Matrices & Determinants Question - None

Question

Let $xC_i$, $x^2C_i$ and $x^3C_i$ ($i \in \{1, 2, 3\}$) be Binomial coefficients, where $x \in \mathbb{N}$

and $$f(x) = \begin{vmatrix} xC_1 & xC_2 & xC_3 \ x^2C_1 & x^2C_2 & x^2C_3 \ x^3C_1 & x^3C_2 & x^3C_3 \end{vmatrix}$$

Then $f(x)$ is a polynomial of degree:

Accepted Answer

Solution:

Taking $x$, $x^2$ and $x^3$ common from $R_1$, $R_2$ and $R_3$ respectively:

$$f(x) = x^4 x^2 x^3 \begin{vmatrix} 1 & (x-1) & (x-1)(x-2) \ 1 & (x^2-1) & (x^2-1)(x^2-2) \ 1 & (x^3-1) & (x^3-1)(x^3-2) \end{vmatrix}$$

Applying $R_1' o R_1 - R_2$ and $R_3' o R_3 - R_2$:

$$f(x) = x^6(x-1)^2 \begin{vmatrix} 0 & -x & x(3-x-x^2) \ 1 & x-1 & (x-1)(x^2-2) \ 0 & x^2 & x^2(x^2-x^3+3) \end{vmatrix}$$

Expanding along $C_1$:

$$f(x) = x^9(x-1)^2(-x^2-x^3+3-3-x-x^2)$$

$$= x^{10}(x-1)^2(x^2-1) = x^{10}(x-1)^3(x+1)$$

Therefore, $f(x)$ is a polynomial of degree $10 + 3 + 1 = 14$.

Question: