$∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a b & \frac{1}{a} + \frac{1}{b} 1 & b c & \frac{1}{b} + \frac{1}{c} 1 & c a & \frac{1}{c} + \frac{1}{a} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ =$

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Solution

$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a b & 1 / a + 1 / b 1 & b c & 1 / b + 1 / c 1 & c a & 1 / c + 1 / a \end{matrix} ∣ ∣ ∣ ∣ ∣$

$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a b & \frac{1}{a} + \frac{1}{b} 1 - 1 & b c - a b & \frac{1}{b} + \frac{1}{c} - \frac{1}{a} - \frac{1}{b} 1 - 1 & c a - a b & \frac{1}{c} + \frac{1}{a} - \frac{1}{a} - \frac{1}{b} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ {\begin{matrix} R_{2} ⟶ R_{2} - R_{1} R_{3} ⟶ R_{3} - R_{1} \end{matrix}}$

$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a b & \frac{1}{a} + \frac{1}{b} 1 & b (c - a) & - \frac{c - a}{a c} 1 & a (c - b) & - \frac{c - b}{b c} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$

$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a b & \frac{1}{a} + \frac{1}{b} 0 & b & - \frac{1}{a c} 0 & a & - \frac{1}{b c} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \times (c - a) (c - b) ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} R_{2} ⟶ \frac{R_{2}}{c - a} R_{3} ⟶ \frac{R_{3}}{c - b} \end{matrix} ⎫ ⎪ ⎬ ⎪ ⎭$

$= (c - a) (c - b) \times {1 \times ((b \times - \frac{1}{b c}) - (a \times - \frac{1}{a c})) + 0 + 0}$

{expanding along 1^st column}

$= (c - a) (c - b) \times (- \frac{1}{c} + \frac{1}{c})$

$= (c - a) (c - b) \times 0 = 0$

$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a b & 1 / c + 1 / b 1 & b c & 1 / b + 1 / c 1 & c a & 1 / c + 1 / a \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

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The determinant $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a b & \frac{1}{a} + \frac{1}{b} 1 & b c & \frac{1}{b} + \frac{1}{c} 1 & c a & \frac{1}{c} + \frac{1}{a} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$ is equal to

Q. If

a, b, c

are non-zero real numbers, then

Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a b & \frac{1}{a} + \frac{1}{b} 1 & b c & \frac{1}{b} + \frac{1}{c} 1 & c a & \frac{1}{c} + \frac{1}{a} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

Q. Represent the following mixed infinite decimal periodic fractions as common fractions:

2 (a + b)^{- 1} (a b)^{1 / 2} {⎛ ⎝ 1 + \frac{1}{4} {(\sqrt{\frac{a}{b}} - \sqrt{\frac{b}{a}})}^{2} ⎞ ⎠}^{1 / 2}

Q. Prove that:

(i)

\frac{a + b + c}{a^{- 1} b^{- 1} + b^{- 1} c^{- 1} + c^{- 1} a^{- 1}} = a b c

(ii)

{(a^{- 1} + b^{- 1})}^{- 1} = \frac{a b}{a + b}

Evaluate tan^-1(a - b)/(1+ab)+ tan^-1(b - c)/(1+bc)+ tan^-1(c - a)/(1+cb) ; ab,bc,ac>-1

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∣∣ ∣ ∣ ∣∣1ab1a+1b1bc1b+1c1ca1c+1a∣∣ ∣ ∣ ∣∣=

$∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a b & \frac{1}{a} + \frac{1}{b} 1 & b c & \frac{1}{b} + \frac{1}{c} 1 & c a & \frac{1}{c} + \frac{1}{a} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ =$