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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
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} = $
  • A
    $0$
  • B
    $25$
  • C
    $625$
  • ✓
    $- 700000$
✓

Answer

Correct option: D.
$- 700000$
d
(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$ .

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