If {Dₚ} = left| {,begin{array}{*{20}{c}}p&{15}&8{{p^2}}&{35}&9{{p^3}}&{25}&amp

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3 and 4 .Determinants and Matrices

hard

If ${D_p} = \left| {\,\begin{array}{*{20}{c}}p&{15}&8\\{{p^2}}&{35}&9\\{{p^3}}&{25}&{10}\end{array}\,} \right|$, then ${D_1} + {D_2} + {D_3} + {D_4} + {D_5} = $

$0$

$25$

$625$

$- 700000$

Solution

(d) ${D_1} = \left| {\,\begin{array}{*{20}{c}}1&{15}&8\\1&{35}&9\\1&{25}&{10}\end{array}\,} \right|,{D_2} = \left| {\,\begin{array}{*{20}{c}}2&{15}&8\\4&{35}&9\\8&{25}&{10}\end{array}\,} \right|$

${D_3} = \left| {\,\begin{array}{*{20}{c}}3&{15}&8\\9&{35}&9\\{27}&{25}&{10}\end{array}\,} \right|,{D_4} = \left| {\,\begin{array}{*{20}{c}}4&{15}&8\\{16}&{35}&9\\{64}&{25}&{10}\end{array}\,} \right|$

${D_5} = \left| {\,\begin{array}{*{20}{c}}5&{15}&8\\{25}&{35}&9\\{125}&{25}&{10}\end{array}\,} \right|$

==> ${D_1} + {D_2} + {D_3} + {D_4} + {D_5} = \left| {\,\begin{array}{*{20}{c}}{15}&{75}&{40}\\{55}&{175}&{45}\\{225}&{125}&{50}\end{array}\,} \right|$

$ = 15(3125) – 75( – 7375) + 40( – 32500)$

$ = 46875 + 553125 – 1300000 = – 700000$ .

Standard 12

Mathematics

If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x – 2}\\{2{x^2} + 3x – 1}&{3x}&{3x – 3}\\{{x^2} + 2x + 3}&{2x – 1}&{2x – 1}\end{array}\,} \right| = Ax – 12$, then the value of $A $ is

medium

(JEE MAIN-2015)

$\left| {\,\begin{array}{*{20}{c}}{a – b}&{b – c}&{c – a}\\{x – y}&{y – z}&{z – x}\\{p – q}&{q – r}&{r – p}\end{array}\,} \right| = $

easy

If $D = \left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}\,} \right|$ for $x \ne 0,y \ne 0$ then $D$ is

medium

(AIEEE-2007)

If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&1&1\\2&{x + 2}&2\\3&3&{x + 3}\end{array}\,} \right| = 0,$ then $x$ is

easy

If the lines $x + 2ay + a = 0, x + 3by + b = 0$ and $x + 4cy + c = 0$ are concurrent, then $a, b$ and $c$ are in :-

normal