# Problems from Rouché–Capelli theorem

Classify the following systems as the number of solutions that they have:

1) $$\left\{ \begin{array} {rcl} x+y-2z & = & 3 \\ 5x+5y-4z &=& 1 \\ 3x+2y-z &=& 1 \end{array}\right.$$

2) $$\left\{ \begin{array} {rcl} x+y-2z & = & 4 \\ 2x+2y-4z &=& 15 \\ x+y-z &=& 1 \end{array}\right.$$

3) $$\left\{ \begin{array} {rcl} x-y+2z & = & 1 \\ x+3y-5z &=& 4 \\ 2x-2y+4z &=& 2 \end{array}\right.$$

See development and solution

### Development:

1)

• 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, \ \$$ $$\Delta_3=\left| \begin{matrix} 1 & 1 & 3\\ 5 & 5 & 1 \\ 3 & 2 & 1 \end{matrix} \right|=-14$$ $$x=\dfrac{12}{6}=2; \ y=-\dfrac{22}{6}=-\dfrac{11}{3}; \ z=-\dfrac{14}{6}=-\dfrac{7}{3}$$$2) • 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. 3) • 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: $$x-\dfrac{7z+3}{4}+2z=1 \Rightarrow x+\dfrac{-7z-3+8z}{4}=1 \Rightarrow$$$ $$x=1-\dfrac{z-3}{4}=\dfrac{4}{4}-\dfrac{z-3}{4}=\dfrac{-z+7}{4}$$\$

### Solution:

1) The system is Determined Compatible and solutions are: $$x=2; \ y=-\dfrac{11}{3}; \ z=-\dfrac{7}{3}$$.

2) The system is Incompatible.

3) The system is Indeterminate Compatible and solutions are: $$x=\dfrac{-z+7}{4}; \ y=\dfrac{7z+3}{4}; \ z=z$$.

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