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Question Number 7911 by ashis786 last updated on 24/Sep/16

If a>0, b>0, c>0 are respectively the p^(th) , q^(th) , r^(th) terms of a GP, then the value of the determinant determinant (((log a),p,1),((log b),q,1),((log c),r,1)) is

$$\mathrm{If}\:{a}>\mathrm{0},\:{b}>\mathrm{0},\:{c}>\mathrm{0}\:\mathrm{are}\:\mathrm{respectively}\:\mathrm{the} \\ $$ $${p}^{\mathrm{th}} ,\:{q}^{\mathrm{th}} ,\:{r}^{\mathrm{th}} \:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{GP},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value} \\ $$ $$\mathrm{of}\:\mathrm{the}\:\mathrm{determinant}\:\begin{vmatrix}{\mathrm{log}\:{a}}&{{p}}&{\mathrm{1}}\\{\mathrm{log}\:{b}}&{{q}}&{\mathrm{1}}\\{\mathrm{log}\:{c}}&{{r}}&{\mathrm{1}}\end{vmatrix}\:\mathrm{is} \\ $$

Answered by prakash jain last updated on 24/Sep/16

Let first term of GP be xy with common ratio y a=xy^p ⇒log a=log x+plog y b=xy^q ⇒log b=log x+qlog y c=xy^r ⇒log c=log x+rlog y determinant (((log a),p,1),((log b),q,1),((log c),r,1))= determinant (((log x+plog y),p,1),((logx+qlog y),q,1),((log x+rlog y),r,1)) = determinant (((log x),p,1),((logx),q,1),((log x),r,1))+ determinant (((plog y),p,1),((qlog y),q,1),((rlog y),r,1)) =log x determinant ((1,p,1),(1,q,1),(1,r,1))+log y determinant ((p,p,1),(q,q,1),(r,r,1)) C1 =C3 (in 1^(st) determinant) and C1=C2 in 2^(nd) =(log x)×0+log y×0=0

$$\mathrm{Let}\:\mathrm{first}\:\mathrm{term}\:\mathrm{of}\:\mathrm{GP}\:\mathrm{be}\:{xy}\:\mathrm{with}\:\mathrm{common}\:\mathrm{ratio}\:{y} \\ $$ $${a}={xy}^{{p}} \Rightarrow\mathrm{log}\:{a}=\mathrm{log}\:{x}+{p}\mathrm{log}\:{y} \\ $$ $${b}={xy}^{{q}} \Rightarrow\mathrm{log}\:{b}=\mathrm{log}\:{x}+{q}\mathrm{log}\:{y} \\ $$ $${c}={xy}^{{r}} \Rightarrow\mathrm{log}\:{c}=\mathrm{log}\:{x}+{r}\mathrm{log}\:{y} \\ $$ $$\begin{vmatrix}{\mathrm{log}\:{a}}&{{p}}&{\mathrm{1}}\\{\mathrm{log}\:{b}}&{{q}}&{\mathrm{1}}\\{\mathrm{log}\:{c}}&{{r}}&{\mathrm{1}}\end{vmatrix}=\begin{vmatrix}{\mathrm{log}\:{x}+{p}\mathrm{log}\:{y}}&{{p}}&{\mathrm{1}}\\{\mathrm{log}{x}+{q}\mathrm{log}\:{y}}&{{q}}&{\mathrm{1}}\\{\mathrm{log}\:{x}+{r}\mathrm{log}\:{y}}&{{r}}&{\mathrm{1}}\end{vmatrix} \\ $$ $$=\begin{vmatrix}{\mathrm{log}\:{x}}&{{p}}&{\mathrm{1}}\\{\mathrm{log}{x}}&{{q}}&{\mathrm{1}}\\{\mathrm{log}\:{x}}&{{r}}&{\mathrm{1}}\end{vmatrix}+\begin{vmatrix}{{p}\mathrm{log}\:{y}}&{{p}}&{\mathrm{1}}\\{{q}\mathrm{log}\:{y}}&{{q}}&{\mathrm{1}}\\{{r}\mathrm{log}\:{y}}&{{r}}&{\mathrm{1}}\end{vmatrix} \\ $$ $$=\mathrm{log}\:{x}\begin{vmatrix}{\mathrm{1}}&{{p}}&{\mathrm{1}}\\{\mathrm{1}}&{{q}}&{\mathrm{1}}\\{\mathrm{1}}&{{r}}&{\mathrm{1}}\end{vmatrix}+\mathrm{log}\:{y}\begin{vmatrix}{{p}}&{{p}}&{\mathrm{1}}\\{{q}}&{{q}}&{\mathrm{1}}\\{{r}}&{{r}}&{\mathrm{1}}\end{vmatrix} \\ $$ $$\mathrm{C1}\:=\mathrm{C3}\:\left(\mathrm{in}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{determinant}\right)\:\:\:\mathrm{and}\:\mathrm{C1}=\mathrm{C2}\:\mathrm{in}\:\mathrm{2}^{\mathrm{nd}} \\ $$ $$=\left(\mathrm{log}\:{x}\right)×\mathrm{0}+\mathrm{log}\:{y}×\mathrm{0}=\mathrm{0} \\ $$

Commented byRasheed Soomro last updated on 24/Sep/16

Also like your way of correction.

$${Also}\:{like}\:{your}\:{way}\:{of}\:{correction}.\:\:\: \\ $$

Commented byashis786 last updated on 24/Sep/16

thank you so much...

$${thank}\:{you}\:{so}\:{much}... \\ $$

Commented bysandy_suhendra last updated on 24/Sep/16

if the first term of GP =x and the ratio=y so a=the p^(th) term of GP = xy^( p−1) isn′t it?

$${if}\:{the}\:{first}\:{term}\:{of}\:{GP}\:={x}\:{and}\:{the}\:{ratio}={y} \\ $$ $${so}\:\:\:{a}={the}\:{p}^{{th}} \:{term}\:{of}\:{GP}\:=\:{xy}^{\:{p}−\mathrm{1}} \: \\ $$ $${isn}'{t}\:{it}? \\ $$

Commented byprakash jain last updated on 24/Sep/16

Thanks. Updated the answer to start with first term xy.

$$\mathrm{Thanks}.\:\mathrm{Updated}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{start}\:\mathrm{with} \\ $$ $$\mathrm{first}\:\mathrm{term}\:{xy}. \\ $$

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