Formulate Cramer's rule for solving systems of linear equations, stating conditions under which the rule is applicable. Formulate Cramer's rule for solving systems of linear equations, stating conditions under which the rule is applicable. Prove Cramer's rule for systems of two equations with two unknowns.
So I just wanted to double check my logic for this question, would something along these lines be correct:
Rule - Only applicable for square matrices that have a non zero determinant.
$$a_1x + b_1y = c_1$$
$$a_2x + b_2y = c_2$$
$$x = \frac{D_1}D,y = \frac {D_2}D$$
Where
\begin{align*} 
D& = \begin{vmatrix}a_1&b_1\\a_2&b_2\\ \end{vmatrix} &
D_1 &= \begin{vmatrix}c_1&b_1\\c_2&b_2\\ \end{vmatrix} &
D_2 &= \begin{vmatrix}a_1&c_1\\a_2&c_2\\ \end{vmatrix}
\end{align*}
 A: This is the correct formulation for the $2\times 2$ case but you still need to prove that these formulas work. So, to finish you must show that
\begin{array}{crcrcrcrcr}
a_1\dfrac{D_1}{D} &+& b_1\dfrac{D_2}{D} &=& c_1 \\
a_2\dfrac{D_1}{D} &+& b_2\dfrac{D_2}{D} &=& c_2
\end{array}
The first of these two equations is proved by
\begin{align*}
a_1\dfrac{D_1}{D} + b_1\dfrac{D_2}{D}
&= a_1\frac{\begin{vmatrix}c_1&b_1\\ c_2&b_2\end{vmatrix}}{\begin{vmatrix}a_1&b_1\\ a_2&b_2\end{vmatrix}}+b_1\frac{\begin{vmatrix}a_1&c_1\\ a_2&c_2\end{vmatrix}}{\begin{vmatrix}a_1&b_1\\ a_2&b_2\end{vmatrix}} \\
&= a_1\frac{c_1b_2-b_1c_2}{a_1b_2-b_1a_2}+b_1\frac{a_1c_2-c_1a_2}{a_1b_2-b_1a_2} \\
&= \frac{a_1c_1b_2-a_1b_1c_2+b_1a_1c_2-b_1c_1a_2}{a_1b_2-b_1a_2} \\
&= \frac{a_1c_1b_2-b_1c_1a_2}{a_1b_2-b_1a_2} \\
&= c_1\frac{a_1b_2-b_1a_2}{a_1b_2-b_1a_2} \\
&= c_1
\end{align*}
Can you prove the second equation?
Cramer's rule in the $n\times n$ case may be stated as follows.
The system $A\vec x=\vec b$ is solved by 
$$
x_j=\frac{\det A_j}{\det A}
$$
where $A_j$ is the $n\times n$ matrix obtained by replacing the $j$th column of $A$ with $\vec b$.
A: So here my answer for $nxn$ case:
$$a_{11}X_1 + a_{12}X_2 + · · · + a_{1n}X_n = b_1$$
$$.........$$
$$a_{n1}X_1 + a_{n2}X_2 + · · · + a_{nn}X_n = b_n$$
Thist can be represented in a form $AX = b$, where
$$A = ∥a_{ij}∥, X = (X_1, . . . X_n)^T$$
$$b = (b_1, . . . b_m)^T$$
For each $i = 1, 2, . . . , n$ denoted by $A_{x_{j}}$
the matrix obtained from $A$ by replacing
the column $j$, i.e., $(a_{1j} , a_{2j} , . . . , a_{nj} )^T$ by $b = (b1, . . . bm)^T$
.
