coefficient of $x$ in a determinant

Question

What is the coefficient of $x$ in the expansion of the determinant$\begin{vmatrix} (1+x)^2 & (1+x)^4 & (1+x)^6 \\ (1+x)^3 & (1+x)^6 & (1+x)^9 \\ (1+x)^4 & (1+x)^8 & (1+x)^{12} \\ \end{vmatrix}$.

I simplified the determinant to be $x^3(1+x)^{16}(x+2)$,how will i find coefficient of $x$?Someone help me in getting answer?

Are you sure you got the correct answer? In case this is true, $x$ has a zero coefficient. — , Aug 02 '15 at 06:05
You multiply $x^3$ by another polynom. There is cannot be terms with $x$. — Michael Galuza, Aug 02 '15 at 06:14

score 1 · Accepted Answer · answered Aug 02 '15 at 06:07

1

$x^3(x+2)(x+1)^{16} = x^4(x+1)^{16}+2x^3(x+1)^{16}= a_1x^3+a_2x^4+\cdots + a_{20}x^{20}$. It appears that the term $bx$ does not exist, as suggested by @BolzWeir so we can take $b = 0$.

answered Aug 02 '15 at 06:07

DeepSea

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coefficient of $x$ in a determinant

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