3 and 4 .Determinants and Matrices
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यदि  किसी समान्तर श्रेणी के $p$ वें, $q$ वें तथा $r$ वें पद क्रमश: $a,b,c$ हों, तो $\left| {\,\begin{array}{*{20}{c}}a&p&1\\b&q&1\\c&r&1\end{array}\,} \right| = $

A

$1$

B

$-1$

C

$0$

D

$pqr$

Solution

(c) माना प्रथम पद  = $A$ तथा सार्वान्तर  =$ D$

 $\therefore a = A + (p – 1)D$, $b = A + (q – 1)D$, $c = A + (r – 1)D$

$\left| {\begin{array}{*{20}{c}}{\,a\,\,\,}&{p\,\,\,}&{1\,}\\{\,b\,\,\,}&{q\,\,\,}&1\\{\,c\,\,\,}&{r\,\,\,}&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{\,A + (p – 1)D\,\,\,}&{p\,\,\,}&{1\,}\\{A + (q – 1)D\,\,\,}&{q\,\,\,}&1\\{A + (r – 1)D\,\,\,}&{r\,\,\,}&1\end{array}} \right|$,

(संक्रिया  ${C_1} \to {C_1} – D{C_2} + D{C_3}$ से)

$ = \,\left| {\begin{array}{*{20}{c}}{\,A\,\,}&{p\,\,}&{1\,}\\{\,A\,\,}&{q\,\,}&1\\{\,A\,\,}&{r\,\,}&1\end{array}} \right| = A\left| {\begin{array}{*{20}{c}}{\,\,1}&{\,\,p}&{\,\,1\,}\\{\,1}&q&1\\{\,1}&r&1\end{array}} \right| = 0$.

Standard 12
Mathematics

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