lambda & x+\lambda \end{array}\right|\left[\mathrm{Applying}{R}_1\to R_2+R_3\right] \\ =\lambda \left|\begin{array}{cc}0& 2x+\lambda \\ -\lambda & x+\lambda \end{array}\right|+x\left|\begin{array}{cc}-\lambda & 0\\ 0& -\lambda \end{array}\right| \\ =\lambda \left[\lambda \right(2x+\lambda \left)\right]+x{\lambda }^2 \\ ={\lambda }^2(2x+\lambda +{\lambda }^{2}x) \\ =3{\lambda }^{2}x+{\lambda }^3 \\ ={\lambda }^2(3x+\lambda ) \end{gathered}$$