# Show that ∣ ∣ ∣ ∣ Y + Z X Y Z + X Z X X + Y Y Z ∣ ∣ ∣ ∣ = ( X + Y + Z ) ( X − Z ) 2 - Mathematics

Sum

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

#### Solution

$Let ∆ =| y + z x y$

$z + x z x$

$x + y y z |$

$\Rightarrow ∆ = | 2\left( x + y + z \right) x + y + z x + y + z$

$z + x z x$

$x + y y z t | \left[ \text{ Applying } R_1 \to R_1 + R_2 + R_3 \right]$

$= \left( x + y + z \right) | 2 1 1$

$z + x z x$

$x + y y z |$

$= \left( x + y + z \right) 0 1 1$

$0 z x$

$x - z y z | \left[ \text{ Applying } C_1 \to C_1 - C_2 - C_3 \right]$

$= \left( x + y + z \right)\left\{ \left( x - z \right) \times \begin{vmatrix}1 & 1 \\ z & x\end{vmatrix} \right\} \left[ \text{ Expanding along } C_1 \right]$

$= \left( x + y + z \right) \left( x - z \right)^2$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 6 Determinants
Exercise 6.2 | Q 43 | Page 61