My first linear algebra proof - is it accurate and written correctly? This is the first proof exercise from "Linear Algebra as an Introduction to Abstract Mathematics." It also happens to be my first proof (outside of proving things like properties of integers and trigonometric identities). I'm hoping someone can critique it and let me know what I did right or what I did wrong.
Problem: Let a, b, c, and d be real numbers. Consider the system of equations (1):
$$ax_1 + bx_2 = 0$$
$$cx_1 + dx_2 = 0$$
Note that $x_1=x_2=0$ is a solution for any choice of a, b, c, and d. 
Prove that, if $ad-bc\neq0$, then $x_1=x_2=0$ is the only solution.
Proof: We will prove by contrapositive. Assume $x_1=x_2=0$ is not the only solution. Assume
$$A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}, X = \begin{bmatrix}x_1 \\ x_2\end{bmatrix}, 0 = \begin{bmatrix}0 \\ 0\end{bmatrix}$$ 
Then the system of equations (1) is equivalent to the expression $AX=0$ (2). For (2) to be true, we must have either $A=0$ or $X=0$. Since we are assuming $x_1=x_2=0$ is not the only solution, we assume $X\neq0$. Then, for (2) to be true, we must have $A=0$, and therefore, $a=b=c=d=0$. If $a=b=c=d=0$, then $ab-cd=0$. $\blacksquare$
 A: Claiming that $AX =0$ implies either $A =0$ or $X= 0$ does not always hold true. Take $A = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$ 
$X = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix} $
Then $AX = 0$ while neither $A$ nor $B$ is actually zero. 
The condition $ad-bc \neq 0$ means that the determinant of $A$ is not zero, which means that $A$ is invertible. This means that $AX = 0 \implies X = IX = A^{-1}AX = A^{-1}O = 0$. It follows that $X = 0$ is the only solution. 
A: Assume that $ad-bc\ne 0$ and that $(x_1,x_2)$ is a solution.
Then in particular
$$0 = d\cdot 0 - b \cdot 0 = d(ax_1+bx_2) - b(cx_1+dx_2) = (ad-bc)x_1 \implies x_1 = 0$$
$$0 = -c \cdot 0 + a \cdot 0 = -c(ax_1+bx_2) +a(cx_1+dx_2) = (ad-bc)x_2 \implies x_2 = 0$$
so $(x_1,x_2) = (0,0)$. Hence $(0,0)$ is the only solution.
A: You can apply the following row operations in order to get that
\begin{align*}
\begin{cases}
ax_{1} + bx_{2} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases} & \Longrightarrow
\begin{cases}
acx_{1} + bcx_{2} = 0\\\\
cx_{1} + dx_{2} = 0 
\end{cases}\\\\
& \Longrightarrow
\begin{cases}
(acx_{1} + bcx_{2}) - (acx_{1} + adx_{2}) = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}\\\\
& \Longrightarrow
\begin{cases}
(bc - ad)x_{2} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}
\end{align*}
Similarly, one has that
\begin{align*}
\begin{cases}
ax_{1} + bx_{2} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases} & \Longrightarrow
\begin{cases}
adx_{1} + bdx_{2} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}\\\\ & \Longrightarrow
\begin{cases}
(adx_{1} + bx_{2}) - (bcx_{1} + bdx_{2}) = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}\\\\ & \Longrightarrow
\begin{cases}
(ad-bc)x_{1} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}
\end{align*}
Since $ad - bc \neq 0$, we get that the solution set $S\subseteq\{(0,0)\}$. 
Once $x_{1} = x_{2} = 0$ is a solution, it is the unique solution.
