Prove that:
Question:

Prove that:

$\left|\begin{array}{ccc}x+4 & x & x \\ x & x+4 & x \\ x & x & x+4\end{array}\right|=16(3 x+4)$

Solution:

Let LHS $=\Delta=\mid x+4 \quad x \quad x$

$\begin{array}{ccc}x & x+4 & x \\ x & x & x+4 \mid\end{array}$

$=\mid 3 x+4 \quad 3 x+4 \quad 3 x+4$

$\begin{array}{cccc}\mathrm{x} & \mathrm{x}+4 & \mathrm{x} & \\ \mathrm{x} & \mathrm{x} & \mathrm{x}+4 & \text { [Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \text { ] }\end{array}$

$=(3 x+4) \mid 1 \quad 1 \quad 1$

$\begin{array}{ccc}x & x+4 & x \\ x & x & x+4 \mid\end{array}$

[Taking out $(3 x+4)$ common from $R_{1}$ ]

$=(3 x+4) \mid 1 \quad 0 \quad 0$

$\begin{array}{ccc}\mathrm{x} & 4 & 0 \\ \mathrm{x} & 0 & 4 \mid\end{array}$              [Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}$ and $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$ ]

$=(3 \mathrm{x}+4)\left(4^{2}\right) \quad\left[\right.$ Expanding along $\left.\mathrm{R}_{1}\right]$

$=16(3 x+4)$

$=$ RHS