Cramer's Rule

Cramer's rule applies when the coefficient matrix of $AX=B$ is square and non-singular; for three equations in three unknowns,

let $\Delta=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}\ne0$. Then

where $\Delta_k$ is $\Delta$ with column $k$ replaced by the constants $b_1,b_2,b_3$ (columns $\ne k$ unchanged). The same pattern gives $x=\Delta_1/\Delta,\ y=\Delta_2/\Delta$ for a two-equation, two-unknown system.

Derivation idea for $x_1$. $x_1\Delta$ equals $\Delta$ with its first column scaled by $x_1$; using the original equations, that scaled column $(a_{11}x_1,a_{21}x_1,a_{31}x_1)$ can be rewritten as $b_i-a_{i2}x_2-a_{i3}x_3$; splitting the determinant by that sum and discarding the two pieces that repeat columns 2 or 3 (hence vanish) leaves exactly $\Delta_1$ -- so $x_1\Delta=\Delta_1$, and dividing by $\Delta\ne0$ gives the rule. $x_2,x_3$ follow identically.

Worked illustration. Solve $x_1-x_2=3,\ 2x_1+3x_2+4x_3=17,\ x_2+2x_3=7$. $\Delta=\begin{vmatrix}1&-1&0\\2&3&4\\0&1&2\end{vmatrix}=6\ne0$; $\Delta_1=\begin{vmatrix}3&-1&0\\17&3&4\\7&1&2\end{vmatrix}=12$; $\Delta_2=\begin{vmatrix}1&3&0\\2&17&4\\0&7&2\end{vmatrix}=-6$; $\Delta_3=\begin{vmatrix}1&-1&3\\2&3&17\\0&1&7\end{vmatrix}=24$. So $x_1=12/6=2,\ x_2=-6/6=-1,\ x_3=24/6=4$.

Word problems. A path $y=ax^2+bx+c$ through three known points, or a scoring/rate/mixture problem, gives a $3\times3$ system in the unknown constants exactly as for matrix inversion; Cramer's rule is convenient here since each unknown is found independently, without computing a full inverse. …