Question

Solve each of the following systems by Gauss-Jordan elimination. (b) X1-2x2+ x3- 4x4=1 X1+3x2 + 7x3 + 2x4=2 -12x2-11x3- 16x4 5 (a) 5x1+2x2 +6x3= 0 -2x1 +x2+3x3 = 0

Answer 1

Answer:

a) The set of solutions is $\{(0,-3x_3,x_3): x_3\; \text{es un real}\}$ y b) the set of solutions is $\{(-6,\frac{-41}{17}-\frac{30}{17}x_4 , \frac{37}{17}+\frac{8}{17} x_4 ,x_4): x_4\;\text{es un real}\}$.

Step-by-step explanation:

a) Let's first find the echelon form of the matrix $\left[\begin{array}{ccc}5&2&6\\-2&1&3\end{array}\right]$.

• We add $\frac{2}{5}$ from row 1 to row 2 and we obtain the matrix $\left[\begin{array}{ccc}5&2&6\\0&\frac{9}{5} &\frac{27}{5}\end{array}\right]$

• From the previous matrix, we multiply row 1 by $\frac{1}{5}$ and the row 2 by $\frac{5}{9}$ and we obtain the matrix $\left[\begin{array}{ccc}1&\frac{2}{5} &\frac{6}{5} \\0&1&3\end{array}\right]$. This matrix is the echelon form of the initial matrix.

The system has a free variable (x3).

• x2+3x3=0, then x2=-3x3

• 0=x1+$\frac{2}{5}$x2+$\frac{6}{5}$x3=

       x1+$\frac{2}{5}$(-3x3)+$\frac{6}{5}$x3=

      x1-$\frac{6}{5}$x3+$\frac{6}{5}$x3

     then x1=0.

The system has infinite solutions of the form (x1,x2,x3)=(0,-3x3,x3), where x3 is a real number.

b) Let's first find the echelon form of the aumented matrix $\left[\begin{array}{ccccc}1&-2&1&-4&1\\1&3&7&2&2\\0&-12&-11&-16&5\end{array}\right]$.

• To row 2 we subtract row 1 and we obtain the matrix $\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&-12&-11&-16&5\end{array}\right]$

• From the previous matrix, we add to row 3, $\frac{12}{5}$ of row 2 and we obtain the matrix $\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&0&\frac{17}{5}&\frac{-8}{5}&\frac{37}{5} \end{array}\right]$.

• From the previous matrix, we multiply row 2 by $\frac{1}{5}$ and the row 3 by $\frac{5}{17}$ and we obtain the matrix $\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&1&\frac{6}{5} &\frac{6}{5}&\frac{1}{5}\\0&0&1&\frac{-8}{17}&\frac{37}{17} \end{array}\right]$. This matrix is the echelon form of the initial matrix.

The system has a free variable (x4).

• x3-$\frac{8}{17}$x4=$\frac{37}{17}$, then x3=$\frac{37}{17}$+ $\frac{8}{17}x4.x2+[tex]\frac{6}{5}$x3+$\frac{6}{5}$x4=$\frac{1}{5}$, x2+$\frac{6}{5}$($\frac{37}{17}$+$\frac{8}{17}x4)+[tex]\frac{6}{5}$x4=$\frac{1}{5}$, then

      x2=$\frac{-41}{17}-\frac{30}{17}$x4.

• x1-2x2+x3-4x4=1, x1+$\frac{82}{17}$+$\frac{60}{17}$x4+$\frac{37}{17}$+$\frac{8}{17}$x4-4x4=1, then x1=$1-\frac{119}{17}=-6$

The system has infinite solutions of the form (x1,x2,x3,x4)=(-6,$\frac{-41}{17}-\frac{30}{17}$x4,$\frac{37}{17}$+ $\frac{8}{17}$x4,x4), where x4 is a real number.