# 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*}

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

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 $\vec b$.

• yes, I see. So without writing it all out, the proof for the secodn equation is near identical to the first. – Daniel Jul 15 '15 at 15:13
• Yes. Do you see how to formulate the statement in the $n\times n$ case? – Brian Fitzpatrick Jul 15 '15 at 15:14
• I think my posted answer is correct? – Daniel Jul 15 '15 at 15:54
• @Jack looks good. I'll add a bit to my answer so you can see how I would phrase it. – Brian Fitzpatrick Jul 15 '15 at 15:55
• Thank you for the help appreciate it :) – Daniel Jul 15 '15 at 16:25

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 $b = (b1, . . . bm)^T$ .