# Derivatives using matrices good

$$\left|\begin{matrix} (1+x)^{a_1b_1} & (1+x)^{a_1b_2} & (1+x)^{a_1b_3} \\ (1+x)^{a_2b_1} & (1+x)^{a_2b_2} & (1+x)^{a_2b_3} \\ (1+x)^{a_3b_1} & (1+x)^{a_3b_2} & (1+x)^{a_3b_3} \\ \end{matrix}\right |$$ Find the coefficient of $x$ in the above determinant. Try using differentiability.

• Ok, so what have you done so far. Did you mean the above is the determinant of the given matrix? May 15, 2016 at 12:51

Denote the determinant by $f(x)$. Then the $x$ coefficient is given by $f'(0)$. Now the derivative of a determinant equals the sum of determinants of the matrices obtained by taking the derivative of each row separately. Hence
\begin{align} f'(0) &= \frac{\mathrm{d}}{\mathrm{d}x} \left| \begin{matrix} (1+x)^{a_1b_1} & (1+x)^{a_1b_2} & (1+x)^{a_1b_3} \\ (1+x)^{a_2b_1} & (1+x)^{a_2b_2} & (1+x)^{a_2b_3} \\ (1+x)^{a_3b_1} & (1+x)^{a_3b_2} & (1+x)^{a_3b_3} \\ \end{matrix} \right|_{x=0} \\ &= \left| \begin{matrix} a_1b_1 & a_1b_2 & a_1b_3 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right| + \left| \begin{matrix} 1 & 1 & 1 \\ a_2b_1 & a_2b_2 & a_2b_3 \\ 1 & 1 & 1 \\ \end{matrix} \right| + \left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ a_3b_1 & a_3b_2 & a_3b_3 \\ \end{matrix} \right| \\ &= 0 \end{align}