This is the first proof exercise of "linear algebra as an introduction to abstract mathematics". This is also my first proof (apart from proving things like the properties of integers and trigonometric identities). I hope someone can criticize it and let me know what I did well 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.

**Evidence**: We will prove by contrapositive. Presume $ x_1 = x_2 = $ 0 is not the only solution. Presume

$$ A = begin {bmatrix} a & b \ c & d end {bmatrix}, X = begin {bmatrix} x_1 \ x_2 end {bmatrix}, 0 = begin {bmatrix} 0 \ 0 end {bmatrix} $$

Then the system of equations (1) is equivalent to the expression $ AX = 0 $. For $ AX = 0 $ to be true we must have either $ A = 0 $ or $ X = 0 $. Since we assume $ x_1 = x_2 = $ 0 is not the only solution, we assume $ X neq0 $. Then for $ AX = 0 $ to be true we must have $ A = 0 $, and so, $ a = b = c = d = 0 $. Yes $ a = b = c = d = 0 $, then $ ab-cd = 0 $. $ blacksquare $