Q137 / 200 · Manipal (MET) 2014 · Mathematics

If $\left|\begin{array}{ccc}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|$ $=k(x+y+z)^3$, then the value of $k$ is

  • A
    1
  • Correct answer
    B
    2
  • C
    4
  • D
    8

Solution

$\begin{aligned} \Delta & =\left|\begin{array}{ccc} x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y \end{array}\right| \\ & =\left|\begin{array}{ccc} 2(x+y+z) & x & y \\ 2(x+y+z) & y+z+2 x & y \\ 2(x+y+z) & x & z+x+2 y \end{array}\right| \end{aligned}$ (using $C_1 \rightarrow C_1+C_2+C_3$ ) Take out $2(x+y+z)$ common from $C_1$, we get $\begin{aligned} & \Delta=2(x+y+z)\left|\begin{array}{ccc} 1 & x & y \\ 1 & y+z+2 x & y \\ 1 & x & z+x+2 y \end{array}\right| \\ & =2(x+y+z)\left|\begin{array}{ccc} 1 & x & y \\ 0 & y+z+x & 0 \\ 0 & 0 & z+x+y \end{array}\right| \end{aligned}$ (using $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$ ) Take out $(x+y+z)$ common from $R_2$ and $R_3$, we get $\begin{aligned} \Delta=2(x+y+z)(x+y+z)(x+y & +z) \\ & \times\left|\begin{array}{lll} 1 & x & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right| \end{aligned}$
Expanding along $R_3$, we get $\begin{aligned} \Delta & =2(x+y+z)^3[(1)(1-0)] \\ & =2(x+y+z)^3=k(x+y+z)^3 \text { (given) } \\ \therefore \quad k & =2 \end{aligned}$

Hence, the correct answer is option: 'B'.

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