Reduced determinant to calculate coefficient of $x$

Question

Question:

Calculate coefficient of $x$ in $$f(x)=\det\begin{pmatrix} x & (1+\sin x)^3 & \cos x \\ 1 & \ln(1+x) & 2 \\ x^2 & 1+x^2 & 0 \end{pmatrix}$$

My attempt:

I know how to do such types of problems. I calculated $f'(x)$ by splitting the given determinant columnwise, and noted that the coefficient of $x$ in $f(x)$ is simply $f'(0)$. I also got the correct answer ($=-2$).

I am curious about the method given in my book. They say that:

Coefficient of $x$ in $f(x)$ is same as the coefficient of $x$ in $g(x)=\det\begin{pmatrix} x & 1 & 1 \\ 1 & x & 2 \\ x^2 & 1 & 0 \end{pmatrix}$

I do not understand how they arrived at the matrix in $g(x)$. I tried relating this to Taylor series, since that is probably the only way to convert $\ln$ or $\sin$ into a polynomial function of $x$, but it did not make sense since:

1. these taylor series only work for $x\to0$, while here we have $\text{Domain}_{f(x)}=\{x:x>-1\}$, and
2. they retained $x$ from $\ln(1+x)$, but dismissed the $3x$ from $(1+\sin x)^3$, though both are having the same degree $1$

I don't know of any other possible method to arrive at $g(x)$ from $f(x)$. Any help is appreciated!

Answer 1

Use small-O notation.

\begin{align} x^2 &= o(x) \\ \sin(x) &= x + o(x) \\ \ln(1+x) &= x + o(x) \\ \cos(x) &= 1 + o(x) \end{align}

Substitute them into $f(x)$.

$$\begin{aligned} f(x) &= \det\begin{pmatrix} x & (1+\sin x)^3 & \cos x \\ 1 & \ln(1+x) & 2 \\ x^2 & 1+x^2 & 0 \end{pmatrix} \\ &= \det\begin{pmatrix} x & (1+x+o(x))^3 & 1+o(x) \\ 1 & x+o(x) & 2 \\ o(x) & 1+o(x) & 0 \end{pmatrix} \\ &= \det\begin{pmatrix} x & 1+3x+o(x) & 1+o(x) \\ 1 & x+o(x) & 2 \\ o(x) & 1+o(x) & 0 \end{pmatrix} \end{aligned}$$

Delete the $o(x)$ from the above determinant to get $$h(x)=\det\begin{pmatrix} x & 1+3x & 1 \\ 1 & x & 2 \\ o(x) & 1 & 0 \end{pmatrix}.$$ Remarks: I wrote $o(x)$ instead of $x^2$ in the lower-left corner since it has no role in the calculation of coefficient of $x$.

You might notice a difference in the top entry $1$ in $g(x)$ and $1+3x$ in $h(x)$, but since the determinant is multilinear,

$$\begin{aligned} h(x) &= \begin{vmatrix} x & 1+3x & 1 \\ 1 & x & 2 \\ o(x) & 1 & 0 \end{vmatrix} \\ &= \begin{vmatrix} x & 1 & 1 \\ 1 & x & 2 \\ o(x) & 1 & 0 \end{vmatrix} + \begin{vmatrix} x & 3x & 1 \\ 1 & 0 & 2 \\ o(x) & 0 & 0 \end{vmatrix} \\ &=\begin{vmatrix} x & 1 & 1 \\ 1 & x & 2 \\ o(x) & 1 & 0 \end{vmatrix} +(-3x) \begin{vmatrix} 1 & 2 \\ o(x) & 0 \end{vmatrix} \\ &=\begin{vmatrix} x & 1 & 1 \\ 1 & x & 2 \\ o(x) & 1 & 0 \end{vmatrix} +(-3x)(-2o(x)) \\ &= g(x) + o(x). \end{aligned}$$

Answer 2

Your suspicion is valid. Consider $f(x)$ with $a_{31}=1$ instead of $x^2$. Then the coefficient will be $3$ and $g(x)$ with $a_{12}=1+3x$ will give $3$, however the book's suggested $g(x)$ will not.