First slide

Introduction to Determinants

Question

If $|\begin{array}{ccc} a & b - c & c + b \\ a + c & b & c - a \\ a - b & a + b & c \end{array}| = 0$ , then the line ax + by + c = 0 passes through the fixed point which is

Moderate

A

(1, 2)
B

(1, 1)
C

(-2, 1)
D

(1, 0)

Solution

Applying $C_{1} \to {aC}_{1}$ and then $C_{1} \to C_{1} + {bC}_{2} + {cC}_{3}$ , and taking $(a^{2} + b^{2} + c^{2})$ common from C₁, we get
$\begin{matrix} Δ = \frac{(a^{2} + b^{2} + c^{2})}{a} |\begin{array}{ccc} 1 & b - c & c + b \\ 1 & b & c - a \\ 1 & b + a & c \end{array}| \\ = \frac{(a^{2} + b^{2} + c^{2})}{a} |\begin{array}{ccc} 1 & b - c & c + b \\ 0 & c & - a - b \\ 0 & a + c & - b \end{array}| (R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}) \\ = \frac{(a^{2} + b^{2} + c^{2})}{a} (- bc + a^{2} + ab + ac + bc) (expanding along C_{1}) \\ = (a^{2} + b^{2} + c^{2}) (a + b + c) \end{matrix}$
Hence, $Δ = 0 \Rightarrow a + b + c = 0$
Therefore, line ax + by + c = 0 passes through the fixed point (1, 1).

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