Find Basis from polynomials I have this question from my course book.

And i have the following answer below?

How do i continue to solve this question i don't understand?
I have tried with Gaussian solver but it shows that there is no answer?
Below:
How should i solve it i have tried with Gaussian and stuff like but i'don't understand??

 A: The vectors $p_{1},p_{2},p_{3}$ are linearly independent. So the only solution that you would find for $ap_{1}+bp_{2}+cp_{3}=0$ is $(a,b,c)=(0,0,0)$. That is what the book or whatever you are reading is trying to prove. This guarantees a unique solution to $ap_{1}+bp_{2}+cp_{3}=p$
What you instead need to do is solve the system :- $AX=B$
$A$ is :-
\begin{bmatrix}
1 & 2 & 3 \\
2 & 9 & 3   \\
1 & 0 & 4  
\end{bmatrix}
and B is  \begin{bmatrix}
2 \\
17\\
-3 
\end{bmatrix}.
Or you need to apply Gaussian Elimination or Gauss-Jordan Method or Cramer's rule whatever method you like to the augmented matrix:-
\begin{bmatrix}
1 & 2 & 3&2 \\
2 & 9 & 3&17   \\
1 & 0 & 4&-3  
\end{bmatrix}
Now you you will get the coefficients $a,b,c$ . You can solve it by hand or use some website or whatever you like. You will get this as the answer:-
$a = 1$
$b= 2$
$c = -1$ seems to work .
So $p=p_{1}+2p_{2}-p_{3}$
A: By definition if we need to prove that  $$\beta:=\{p_{1},p_{2},p_{3}\}=\{1+2x+x^{2},2+9x,3+3x+4x^{2}\}\subseteq P_{2}(\mathbf{R})$$
is a basis for the vector space $P_{2}(\mathbf{R})$ so we need to show that

*

*$\beta$ is linearly independent.

*${\rm span}(\beta)=P(\mathbf{R})$.

Claim 1: $\beta$ is a linearly independent set.
Let $\alpha_{1},\alpha_{2}$ and $\alpha_{3}\in \mathbf{R}$ and
$$0+0x+0x^2=\alpha_{1}p_{1}+\alpha_{2}p_{2}+\alpha_{3}p_{3}$$
then we have that
$$0+0x+0x^{2}=\alpha_{1}(1+2x+x^{2})+\alpha_{2}(2+9x)+\alpha_{3}(3+3x+4x^{2})$$
$$0+0x+0x^{2}=(\alpha_{1}+2\alpha_{2}+3\alpha_{3})+(2\alpha_{1}+9\alpha_{2}+3\alpha_{3})x+
(\alpha_{1}+0\alpha_{2}+4\alpha_{3})x^{2}$$
The linear system is given by
$$\begin{cases}\alpha_{1}+2\alpha_{2}+3\alpha_{3}=0\\ 2\alpha_{1}+9\alpha_{2}+3\alpha_{3}=0\\ \alpha_{1}+0\alpha_{2}+4\alpha_{3}=0 \end{cases}$$
and the solution for the linear system using Gauss's elimination is $$\alpha_{1}=0, \quad \alpha_{2}=0 \quad \text{and} \quad \alpha_{3}=0.$$
Hence $\beta$ is linearly independent set in $P_{2}(\mathbf{R})$.
Claim 2: ${\rm span}(\beta)=P_{2}(\mathbf{R})$.
Let $a+bx+cx^{2}\in P_{2}(\mathbf{R})$ and $$a+bx+cx^{2}=\alpha_{1}p_{1}+\alpha_{2}p_{2}+\alpha_{3}p_{3}$$
then we have the linear system
$$\begin{cases}\alpha_{1}+2\alpha_{2}+3\alpha_{3}=a\\ 2\alpha_{1}+9\alpha_{2}+3\alpha_{3}=b\\ \alpha_{1}+0\alpha_{2}+4\alpha_{3}=c\end{cases}$$
and using the Gauss's elimination we have that
$$\cdots \sim \begin{pmatrix} 1 & 2 & 3 & | & a \\ 0 & 5 & -3 & | & -2a+b \\ 0 & 0 & -\frac{1}{5} & | & \frac{-9a+2b+5c}{5} \end{pmatrix}$$
Hence the linear system is consistent for all $a,b$ and $c\in \mathbf{R}$ and we can conclude that ${\rm span}(\beta)=P_{2}(\mathbf{R})$.
By claim 1 and claim 2 we have that $\beta$ is a basis for the vector space $P_{2}(\mathbf{R})$.
Finally, we can write $$2+17x-3x^{2}=\alpha_{1}p_{1}+\alpha_{2}p_{2}+\alpha_{3}p_{3}$$for some $\alpha_{1},\alpha_{2}$ and $\alpha_{3}$ in $\mathbf{R}$. Indeed, we have
$$\alpha_{1}=1, \quad \alpha_{2}=2 \quad \text{and} \quad \alpha_{3}=-1.$$
because $$2+17-3x^{2}=1(1+2x+x^{2})+2(2+9x)+(-1)(3+3x+4x^{2}).$$
