Cramer's Rule.

Let's consider a system of $n$ linear equations with $n$ unknowns of the general form.

$$ \left\{\begin{array}{lcl}a_{11}x_1+a_{12}x_2+...+a_{1n}x_n=b_1\\a_{21}x_1+a_{22}x_2+...+a_{2n}x_n=b_2\\..............................\\a_{n1}x_1+a_{n2}x_2+...+a_{nn}x_n=b_n\end{array}\right. ,\quad\quad (1)$$ or, in matrix form, $AX=B,$ where

$A=\begin{pmatrix}a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\...&...&...&...\\a_{n1}&a_{n2}&...&a_{nn}\end{pmatrix};$ $X=\begin{pmatrix}x_1\\x_2\\...\\x_n\end{pmatrix};$ $B=\begin{pmatrix}b_1\\b_2\\...\\b_n\end{pmatrix}.$

Cramer's Rule: If $\det A = \Delta \neq 0$, meaning matrix $A$ has an inverse, then system (1) has a unique solution given by $x_i = \frac{\Delta_i}{\Delta}$, $i = 1, 2, \ldots, n$, where $\Delta_i$ is the determinant obtained from $\Delta$ by replacing the $i$-th column with the column of constants.

Examples:

Solve the following systems using Cramer's Rule:

1. $$\left\{\begin{array}{lcl}3x-5y=13\\2x+7y=81\end{array}\right.$$ Solution.

The matrix $A=\begin{pmatrix}3&-5\\2&7\end{pmatrix}$ is non-singular because

$\det A=\begin{vmatrix}3&-5\\2&7\end{vmatrix}=21+10=31\neq 0.$ Therefore, the system has a unique solution.

$\Delta=\det A=21;$

$\Delta_1=\begin{vmatrix} 13&-5\\81&7\end{vmatrix}=91+405=496;$

$\Delta_2=\begin{vmatrix} 3&13\\2&81\end{vmatrix}=243-26=217.$

From here, we obtain $x=\frac{\Delta_1}{\Delta}=\frac{496}{31}=16;$ $y=\frac{\Delta_2}{\Delta}=\frac{217}{31}=7.$

Answer: $x=16;$ $y=7.$

2. $$\left\{\begin{array}{lcl}7x+2y+3z=15\\ 5x-3y+2z=15\\10x-11y+5z=36\end{array}\right.$$

Solution.

The matrix $A=\begin{pmatrix}7&2&3\\5&-3&2\\10&-11&5\end{pmatrix}$ is non-singular because

$\det A=\begin{vmatrix}7&2&3\\5&-3&2\\10&-11&5\end{vmatrix}=-105-165+40+90+154-50=-36\neq 0.$ Thus, the system has a unique solution.

$\Delta=\det A=-36;$

$\Delta_1=\begin{vmatrix}15&2&3\\15&-3&2\\36&-11&5\end{vmatrix}=-225-495+144+324+330-150=-72.$

$\Delta_2=\begin{vmatrix}7&15&3\\5&15&2\\10&36&5\end{vmatrix}=525+540+300-450-504-375=36\neq 0.$

$\Delta_3=\begin{vmatrix}7&2&15\\5&-3&15\\10&-11&36\end{vmatrix}=-756-825+300+450+1155-360=-36.$

Hence, we obtain: $x=\frac{\Delta_1}{\Delta}=\frac{-72}{-36}=2;$ $y=\frac{\Delta_2}{\Delta}=\frac{36}{-36}=-1;$ $z=\frac{\Delta_3}{\Delta}=\frac{-36}{-36}=1.$

Answer:$x=2;$ $y=-1;$ $z=1.$

3. $$\left\{\begin{array}{lcl}x+y-2z=6\\ 2x+3y-7z=16\\5x+2y+z=16\end{array}\right.$$

Solution.

The matrix $A=\begin{pmatrix}1&1&-2\\2&3&-7\\5&2&1\end{pmatrix}$ is non-degenerate, since

$\det A=\begin{vmatrix}1&1&-2\\2&3&-7\\5&2&1\end{vmatrix}=3-8-35+30+14-2=2\neq 0.$ Therefore, the system has a unique solution.

$\Delta=\det A=2;$

$\Delta_1=\begin{vmatrix}6&1&-2\\16&3&-7\\16&2&1\end{vmatrix}=18-64-112+96+84-16=6;$

$\Delta_2=\begin{vmatrix}\mathrm{l}&6&-2\\2&16&-7\\5&16&1\end{vmatrix}=16-64-210+160+112-12=2;$

$\Delta_3=\begin{vmatrix}1&1&6\\2&3&16\\5&2&16\end{vmatrix}=48+24+80-90-32-32=-2;$

Hence, we obtain: $x=\frac{\Delta_1}{\Delta}=\frac{6}{2}=3;$ $y=\frac{\Delta_2}{\Delta}=\frac{2}{2}=1;$ $z=\frac{\Delta_3}{\Delta}=\frac{-2}{2}=-1.$

Answer: $x=3;$ $y=1;$ $z=-1.$

Homework:

Solve the following systems of equations using Cramer's rule.

1. $\left\{\begin{array}{lcl}3x-4y=-6\\3x+4y=18.\end{array}\right. $

Answer: $x=2; y=3.$

2. $\left\{\begin{array}{lcl}2x+y=5\\x+3z=16\\5y-z=10.\end{array}\right. $

Answer: $x=1; y=3; z=5.$

3. $\left\{\begin{array}{lcl}4x_1+4x_2+5x_3+5x_4=0\\ 2x_1+3x_3-x_4=10\\x_1+x_2-5x_3=-10\\3x_2+2x_3=1.\end{array}\right. $

Answer: $x_1=1; x_2=-1; x_3=2; x_4=-2.$

Tags: Cramer's Rule, linear algebra, linear equations, matrix