Problems from Rouché–Capelli theorem

Development:

1)

• We take the coefficient matrix and its range. $$A=\left(\begin{array}{ccc}1& 1& -2\\ 5& 5& -4\\ 3& 2& -1\end{array}\right)$$ We calculate the range $$|1|\ne 0;\left|\begin{array}{cc}1& -2\\ 5& -4\end{array}\right|=6\ne 0;\left|\begin{array}{ccc}1& 1& -2\\ 5& 5& -4\\ 3& 2& -1\end{array}\right|=6\ne 0$$

so $r(A)=3$.

• We find the range of the augmented matrix. $$A^{\prime }=\left(\begin{array}{cccc}1& 1& -2& 3\\ 5& 5& -4& 1\\ 3& 2& -1& 1\end{array}\right)$$ Until $3\times 3$ we have different from zero $$\left|\begin{array}{ccc}1& 1& -2\\ 5& 5& -4\\ 3& 2& -1\end{array}\right|=6\ne 0$$ and it's not possible $4\times 4$ so $r(A^{\prime })=3$.

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

$r=r^{\prime }=n$, Determinate Compatible System.

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

${\mathrm{\Delta }}_1=\left|\begin{array}{ccc}3& 1& -2\\ 1& 5& -4\\ 1& 2& -1\end{array}\right|=12,$ ${\mathrm{\Delta }}_2=\left|\begin{array}{ccc}1& 3& -2\\ 5& 1& -4\\ 3& 1& -1\end{array}\right|=-22,$ ${\mathrm{\Delta }}_3=\left|\begin{array}{ccc}1& 1& 3\\ 5& 5& 1\\ 3& 2& 1\end{array}\right|=-14$ $$x={\displaystyle \frac{12}{6}}=2;y=-{\displaystyle \frac{22}{6}}=-{\displaystyle \frac{11}{3}};z=-{\displaystyle \frac{14}{6}}=-{\displaystyle \frac{7}{3}}$$

2)

• We take the coefficient matrix and its range. $$A=\left(\begin{array}{ccc}1& 1& -2\\ 2& 2& -4\\ 1& 1& -1\end{array}\right)$$ We calculate the range $$|1|\ne 0;\left|\begin{array}{cc}1& -2\\ 2& -4\end{array}\right|=-2\ne 0;\left|\begin{array}{ccc}1& 1& -2\\ 2& 2& -4\\ 1& 1& -1\end{array}\right|=0$$

so $r(A)=2$.

• We find the range of the augmented matrix. $$A^{\prime }=\left(\begin{array}{cccc}1& 1& -2& 4\\ 2& 2& -4& 15\\ 1& 1& -1& 1\end{array}\right)$$ We check order $3\times 3$ because until $2\times 2$ we have different from zero: $$\left|\begin{array}{ccc}1& -2& 4\\ 2& -4& 15\\ 1& -1& 1\end{array}\right|=-7\ne 0$$ then $r(A^{\prime })=3$.

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

$r\ne r^{\prime }$, Incompatible System.

3)

• We take the coefficient matrix and its range. $$A=\left(\begin{array}{ccc}1& -1& 2\\ 1& 3& -5\\ 2& -2& 4\end{array}\right)$$ We calculate the range $$|1|\ne 0;\left|\begin{array}{cc}1& -1\\ 1& 3\end{array}\right|=4\ne 0;\left|\begin{array}{ccc}1& -1& 2\\ 1& 3& -5\\ 2& -2& 4\end{array}\right|=0$$

so $r(A)=2$.

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

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

$r=r^{\prime }\ne n$, Indeterminate Compatible System.

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

$$\begin{gathered} z=z \\ -4y+7z=-3\Rightarrow -4y=-7z-3\Rightarrow y={\displaystyle \frac{7z+3}{4}} \end{gathered}$$

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

Solution:

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

2) The system is Incompatible.

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

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