Show that

Question:

Show that $\left|\begin{array}{lll}y+z & x & y \\ z+x & z & x \\ x+y & y & z\end{array}\right|=(x+y+z)(x-z)^{2}$

Solution:

Let $\Delta=\mid y+z \quad x \quad y$

$z+x \quad z \quad x$

$x+y \quad y \quad z \mid$

$\Rightarrow \Delta=\mid 2(x+y+z) \quad x+y+z \quad x+y+z$

$z+x \quad z \quad x$

$\begin{array}{llll}x+y & y & z \mid & \text { [Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \text { ] }\end{array}$

$=(x+y+z) \mid \begin{array}{lll}2 & 1 & 1\end{array}$

$\begin{array}{ccc}z+x & z & x \\ x+y & y & z \mid\end{array}$

$=(\mathrm{x}+y+\mathrm{z}) \mid 0 \quad 1 \quad 1$

$0 \quad \mathrm{z} \quad \mathrm{x}$

$\begin{array}{lll}x-z & \mathrm{y} & \mathrm{z}\end{array}$      [Applying $C_{1} \rightarrow C_{1}-C_{2}-C_{3}$ ]

$=(\mathrm{x}+y+\mathrm{z})\left\{(x-z) \times\left|\begin{array}{ll}1 & 1 \\ z & x\end{array}\right|\right\}$      [Expanding along $C_{1}$ ]

$=(\mathrm{x}+y+\mathrm{z})(x-z)^{2}$

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