If x,y and z be the pth, qth and rth terms of an AP, show that determinant ((p,q,r),(x,y,z),(1,1,1))=0


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Question Number 175838 by Rasheed.Sindhi last updated on 08/Sep/22

$I f x, y a n d z b e t h e p t h, q t h a n d r t h$ $t e r m s o f a n A P, s h o w t h a t$ $\| \begin{matrix} p & q & r \\ x & y & z \\ 1 & 1 & 1 \end{matrix} \| = 0$
	Answered by Rasheed.Sindhi last updated on 08/Sep/22

	$\| \begin{matrix} p & q & r \\ x & y & z \\ 1 & 1 & 1 \end{matrix} \| = 0$ $\| \begin{matrix} p & q & r \\ a + (p - 1) d & a + (q - 1) d & a + (q - 1) d \\ 1 & 1 & 1 \end{matrix} \|$ $= \| \begin{matrix} p & q & r \\ a - d + p d & a - d + q d & a - d + q d \\ 1 & 1 & 1 \end{matrix} \|$ $= \| \begin{matrix} p & q & r \\ a - d & a - d & a - d \\ 1 & 1 & 1 \end{matrix} \| [R 2 - R 1 \cdot d]$ $= (a - d) \| \begin{matrix} p & q & r \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \| = (a - d) \cdot 0 = 0$ $[T w o r o w s a r e i d e n t i c a l]$ $P r o v e d$
	Answered by som(math1967) last updated on 08/Sep/22

	$\| \begin{matrix} p & q & r \\ a + p d - d & a + q d - d & a + r d - d \\ 1 & 1 & 1 \end{matrix} \|$ $C_{1}^{l} \to C_{1} - C_{2}, C_{2}^{l} \to C_{2} - C_{3}$ $\| \begin{matrix} p - q & q - r & r \\ (p - q) d & (q - r) d & a + r d - d \\ 0 & 0 & 1 \end{matrix} \|$ $(p - q) (q - r) \| \begin{matrix} 1 & 1 & r \\ d & d & a + r d - d \\ 0 & 0 & 1 \end{matrix} \|$ $= (p - q) (q - r) \times 0 [C_{1}, C_{2} i d e n t i c a l]$ $= 0$
	Commented by Rasheed.Sindhi last updated on 08/Sep/22

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