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

Sum

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

 
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Solution

\[Let  ∆ =|   y + z   x   y\]

\[ z + x  z  x \]

\[x + y y z |\]

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

\[ z + x z x \]

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

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

\[ z + x z x \]

\[ x + y y z  | \]

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

\[0 z x\]

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

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

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

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APPEARS IN

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