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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
$\text{A(adj. A) = (adj.A)A} | \text{A} = | \text{I}_{3}.$
If a, b and c are all non-zero and $ \begin{vmatrix} \text{1 + a} & 1 & 1 \\ 1 & \text{1 + b} & 1 \\ 1 & 1 & \text{1 + c} \end{vmatrix} = 0,$ then prove that $\frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} 1 = 0.$
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Answer

$\text{abc} \begin{vmatrix} \frac{1}{\text{a}} + 1 & \frac{1}{\text{b}} & \frac{1}{\text{c}} \\ \frac{1}{\text{a}} & \frac{1}{\text{b}} + 1 & \frac{1}{\text{c}} \\ \frac{1}{\text{a}} & \frac{1}{\text{b}} & \frac{1}{\text{c}} + 1 \end{vmatrix} = 0$
$ \text{C}_{1} \rightarrow \text{C}_{1} + \text{C}_{2} + \text{C}_{3}$
$\Rightarrow \text{abc} \begin{vmatrix} 1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} & \frac{1}{\text{b}}& \frac{1}{\text{c}} \\ 1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} & \frac{1}{\text{b}} + 1& \frac{1}{\text{c}} \\ 1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} & \frac{1}{\text{b}}& \frac{1}{\text{c}} + 1 \end{vmatrix} = 0$
$\Rightarrow \text{abc} \bigg(1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}}\bigg) \begin{vmatrix} 1 & \frac{1}{\text{b}} & \frac{1}{\text{c}} \\ 1 & \frac{1}{\text{b}} + 1 & \frac{1}{\text{c}} \\ 1 & \frac{1}{\text{b}} & \frac{1}{\text{c}} + 1 \end{vmatrix} = 0$
$\text{R}_{2} \rightarrow - \text{R}_{2} - \text{R}_{1}. \text{R}_{3} \rightarrow \text{R}_{3}- \text{R}_{1}$
$\Rightarrow \text{abc} \bigg(1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}}\bigg) = 0$
$\because \text{a, b, c,}\neq 0$
$\therefore 1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} = 0$

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