Rouché–Capelli theorem

• We take the coefficient matrix and its range. $$A=\begin{pmatrix} 1 & 1 & -2 \\ 5 & 5 & -4 \\ 3 & 2 & -1 \end{pmatrix}$$ We calculate the range $$ |1|\neq0; \ \ \ \left| \begin{matrix} 1 & -2\\ 5 & -4 \end{matrix} \right|=6\neq0; \ \ \ \left| \begin{matrix} 1 & 1 & -2\\ 5 & 5 & -4 \\ 3 & 2 & -1 \end{matrix} \right|=6\neq0$$

so $r(A)=3$.

• We find the range of the augmented matrix. $$A'=\begin{pmatrix} 1 & 1 & -2 & 3 \\ 5 & 5 & -4 & 1 \\ 3 & 2 & -1 & 1 \end{pmatrix}$$ Until $3\times3$ we have different from zero $$\left| \begin{matrix} 1 & 1 & -2\\ 5 & 5 & -4 \\ 3 & 2 & -1 \end{matrix} \right|=6\neq0$$ and it's not possible $4\times4$ so $r(A')=3$.

• We apply the theorem of Rouché, we have $n=3$ (number of unknowns) and $r(A)=r(A')=3$, we are on the case:

$r=r'=n$, Determinate Compatible System.

• Finally we solve the compatible system. It can be done using the Gauss method or by using Cramer's rule.

$\Delta_1=\left| \begin{matrix} 3 & 1 & -2\\ 1 & 5 & -4 \\ 1 & 2 & -1 \end{matrix} \right|=12, \ \ $ $\Delta_2=\left| \begin{matrix} 1 & 3 & -2\\ 5 & 1 & -4 \\ 3 & 1 & -1 \end{matrix} \right|=-22, \ \ $ $\Delta_3=\left| \begin{matrix} 1 & 1 & 3\\ 5 & 5 & 1 \\ 3 & 2 & 1 \end{matrix} \right|=-14$ $$x=\dfrac{12}{6}=2; \ y=-\dfrac{22}{6}=-\dfrac{11}{3}; \ z=-\dfrac{14}{6}=-\dfrac{7}{3}$$

• We take the coefficient matrix and its range. $$A=\begin{pmatrix} 1 & 1 & -2 \\ 2 & 2 & -4 \\ 1 & 1 & -1 \end{pmatrix}$$ We calculate the range $$ |1|\neq0; \ \ \ \left| \begin{matrix} 1 & -2\\ 2 & -4 \end{matrix} \right|=-2\neq0; \ \ \ \left| \begin{matrix} 1 & 1 & -2\\ 2 & 2 & -4 \\ 1 & 1 & -1 \end{matrix} \right|=0$$

so $r(A)=2$.

• We find the range of the augmented matrix. $$A'=\begin{pmatrix} 1 & 1 & -2 & 4 \\ 2 & 2 & -4 & 15 \\ 1 & 1 & -1 & 1 \end{pmatrix}$$ We check order $3\times3$ because until $2\times2$ we have different from zero: $$\left| \begin{matrix} 1 & -2 & 4\\ 2 & -4 & 15 \\ 1 & -1 & 1 \end{matrix} \right|=-7\neq0$$ then $r(A')=3$.

• We apply the theorem of Rouché, we have $n=3$ (number of unknowns) and $r(A)=2$, $r(A')=3$, we are on the case:

$r\neq r'$, Incompatible System.

• We take the coefficient matrix and its range. $$A=\begin{pmatrix} 1 & -1 & 2 \\ 1 & 3 & -5 \\ 2 & -2 & 4 \end{pmatrix}$$ We calculate the range $$ |1|\neq0; \ \ \ \left| \begin{matrix} 1 & -1\\ 1 & 3 \end{matrix} \right|=4\neq0; \ \ \ \left| \begin{matrix} 1 & -1 & 2\\ 1 & 3 & -5 \\ 2 & -2 & 4 \end{matrix} \right|=0$$

so $r(A)=2$.

• We find the range of the augmented matrix. $$A'=\begin{pmatrix} 1 & -1 & 2 & 1 \\ 1 & 3 & -5 & 4 \\ 2 & -2 & 4 & 2 \end{pmatrix}$$ We check order $3\times3$ because until $2\times2$ we have different from zero: $$\left| \begin{matrix} 1 & -1 & 1\\ 1 & 3 & 4 \\ 2 & -2 & 2 \end{matrix} \right|=0; \ \ \left| \begin{matrix} 1 & 2 & 1\\ 1 & -5 & 4 \\ 2 & 4 & 2 \end{matrix} \right|=0; \left| \begin{matrix} -1 & 2 & 1\\ 3 & -5 & 4 \\ -2 & 4 & 2 \end{matrix} \right|=0; $$ then $r(A')=2$.

• We apply the theorem of Rouché, we have $n=3$ (number of unknowns) and $r(A)=r(A')=2$, we are on the case:

$r=r'\neq n$, Indeterminate Compatible System.

• Finally we solve the compatible system. It can be done by the method of Gauss. $$\begin{pmatrix} 1 & -1 & 2 & | & 1 \\ 1 & 3 & -5 & | & 4 \\ 2 & -2 & 4 & | & 2 \end{pmatrix} \rightarrow \left\{ \begin{array}{c} r1-r2 \rightarrow r2 \\ 2r1-r3\rightarrow r3 \end{array} \right. \rightarrow \begin{pmatrix} 1 & -1 & 2 & | & 1 \\ 0 & -4 & 7 & | & -3 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$$

$$z=z \\ -4y+7z=-3 \Rightarrow -4y=-7z-3 \Rightarrow y=\dfrac{7z+3}{4}$$

Replacing in the first equation: $$x-\dfrac{7z+3}{4}+2z=1 \Rightarrow x+\dfrac{-7z-3+8z}{4}=1 \Rightarrow $$ $$x=1-\dfrac{z-3}{4}=\dfrac{4}{4}-\dfrac{z-3}{4}=\dfrac{-z+7}{4}$$