Evaluate

Question:

Evaluate $\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|$

Solution:

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

Applying $R, \rightarrow R, R, R$, we have:

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

$=2(x+y)\left|\begin{array}{ccc}1 & 1 & 1 \\ y & x+y & x \\ x+y & x & y\end{array}\right|$

Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}$ and $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$, we have:

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

Expanding along $R_{1}$, we have:

$\begin{aligned} \Delta &=2(x+y)\left[-x^{2}+y(x-y)\right] \\ &=-2(x+y)\left(x^{2}+y^{2}-y x\right) \\ &=-2\left(x^{3}+y^{3}\right) \end{aligned}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now