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### Answers (1)

Answer:

Correct option (b)

Hint:

Solve determinant by applying row and column operation.

Given:

\begin{aligned} &\text { Let } \Delta=\left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right| \end{aligned}

We have to find the value of $\Delta$

Solution:

\begin{aligned} &\text { Here } \: \: \Delta=\left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right| \end{aligned}

Applying R1 → R1 - R2

$\Rightarrow \Delta =\left|\begin{array}{ccc} -2y & y &y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right|$

Applying R3 → R3 - R2

$\Rightarrow \Delta =\left|\begin{array}{ccc} -2y & y &y \\ x+2 y & x & x+y \\ -y & 2 y & -y \end{array}\right|$

Taking common y from R1 and R3

$\Rightarrow \Delta =y^{2}\left|\begin{array}{ccc} -2 & 1 &1 \\ x+2 y & x & x+y \\ -1 & 2 & -1 \end{array}\right|$

Applying C2 → C2 + 2C1; C3 → C3 -C1

$\Rightarrow \Delta =y^{2}\left|\begin{array}{ccc} -2 & -3 &3 \\ x+2 y & x & x+y \\ -1 & 0 & 0 \end{array}\right|$

Expanding along R3, we get

\begin{aligned} &\Delta=y^{2}[(-1)(3 y-9 x-12 y)] \\ &\Delta=y^{2}[9 x+9 y] \end{aligned}

\begin{aligned} &\Delta=9y^{2}[ x+ y] \end{aligned}

\begin{aligned} &\text { Hence } \left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right| \end{aligned}=9y^{2}[x+y]

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