Why is this set of polynomials linearly dependent?

Hint

You represent $1 + 2t + t^2$ as a column vector \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} and you represent $3 -9t^2$ as a column vector \begin{bmatrix} 3 \\ 0 \\ -9 \end{bmatrix} and you represent $1 + 4t + 5t^2$ as a column vector \begin{bmatrix} 1 \\ 4 \\ 5 \end{bmatrix}

Now you have a matrix

\begin{bmatrix} 1 & 3 & 1 \\ 2 & 0 &4 \\1 & -9 & 5 \end{bmatrix}

Now I believe You got it from here

$$6\cdot (1+2t+t^2) - 3\cdot (1 + 4t + 5t^2) = 3 - 9t^2.$$ We can write one of the polynomials as a linear combination of the two other polynomials and therefore they are linearly dependent.

An alternative way to solve this problem is by using the Wronskian. Put \begin{align*} f(t) &= 1 + 2\,t+ t^2 & g(t) &= 3-9\,t^2 & h(t) &= 1 + 4\,t + 5\,t^2 \end{align*} and define $$ W(t)= \begin{bmatrix} f(t) & g(t) & h(t) \\ f^\prime(t) &g^\prime(t) & h^\prime(t) \\ f^{\prime\prime}(t) & g^{\prime\prime}(t) & h^{\prime\prime}(t) \end{bmatrix} = \begin{bmatrix} 1+2\,t+t^2 & 3-9\,t^2 & 1+4\,t+5\,t^2 \\ 2+2\,t & -18\,t & 4+10\,t \\ 2 & -18 & 10 \end{bmatrix} $$ If there exists a $t_0$ such that $\det W(t_0)\neq 0$, then $\{f,g,h\}$ is linearly independent. Since each of $f$, $g$, and $h$ is analytic, if $\det W(t)=0$ for all $t\in\Bbb R$, then $\{f,g,h\}$ is linearly dependent.

Now, note that adding $\DeclareMathOperator{Col}{Col}9\cdot\Col_1$ to $\Col_2$ and subtracting $5\cdot\Col_1$ from $\Col_3$ gives $$ \det W(t) = \begin{vmatrix} 1+2\,t+t^2 &12+18\,t& -4-6\,t \\ 2+2\,t&18&-6 \\ 2&0&0 \end{vmatrix} $$ Then adding $3\cdot\Col_3$ to $\Col_2$ gives $$ \det W(t) = \begin{vmatrix} 1+2\,t+t^2 & 0 & -4-6\,t \\ 2+2\,t & 0 & -6 \\ 2 & 0 & 0 \end{vmatrix} = 0 $$ This implies that $\{f,g,h\}$ is linearly dependent.