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Worked Examples · Example 28

Q.Evaluate Δ=∣1ab+c1ba+c1ca+b∣\Delta = \begin{vmatrix} 1 & a & b+c \\ 1 & b & a+c \\ 1 & c & a+b \end{vmatrix}.

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✓ Free question

Applying the column operation C3→C3+C2C_3 \to C_3 + C_2 makes column 3 equal to (a+b+c)(a+b+c) times column 1, so two columns become proportional and Δ=0\Delta = 0.

Column operation Ci→Ci+Cj leaves Δ unchanged; if two columns are proportional, Δ=0.\text{Column operation } C_i \to C_i + C_j \text{ leaves } \Delta \text{ unchanged; if two columns are proportional, } \Delta = 0.

  1. Start with Δ=∣1ab+c1ba+c1ca+b∣\Delta = \begin{vmatrix} 1 & a & b+c \\ 1 & b & a+c \\ 1 & c & a+b \end{vmatrix}.
  2. Apply C3→C3+C2C_3 \to C_3 + C_2 (add column 2 to column 3):

Δ=∣1aa+b+c1ba+b+c1ca+b+c∣.\Delta = \begin{vmatrix} 1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c \end{vmatrix}.

  1. Take the common factor (a+b+c)(a+b+c) out of column 3:

Δ=(a+b+c)∣1a11b11c1∣.\Delta = (a+b+c)\begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix}.

  1. Column 1 and column 3 are identical, so the determinant of the 3×33\times3 array is 00.
  2. Hence Δ=(a+b+c)×0=0.\Delta = (a+b+c)\times 0 = 0.
✓Final answer

Δ=0\Delta = 0.

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