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∣ ∣ ∣ ∣ B + C a A B C + a B C C a + B ∣ ∣ ∣ ∣ = 4 a B C - Mathematics

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Question

\[\begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix} = 4abc\]

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Solution

\[∆ = \begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix}\]

\[ = \begin{vmatrix}0 & - 2c & - 2b \\ b & c + a & b \\ c & c & a + b\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 - ( R_2 + R_3 ) \right]\]

\[ = \begin{vmatrix}0 & - 2c & - 2b \\ b & c + a - b & 0 \\ c & 0 & a + b - c\end{vmatrix} \left[\text{ Applying }C_2 \text{ to }C_2 - C_1\text{ and }C_3 \text{ to }C_3 - C_1 \right]\]

\[ = 0\begin{vmatrix}c + a - b & 0 \\ 0 & a + b - c\end{vmatrix} - ( - 2c)\begin{vmatrix}b & 0 \\ c & a + b - c\end{vmatrix} - 2b\begin{vmatrix}b & c + a - b \\ c & 0\end{vmatrix} \left[\text{ Expanding along }R_1 \right]\]

\[ = 2c[b(a + b - c) - 0] - 2b[0 - c(c + a - b)]\]

\[ = 2bc[a + b - c] - 2bc[b - c - a]\]

\[ = 2bc[(a + b - c) - (b - c - a)]\]

\[ = 4abc\]

Hence proved.

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Chapter 6: Determinants - Exercise 6.2 [Page 60]

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RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.2 | Q 32 | Page 60

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