Question

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

• A. $1$
• B. $2$
• C. $4$
• D. $8$

Answer 1

The correct option is B

$2$

Explanation for the correct option

Step 1: Simplify the given matrix by performing column operation

It is given that $\left|\begin{array}{ccc}x+y+2z& x& y\\ z& y+z+2x& y\\ z& x& z+x+2y\end{array}\right|=k{\left(x+y+z\right)}^3$.

Apply $C_1\to C_1+C_2+C_3$.

$$\begin{gathered} \Rightarrow \left|\begin{array}{ccc}x+y+2z+x+y& x& y\\ z+y+z+2x+y& y+z+2x& y\\ z+x+z+x+2y& x& z+x+2y\end{array}\right|=k{\left(x+y+z\right)}^3 \\ \Rightarrow \left|\begin{array}{ccc}2x+2y+2z& x& y\\ 2z+2y+2z& y+z+2x& y\\ 2z+2x+2y& x& z+x+2y\end{array}\right|=k{\left(x+y+z\right)}^3 \\ \Rightarrow \left|\begin{array}{ccc}2\left(x+y+z\right)& x& y\\ 2\left(x+y+z\right)& y+z+2x& y\\ 2\left(x+y+z\right)& x& z+x+2y\end{array}\right|=k{\left(x+y+z\right)}^3 \\ \Rightarrow 2\left(x+y+z\right)\left|\begin{array}{ccc}1& x& y\\ 1& y+z+2x& y\\ 1& x& z+x+2y\end{array}\right|=k{\left(x+y+z\right)}^3 \end{gathered}$$

Step 2: Simplify the matrix further by performing row operation

Apply $R_2\to R_2-R_1$.

$$\begin{gathered} \Rightarrow 2\left(x+y+z\right)\left|\begin{array}{ccc}1& x& y\\ 1-1& y+z+2x-x& y-y\\ 1& x& z+x+2y\end{array}\right|=k{\left(x+y+z\right)}^3 \\ \Rightarrow 2\left(x+y+z\right)\left|\begin{array}{ccc}1& x& y\\ 0& y+z+x& 0\\ 1& x& z+x+2y\end{array}\right|=k{\left(x+y+z\right)}^3 \\ \Rightarrow 2\left(x+y+z\right)\left(x+y+z\right)\left|\begin{array}{ccc}1& x& y\\ 0& 1& 0\\ 1& x& z+x+2y\end{array}\right|=k{\left(x+y+z\right)}^3 \end{gathered}$$

Step 3: Solve for the required value

Apply $R_3\to R_3-R_1$

.$$\begin{gathered} \Rightarrow 2{\left(x+y+z\right)}^2\left|\begin{array}{ccc}1& x& y\\ 0& 1& 0\\ 1-1& x-x& z+x+2y-y\end{array}\right|=k{\left(x+y+z\right)}^3 \\ \Rightarrow 2{\left(x+y+z\right)}^2\left|\begin{array}{ccc}1& x& y\\ 0& 1& 0\\ 0& 0& z+x+y\end{array}\right|=k{\left(x+y+z\right)}^3 \\ \Rightarrow 2{\left(x+y+z\right)}^2\left(x+y+z\right)\left|\begin{array}{ccc}1& x& y\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|=k{\left(x+y+z\right)}^3 \\ \Rightarrow 2{\left(x+y+z\right)}^2\left(x+y+z\right)\left[1\left\{\left(1\times 1\right)-\left(0\times 0\right)\right\}-x\left\{\left(0\times 1\right)-\left(0\times 0\right)\right\}+y\left\{\left(0\times 0\right)-\left(0\times 1\right)\right\}\right]=k{\left(x+y+z\right)}^3 \\ \Rightarrow 2{\left(x+y+z\right)}^3=k{\left(x+y+z\right)}^3 \\ \Rightarrow k=2 \end{gathered}$$

Therefore, the value of $k$ is $2$.

Hence, option(B) is the correct option i.e. $2$