Question

The value of $\left|\begin{array}{ccc}3x& -x+y& -x+z\\ x-y& 3y& z-y\\ x-z& y-z& 3z\end{array}\right|$ equals to

• A. $6(x+y+z)$
• B. $3(x+y+z)(xy+yz+zx)$
• C. $6(x+y+z)(xy+yz+zx)$
• D. $9(x+y)(xy+2yz)$

Answer 1

The correct option is B $3(x+y+z)(xy+yz+zx)$

We have $\left|\begin{array}{ccc}3x& -x+y& -x+z\\ x-y& 3y& z-y\\ x-z& y-z& 3z\end{array}\right|$
Applying $C_1\to C_1+C_2+C_3$
$=\left|\begin{array}{ccc}x+y+z& -x+y& -x+z\\ x+y+z& 3y& z-y\\ x+y+z& y-z& 3z\end{array}\right|$ Taking $(x+y+z)$ common from $C_1$
$=(x+y+z)\left|\begin{array}{ccc}1& -x+y& -x+z\\ 1& 3y& z-y\\ 1& y-z& 3z\end{array}\right|$
Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$=(x+y+z)\left|\begin{array}{ccc}1& -x+y& -x+z\\ 0& 2y+x& x-y\\ 0& x-z& 2z+x\end{array}\right|$
Applying $C_2\to C_2-C_3$
$=(x+y+z)\left|\begin{array}{ccc}1& y-z& -x+z\\ 0& 3y& x-y\\ 0& -3z& 2z+x\end{array}\right|$

Expanding along first column, we get
$(x+y+z)\cdot 1\left[3y\right(2z+x)+(3z\left)\right(x-y\left)\right]=(x+y+z)(3yz+3yx+3xz)=3(x+y+z)(xy+yz+zx)$