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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*}

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  • $\begingroup$ can you formulate your for a general system $Ax=b$ with $A$ a $n\times n$ matrix? $\endgroup$ – user190080 Jul 15 '15 at 13:21
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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 $\vec b$.

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  • $\begingroup$ yes, I see. So without writing it all out, the proof for the secodn equation is near identical to the first. $\endgroup$ – Daniel Jul 15 '15 at 15:13
  • $\begingroup$ Yes. Do you see how to formulate the statement in the $n\times n$ case? $\endgroup$ – Brian Fitzpatrick Jul 15 '15 at 15:14
  • $\begingroup$ I think my posted answer is correct? $\endgroup$ – Daniel Jul 15 '15 at 15:54
  • $\begingroup$ @Jack looks good. I'll add a bit to my answer so you can see how I would phrase it. $\endgroup$ – Brian Fitzpatrick Jul 15 '15 at 15:55
  • $\begingroup$ Thank you for the help appreciate it :) $\endgroup$ – Daniel Jul 15 '15 at 16:25
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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 $b = (b1, . . . bm)^T$ .

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