I have been for the entirety of this and you're right I should have posted my work. A lot could have been gleaned from it. I also just seem to not understand forums and am reluctant to use latex, but I am improving.
Here is the question written in latex
Consider the [imath](2\times2)[/imath] system
[math]a_{11}x_{1}+a_{12}x_{2}=b_{1}\\a_{21}x_{1}+a_{22}x_{2}=b_{2}[/math]Show that if [imath]a_{11}a_{22}-a_{12}a_{21}\neq0[/imath], then this system
is equivalent to a system of the form
[math]c_{11}x_1+c_{12}x_2=d_1[/math][math]c_{22}x_2=d_2,[/math]where [imath]c_{11}\neq0[/imath] and [imath]c_{22}\neq0.[/imath] Note that the second
system always has a solution. [Hint: First suppose that [imath]a_{11}\neq0[/imath], and then consider the special case in which [imath]a_{11}=0[/imath].]
Here is my work.
1.) I create an augmented matrix of the first system
[math]\begin{bmatrix} a_{11} & a_{12} & b_{1}\\ a_{21} & a_{22} & b_{2}\\ \end{bmatrix}[/math]
2. ) I have been learning how to do Gauss-Jordan Elimination, so I assume that is what I am to do and do as such.
I assume [imath]a_{11}[/imath]is not zero, and I want to get [imath]a_{11}[/imath] via Gauss-Jordan method, so I use the following elementary row operations to do such.
[math]\frac{R_{1}}{a_{11}} \longrightarrow R_{1}[/math]
Which gives us
[math]\begin{bmatrix} 1 & \frac{a_{12}}{a_{11}} & \frac{b_1}{a_{11}}\\ a_{21} & a_{22} & b_2 \end{bmatrix}[/math]
2.) Now that we have 1 for the [imath]a_{11}[/imath] entry I want to put zero in the rest of its column and I use the elementary row operations to do such.
[math]R_{2} - a_{11}*R_{1}\longrightarrow R_{2}[/math]
Which gives us
[math]\begin{bmatrix} 1 & \frac{a_{12}}{a_{11}} & \frac{b_1}{a_{11}}\\ 0 & a_{22}-\frac{a_{12}a_{21}}{a_{11}} & b_2 - \frac{b_1a_{21}}{a_{11}} \end{bmatrix}[/math]
3.) I recognize some familiar algebra in the [imath]a_{22}[/imath] entry [math]a_{22}-\frac{a_{12}a_{21}}{a_{11}}[/math] and with algebraic manipulation rearrange it to this [math]\frac{a_{11}a_{22}-a_{12}a_{21}}{a_{11}}[/math]
Which the problem tells us to assume that [math]a_{11}a_{22}-a_{12}a_{21} \not= 0[/math]
4.) Okay, so we finally got to a point in the problem where the prompt seems familiar enough to lets us know we are on the right track. That we have our first system of linear equation modified enough that we can start making an analysis of row equivalence within the constraints of the prompt.
5.) I create an augmented matrix for the second set of linear equations
[math]\begin{bmatrix} c_{11} & c_{12} & d_{1} \\ c_{21} & c_{22} & d_{2} \\ \end{bmatrix}[/math]
6.) I remember that I am told to assume that [imath]c_{11}\neq0[/imath] and [imath]c_{22}\neq0.[/imath], I and remember that [imath]a_{11}a_{22}-a_{12}a_{21} \not= 0[/imath], and we made our own assumption via hint that [imath]a_{11} \not=[/imath]. Meaning that all of these do not equal zero. So with all of that in mind and the fact we were able to use elementary row operations, we have hopefully shown that both systems are row equivalent when making these assumptions.
Edit: I forgot to convert a row back in this where it would return it to the original augmented form. I'm aware I didn't but it doesn't change the rest of the steps.