Question

Using properties of determinants, prove that $\left|\begin{array}{ccc}1& 1& 1+3x\\ 1+3y& 1& 1\\ 1& 1+3z& 1\end{array}\right|=9(3xyz+xy+yz+zx)$

Answer 1

Let $\Delta =\left|\begin{array}{ccc}1& 1& 1+3x\\ 1+3y& 1& 1\\ 1& 1+3z& 1\end{array}\right|$
Taking $x,y,z$ common from $R_1,R_2,R_3$ respectively
= $xyz\left|\begin{array}{ccc}\frac{1}{x}& \frac{1}{x}& \frac{1}{x}+3\\ \frac{1}{y}+3& \frac{1}{y}& \frac{1}{y}\\ \frac{1}{z}& \frac{1}{z}+3& \frac{1}{z}\end{array}\right|\text{Operating}{R}_1\to R_1+R_2+R_3=xyz\left|\begin{array}{ccc}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3& \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3& \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3\\ \frac{1}{y}+3& \frac{1}{y}& \frac{1}{y}\\ \frac{1}{z}& \frac{1}{z}+3& \frac{1}{z}\end{array}\right|\text{Taking}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3\text{common from}{R}_1=xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3)\left|\begin{array}{ccc}1& 1& 1\\ \frac{1}{y}+3& \frac{1}{y}& \frac{1}{y}\\ \frac{1}{z}& \frac{1}{z}+3& \frac{1}{z}\end{array}\right|$
Operating $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$
$=(yz+zx+xy+3xyz)\left|\begin{array}{ccc}1& 0& 0\\ \frac{1}{y}+3& -3& -3\\ \frac{1}{z}& 3& 0\end{array}\right|$
Expanding along $R_1$, we get-
$=(xy+yz+zx+3xyz)(0+9)=9(xy+yz+zx+3xyz)$