Question

Solve the following system of linear equations: 3x1+6x2−9x3+6x4 = 6 −x1−2x2+8x3+3x4 = −17 2x1+4x2−3x3+7x4 = −4 If the system has no solution, demonstrate this by giving a row-echelon form of the augmented matrix for the system. You can resize a matrix (when appropriate) by clicking and dragging the bottom-right corner of the matrix.

Answer 1

$-x_1-2x_2+8x_3+3x_4=-17$

$2x_1+4x_2-3x_3+7x_4=-4$

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$\begin{pmatrix}3&6&-9&6\\ -1&-2&8&3\\ 2&4&-3&7\end{pmatrix}\begin{pmatrix}x_1\\ x_2\\ x_3\\ x_4\end{pmatrix}=\begin{pmatrix}6\\ -17\\ -4\end{pmatrix}$

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augmented matrix is

$\begin{pmatrix}3&6&-9&6&6\\ -1&-2&8&3&-17\\ 2&4&-3&7&-4\end{pmatrix}$

$R_2\:\leftarrow \:R_2+\frac{1}{3}\cdot \:R_1$

$=\begin{pmatrix}3&6&-9&6&6\\ 0&0&5&5&-15\\ 2&4&-3&7&-4\end{pmatrix}$

$\:R_3\:\leftarrow \:R_3-\frac{2}{3}\cdot \:R_1$

$=\begin{pmatrix}3&6&-9&6&6\\ 0&0&5&5&-15\\ 0&0&3&3&-8\end{pmatrix}$

$R_3\:\leftarrow \frac{1}{3}\cdot \:R_3$

$=\begin{pmatrix}3&6&-9&6&6\\ 0&0&5&5&-15\\ 0&0&1&1&-\frac{8}{3}\end{pmatrix}$

$R_2\:\leftarrow \:R_2-5\cdot \:R_3$

$=\begin{pmatrix}3&6&-9&6&6\\ 0&0&0&0&-\frac{5}{3}\\ 0&0&1&1&-\frac{8}{3}\end{pmatrix}$

$R_1\:\leftarrow \:R_1+9\cdot \:R_3$

$=\begin{pmatrix}3&6&0&15&-18\\ 0&0&0&0&-\frac{5}{3}\\ 0&0&1&1&-\frac{8}{3}\end{pmatrix}$

$R_2\:\leftarrow \:-\frac{3}{5}\cdot \:R_2$

$=\begin{pmatrix}3&6&0&15&-18\\ 0&0&0&0&1\\ 0&0&1&1&-\frac{8}{3}\end{pmatrix}$

$R_1\:\leftarrow \:R_1+18\cdot \:R_2$

$=\begin{pmatrix}3&6&0&15&0\\ 0&0&0&0&1\\ 0&0&1&1&-\frac{8}{3}\end{pmatrix}$

$R_1\:\leftarrow \frac{1}{3}\cdot \:R_1$

$=\begin{pmatrix}1&2&0&5&0\\ 0&0&0&0&1\\ 0&0&1&1&-\frac{8}{3}\end{pmatrix}$

from the second row we can write equation

$0x_1+0x_2+0x_3+0x_4=1$

which is not possible

hence system has no solution