Determinants can be tiresome

Algebra Level 3

If ∣ x 2 − 2 x + 3 7 x + 2 x + 4 2 x + 7 x 2 − x + 2 3 x 3 2 x − 1 x 2 − 4 x + 7 ∣ = a x 6 + b x 5 + c x 4 + d x 3 + e x 2 + f x + g \left | \begin{array}{ccc} x^{2}-2x+3 & 7x+2 & x+4 \\ 2x+7 & x^{2}-x+2& 3x \\ 3 & 2x-1 & x^{2}-4x+7 \\ \end{array} \right | =ax^{6}+bx^{5}+cx^{4}+dx^{3}+ex^{2}+fx+g then the value of g g is

None of these. -2 4 12 1 6

Rohit Udaiwal by Rohit Udaiwal , 20, India

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1 solution

Akhil Bansal
Nov 2, 2015

⇒ ∣ x 2 − 2 x + 3 7 x + 2 x + 4 2 x + 7 x 2 − x + 2 3 x 3 2 x − 1 x 2 − 4 x + 7 ∣ = a x 6 + b x 5 + c x 4 + d x 3 + e x 2 + f x + g \Rightarrow \left| \begin{array}{ccc} x^{2}-2x+3 & 7x+2 & x+4 \\ 2x+7 & x^{2}-x+2 & 3x \\ 3 & 2x-1 & x^{2}-4x+7 \\ \end{array} \right | =ax^{6}+bx^{5}+cx^{4}+dx^{3}+ex^{2}+fx+\color{#3D99F6}g

Above equation is an identity. Hence, it must be true for all real x,
Substituting x = 0 x = 0 .

⇒ ∣ 3 2 4 7 2 0 3 − 1 7 ∣ = g \Rightarrow \left| \begin{array}{ccc} 3 & 2 & 4 \\ 7 & 2 & 0 \\ 3 & -1 & 7 \\ \end{array} \right | = \color{#3D99F6}g

⇒ 3 ∗ ( 14 − 0 ) − 2 ∗ ( 49 − 0 ) + 4 ∗ ( − 7 − 6 ) = g \Rightarrow 3 * (14 - 0) - 2* ( 49 - 0) + 4 * (-7 - 6) = \color{#3D99F6}g ⇒ g = − 108 \Rightarrow \color{#3D99F6}g = -108

2 Helpful 0 Interesting 0 Brilliant 0 Confused

edit it, as value of determinant comes out to be -108 not -134 , but still answer is none of these :)

RAJ RAJPUT - 5 years, 7 months ago

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Thanks, bcos i solved the whole question in my mind, i made mistake in calculating value of determinant.

Akhil Bansal - 5 years, 7 months ago

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