First slide

Introduction to Determinants

Question

$If |\begin{array}{lll} a b 1 \\ b c 1 \\ c a 1 \end{array}| = 2020 and if |\begin{array}{ccc} c - a & c - b & a b \\ a - b & a - c & b c \\ b - c & b - a & c a \end{array}| - |\begin{array}{ccc} c - a & c - b & c^{2} \\ a - b & a - c & a^{2} \\ b - c & b - a & b^{2} \end{array}| = P, then the number of$
$positive divisors of P is$

Difficult

A

36
B

45
C

55
D

39

Solution

$|\begin{matrix} a & b & 1 \\ b & c & 1 \\ c & a & 1 \end{matrix}| = a b + b c + c a - a^{2} - b^{2} - c^{2} = 2020 \Rightarrow g i v e n d e t e r m i n e n t P = |\begin{matrix} c - a & c - b & a b - c^{2} \\ a - b & a - c & b c - a^{2} \\ b - c & b - a & c a - b^{2} \end{matrix}|$
$n o w R_{1} \to R_{1} + R_{2} + R_{3} \Rightarrow |\begin{matrix} 0 & 0 & 2020 \\ a - b & a - c & b c - a^{2} \\ b - c & b - a & c a - b^{2} \end{matrix}| = (2020) (a b + b c + c a - a^{2} - b^{2} - c^{2})$
$= {(2020)}^{2} =$ $2^{4} \times 5^{2} \times 101^{2} n u m b e r p o s i t i v e d i v i s o r s = (4 + 1) (2 + 1) (2 + 1) = 45$

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