Question

Using properties of determinants, prove that

$|\begin{matrix} 1 & b c + a d & b^{2} c^{2} + a^{2} d^{2} \\ 1 & c a + b d & c^{2} a^{2} + b^{2} d^{2} \\ 1 & a b + c d & a^{2} b^{2} + c^{2} d^{2} \end{matrix}| = (a - b) (a - c) (a - d) (b - c) (b - d) (c - d)$ .

Accepted Answer

We have to prove that

$|\begin{matrix} 1 & b c + a d & b^{2} c^{2} + a^{2} d^{2} \\ 1 & c a + b d & c^{2} a^{2} + b^{2} d^{2} \\ 1 & a b + c d & a^{2} b^{2} + c^{2} d^{2} \end{matrix}| = (a - b) (a - c) (a - d) (b - c) (b - d) (c - d)$

$L . H . S . = |\begin{matrix} 1 & b c + a d & b^{2} c^{2} + a^{2} d^{2} \\ 1 & c a + b d & c^{2} a^{2} + b^{2} d^{2} \\ 1 & a b + c d & a^{2} b^{2} + c^{2} d^{2} \end{matrix}|$

$= |\begin{matrix} 1 & b c + a d & b^{2} c^{2} + a^{2} d^{2} \\ 0 & c a + b d - b c - a d & c^{2} a^{2} + b^{2} d^{2} - b^{2} c^{2} - a^{2} d^{2} \\ 0 & a b + c d - b c - a d & a^{2} b^{2} + c^{2} d^{2} - b^{2} c^{2} - a^{2} d^{2} \end{matrix}|$ [Applying $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ ]

$= |\begin{matrix} 1 & b c + a d & b^{2} c^{2} + a^{2} d^{2} \\ 0 & c (a - b) - d (a - b) & c^{2} (a^{2} - b^{2}) - d^{2} (a^{2} - b^{2}) \\ 0 & b (a - c) - d (a - c) & b^{2} (a^{2} - c^{2}) - d^{2} (a^{2} - c^{2}) \end{matrix}|$

$= |\begin{matrix} 1 & b c + a d & b^{2} c^{2} + a^{2} d^{2} \\ 0 & (a - b) (c - d) & (a^{2} - b^{2}) (c^{2} - d^{2}) \\ 0 & (a - c) (b - d) & (a^{2} - c^{2}) (b^{2} - d^{2}) \end{matrix}|$

$= (a - b) (c - d) (a - c) (b - d) |\begin{matrix} 1 & b c + a d & b^{2} c^{2} + a^{2} d^{2} \\ 0 & 1 & (a + b) (c + d) \\ 0 & 1 & (a + c) (b + d) \end{matrix}|$ [Taking out $(a - b) (c - d)$ Common from second row and $(a - c) (b - d)$ From third row]

$= (a - b) (c - d) (a - c) (b - d) |\begin{matrix} 1 & (a + b) (c + d) \\ 1 & (a + c) (b + d) \end{matrix}|$ [Expanding along first column]

$= (a - b) (c - d) (a - c) (b - d) 1 [(a + c) (b + d) - (a + b) (c + d)]$

$= (a - b) (c - d) (a - c) (b - d) [a b + a d + c b + c d - a c - a d - b c - b d]$

$= (a - b) (c - d) (a - c) (b - d) [a b + c d - a c - b d]$

$= (a - b) (c - d) (a - c) (b - d) [a (b - c) - d (b - c)]$

$= (a - b) (a - c) (a - d) (b - c) (b - d) (c - d) = R . H . S .$

Hence, The given result is proved.

M L Aggarwal Solutions for Chapter: Determinants, Exercise 2: EXERCISE 4.2

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Determinants, Exercise 2: EXERCISE 4.2

Questions from M L Aggarwal Solutions for Chapter: Determinants, Exercise 2: EXERCISE 4.2 with Hints & Solutions

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