Question

Evaluate:
$\left|\begin{array}{ccc}x+4& x& x\\ x& x+4& x\\ x& x& x+4\end{array}\right|$

Answer 1

$\left|\begin{array}{ccc}x+4& x& x\\ x& x+4& x\\ x& x& x+4\end{array}\right|=\left|\begin{array}{ccc}2x+4& 2x+4& 2x\\ x& x+4& x\\ x& x& x+4\end{array}\right|$ $[\because R_1\to R_1+R_2]$
$=\left|\begin{array}{ccc}2x& 2x& 2x\\ x& x+4& x\\ x& x& x+4\end{array}\right|+\left|\begin{array}{ccc}4& 4& 0\\ x& x+4& x\\ x& x& x+4\end{array}\right|$
[Here, given determinant is expressed as sum of two determinants]
$=2x\left|\begin{array}{ccc}1& 1& 1\\ x& x+4& x\\ x& x& x+4\end{array}\right|+4\left|\begin{array}{ccc}1& 1& 0\\ x& x+4& x\\ x& x& x+4\end{array}\right|$
[taking 2x common from first row of first determinant and 4 from first row of second determinant]
Applying $C_1\to C_1-C_3$ and $C_2\to C_2-C_3$ in first and applying $C_1\to C_1-C_2$ in second, we get
$=2x\left|\begin{array}{ccc}0& 0& 1\\ 0& 4& x\\ -4& -4& x+4\end{array}\right|+4\left|\begin{array}{ccc}0& 1& 0\\ -4& x+4& x\\ 0& x& x+4\end{array}\right|$
Expanding both the along first column, we get
$2x[-4(-4)+4[4(x+4-0)\left]\right]=2x\times 16+16(x+4)=32x+16x+64=16(3x+4)$