Theorem There is a conic through any five $\color{red}{\textit{pairwise distinct }}$ points in a real plane. $(1)$ It is unique if and only if no four of the points are colinear; $(2)$ It is nondegenerate if and only if no three of the points are collinear .
How to prove this theorem ,only applying some theories of linear algebra(including matrix)？
Let $\textbf{p}_{i}=(x_{i},y_{i}),i=1,2,3,4,5.$ be the five points in the plane .
Consider a homogeneous system $\textbf{Ax}=\textbf{0}$ whose coefficient matrix is
$\textbf{A}= \begin{pmatrix} x_{1}^{2}& x_{1}y_{1}& y_{1}^{2}& x_{1}& y_{1}& 1\\ x_{2}^{2}& x_{2}y_{2}& y_{2}^{2}& x_{2}& y_{2}& 1 \\ x_{3}^{2}& x_{3}y_{3}& y_{3}^{2}& x_{3}& y_{3}& 1 \\ x_{4}^{2}& x_{4}y_{4}& y_{4}^{2}& x_{4}& y_{4}& 1\\ x_{5}^{2}& x_{5}y_{5}& y_{5}^{2}& x_{5}& y_{5}& 1 \end{pmatrix}; $ $\qquad\qquad\textbf{x}$ = $\begin{pmatrix} A\\ B\\ C\\ D\\ E\\ F \end{pmatrix};$ $\qquad\qquad\textbf{0}$=$\begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 0 \end{pmatrix}.$
$\textbf{A}_{i}(i\in\{1,2,3,4,5,6\})$ is the $5\times 5$ square matrix obtained by deleting the $i^{th}$ column of $\textbf{A}.$ $\\$
Then the above theorem is equalvlity of the following statement from the algebraic point of view.
$(1){'} $ $rank(A)=5$ and at least one of $\textbf{A}_{1},\textbf{A}_{2},\textbf{A}_{3}$ is nonsingular. $\Longleftrightarrow $ $\forall\quad k_{1},k_{2},k_{3},k_{4}\in \{1,2,3,4,5\} $and $k_{1}<k_{2}<k_{3}<k_{4}$,$ rank(\begin{pmatrix} x_{k_{1}}& y_{k_{1}}& 1\\ x_{k_{2}}& y_{k_{2}}& 1 \\ x_{k_{3}}& y_{k_{3}}& 1\\ x_{k_{4}}& y_{k_{4}}& 1 \end{pmatrix})=3.$
$(2){'}$ A conic through $\textbf{p}_{i}=(x_{i},y_{i}),i=1,2,3,4,5$ is nondegenerate.$\Longleftrightarrow $ $ \forall \quad k_{1},k_{2},k_{3}\in \{1,2,3,4,5\} $and $k_{1}<k_{2}<k_{3}$,$ rank(\begin{pmatrix} x_{k_{1}}& y_{k_{1}}& 1\\ x_{k_{2}}& y_{k_{2}}& 1 \\ x_{k_{3}}& y_{k_{3}}& 1 \end{pmatrix})=3.$
Until now, how to prove $(1){'}$,$(2){'}$ hold,only using some theories of linear algebra(including matrix)？
Start with (2). Assume that there are three collinear points on the conic. Or in other words, there exists a line which intersects the conic in (at least) three positions. The line might be described as $\{p+\lambda v\mid\lambda\in\mathbb R\}$ (or perhaps field other than $\mathbb R$), so there are three values $\lambda$ for which the equation of the conic is satisfied. That equation however is only quadratic in $\lambda$. The only quadratic polynomial with more than two zeros is the null polynomial. So the whole line through the three points belongs to the conic. If you defined a degenerate conic as one containing a straight line, you are done. Otherwise you may need to link this to your definition of a degenerate conic.
Conversely, every degenerate conic consists of two (not necessarily distinct) lines. If you pick five points from these, you can't avoid at least three of them being collinear. If no three defining points are collinear, the resulting conic cannot be degenerate.
Update: The previous paragraph misses the case of a degenerate conic consisting of just a single point. That may be seen as a pair of lines over the complex numbers, but over the reals you have even more difficulty to even pick distinct points, let alone non-collinear ones. So the same argument applies there.
With this established, you can consider different cases for (1). For the time being better restrict the statement to pairwise distinct points, but feel free to revisit the non-distinct case afterwards, even though you know from your comment that the full conclusion won't hold there.
(a) Assume you had two different conics $\mathcal C_1$ and $\mathcal C_2$ passing through $\mathbf p_1$ through $\mathbf p_5$. Pick a point $\mathbf p_6$ arbitrarily on the line between $\mathbf p_1$ and $\mathbf p_2$. Show that there exists a linear combination of (the parameter vectors of) $\mathcal C_1$ and $\mathcal C_2$. That linear combination will pass through $\mathbf p_1$ through $\mathbf p_6$, which includes three collinear points. Rest left as an exercise.
(b) Picture three points on one line, two on another, and turn that into algebra.
(c) Picture four points on one line, and see how the second line through the remaining point can be chosen arbitrarily. Turn this into algebra.
I hope this gave you some pointers on how to tackle the problem. I tried to be verbose at the beginning to get you started, and more sketchy towards the end to give you room to find your own approach.