Cramer's Rule.

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

$$\{\begin{array}{l}{a}_{11}{x}_1+a_{12}{x}_2+...+a_{1n}{x}_n=b_1\\ a_{21}{x}_1+a_{22}{x}_2+...+a_{2n}{x}_n=b_2\\ ..............................\\ a_{n1}{x}_1+a_{n2}{x}_2+...+a_{nn}{x}_n=b_n\end{array},(1)$$ or, in matrix form, $AX=B,$ where

$A=\left(\begin{array}{cccc}{a}_{11}& a_{12}& ...& a_{1n}\\ a_{21}& a_{22}& ...& a_{2n}\\ ...& ...& ...& ...\\ a_{n1}& a_{n2}& ...& a_{nn}\end{array}\right);$ $X=\left(\begin{array}{c}{x}_1\\ x_2\\ ...\\ x_n\end{array}\right);$ $B=\left(\begin{array}{c}{b}_1\\ b_2\\ ...\\ b_n\end{array}\right).$

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

Examples:

Solve the following systems using Cramer's Rule:

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

The matrix $A=\left(\begin{array}{cc}3& -5\\ 2& 7\end{array}\right)$ is non-singular because

$detA=\left|\begin{array}{cc}3& -5\\ 2& 7\end{array}\right|=21+10=31\ne 0.$ Therefore, the system has a unique solution.

$\mathrm{\Delta }=detA=21;$

${\mathrm{\Delta }}_1=\left|\begin{array}{cc}13& -5\\ 81& 7\end{array}\right|=91+405=496;$

${\mathrm{\Delta }}_2=\left|\begin{array}{cc}3& 13\\ 2& 81\end{array}\right|=243-26=217.$

From here, we obtain $x=\frac{{\mathrm{\Delta }}_1}{\mathrm{\Delta }}=\frac{496}{31}=16;$ $y=\frac{{\mathrm{\Delta }}_2}{\mathrm{\Delta }}=\frac{217}{31}=7.$

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

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

Solution.

The matrix $A=\left(\begin{array}{ccc}7& 2& 3\\ 5& -3& 2\\ 10& -11& 5\end{array}\right)$ is non-singular because

$detA=\left|\begin{array}{ccc}7& 2& 3\\ 5& -3& 2\\ 10& -11& 5\end{array}\right|=-105-165+40+90+154-50=-36\ne 0.$ Thus, the system has a unique solution.

$\mathrm{\Delta }=detA=-36;$

${\mathrm{\Delta }}_1=\left|\begin{array}{ccc}15& 2& 3\\ 15& -3& 2\\ 36& -11& 5\end{array}\right|=-225-495+144+324+330-150=-72.$

${\mathrm{\Delta }}_2=\left|\begin{array}{ccc}7& 15& 3\\ 5& 15& 2\\ 10& 36& 5\end{array}\right|=525+540+300-450-504-375=36\ne 0.$

${\mathrm{\Delta }}_3=\left|\begin{array}{ccc}7& 2& 15\\ 5& -3& 15\\ 10& -11& 36\end{array}\right|=-756-825+300+450+1155-360=-36.$

Hence, we obtain: $x=\frac{{\mathrm{\Delta }}_1}{\mathrm{\Delta }}=\frac{-72}{-36}=2;$ $y=\frac{{\mathrm{\Delta }}_2}{\mathrm{\Delta }}=\frac{36}{-36}=-1;$ $z=\frac{{\mathrm{\Delta }}_3}{\mathrm{\Delta }}=\frac{-36}{-36}=1.$

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

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

Solution.

The matrix $A=\left(\begin{array}{ccc}1& 1& -2\\ 2& 3& -7\\ 5& 2& 1\end{array}\right)$ is non-degenerate, since

$detA=\left|\begin{array}{ccc}1& 1& -2\\ 2& 3& -7\\ 5& 2& 1\end{array}\right|=3-8-35+30+14-2=2\ne 0.$ Therefore, the system has a unique solution.

$\mathrm{\Delta }=detA=2;$

${\mathrm{\Delta }}_1=\left|\begin{array}{ccc}6& 1& -2\\ 16& 3& -7\\ 16& 2& 1\end{array}\right|=18-64-112+96+84-16=6;$

${\mathrm{\Delta }}_2=\left|\begin{array}{ccc}\mathrm{l}& 6& -2\\ 2& 16& -7\\ 5& 16& 1\end{array}\right|=16-64-210+160+112-12=2;$

${\mathrm{\Delta }}_3=\left|\begin{array}{ccc}1& 1& 6\\ 2& 3& 16\\ 5& 2& 16\end{array}\right|=48+24+80-90-32-32=-2;$

Hence, we obtain: $x=\frac{{\mathrm{\Delta }}_1}{\mathrm{\Delta }}=\frac{6}{2}=3;$ $y=\frac{{\mathrm{\Delta }}_2}{\mathrm{\Delta }}=\frac{2}{2}=1;$ $z=\frac{{\mathrm{\Delta }}_3}{\mathrm{\Delta }}=\frac{-2}{2}=-1.$

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

Homework:

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

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

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

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

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

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

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

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