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Prove the Following Identities: ∣ ∣ ∣ ∣ Y + Z Z Y Z Z + X X Y X X + Y ∣ ∣ ∣ ∣ = 4 X Y Z - Mathematics

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Question

Prove the following identities:

\[\begin{vmatrix}y + z & z & y \\ z & z + x & x \\ y & x & x + y\end{vmatrix} = 4xyz\]

Solution

\[LHS: \]

\[\begin{vmatrix}y + z & z & y \\ z & z + x & x \\ y & x & x + y\end{vmatrix}\]

\[ = \begin{vmatrix}y + z - z - y & z - z - x - x & y - x - x - y \\ z & z + x & x \\ y & x & x + y\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 - R_2 - R_3 \right]\]

\[ = \begin{vmatrix}0 & - 2x & - 2x \\ z & z + x & x \\ y & x & x + y\end{vmatrix}\]

\[ = - 2x\begin{vmatrix}0 & 1 & 1 \\ z & z + x & x \\ y & x & x + y\end{vmatrix} \left[\text{ Taking }- 2x\text{ common from }R_1 \right]\]

\[ = - 2x\begin{vmatrix}0 & 0 & 1 \\ z & z & x \\ y & - y & x + y\end{vmatrix} \left[\text{ Applying }C_2 \text{ to }C_2 - C_3 \right]\]

\[ = - 2x\left( - zy - zy \right) \left[\text{ Expanding along first row }\right]\]

\[ = 4xyz\]

\[ = RHS\]

\[ \therefore \begin{vmatrix}y + z & z & y \\ z & z + x & x \\ y & x & x + y\end{vmatrix} = 4xyz\]

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Chapter 6: Determinants - Exercise 6.2 [Page 60]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.2 | Q 39 | Page 60

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