Question

PROBLEM 3 Find all solutions of the equation Ax = b, when 1 1-(424). --( 1 2 2 2 A, BER.

Answer 1

for simplicity use

a = a. B=b

system Ax=b is

= 0 71. NT

augmented matrix is

1 2 -1 -1 2 1 -1 a 2 2 1 b b)

R1 + R2

$=\begin{pmatrix}2&-1&a&2\\ 1&-1&2&1\\ -1&2&1&b\end{pmatrix}$

$R_2\:\leftarrow \:R_2-\frac{1}{2}\cdot \:R_1$

$R_3\:\leftarrow \:R_3+\frac{1}{2}\cdot \:R_1$

$=\begin{pmatrix}2&-1&a&2\\ 0&-\frac{1}{2}&\frac{4-a}{2}&0\\ 0&\frac{3}{2}&\frac{2+a}{2}&b+1\end{pmatrix}$

$R_2\:\leftrightarrow \:R_3$

$=\begin{pmatrix}2&-1&a&2\\ 0&\frac{3}{2}&\frac{2+a}{2}&b+1\\ 0&-\frac{1}{2}&\frac{4-a}{2}&0\end{pmatrix}$

$R_3\:\leftarrow \:R_3+\frac{1}{3}\cdot \:R_2$

$=\begin{pmatrix}2&-1&a&2\\ 0&\frac{3}{2}&\frac{2+a}{2}&b+1\\ 0&0&\frac{-a+7}{3}&\frac{b+1}{3}\end{pmatrix}$

$R_3\:\leftarrow \frac{3}{-a+7}\cdot \:R_3$

$=\begin{pmatrix}2&-1&a&2\\ 0&\frac{3}{2}&\frac{2+a}{2}&b+1\\ 0&0&1&\frac{b+1}{-a+7}\end{pmatrix}$

$R_2\:\leftarrow \:R_2-\frac{2+a}{2}\cdot \:R_3$

$R_1\:\leftarrow \:R_1-a\cdot \:R_3$

$=\begin{pmatrix}2&-1&0&\frac{-ab-3a+14}{-a+7}\\ 0&\frac{3}{2}&0&\frac{-3ab-3a+12b+12}{2\left(-a+7\right)}\\ 0&0&1&\frac{b+1}{-a+7}\end{pmatrix}$

$\:R_2\:\leftarrow \frac{2}{3}\cdot \:R_2$

$=\begin{pmatrix}2&-1&0&\frac{-ab-3a+14}{-a+7}\\ 0&1&0&\frac{\left(b+1\right)\left(-a+4\right)}{-a+7}\\ 0&0&1&\frac{b+1}{-a+7}\end{pmatrix}$

$R_1\:\leftarrow \:R_1+1\cdot \:R_2$

$=\begin{pmatrix}2&0&0&\frac{-2ab-4a+4b+18}{-a+7}\\ 0&1&0&\frac{\left(b+1\right)\left(-a+4\right)}{-a+7}\\ 0&0&1&\frac{b+1}{-a+7}\end{pmatrix}$

$R_1\:\leftarrow \frac{1}{2}\cdot \:R_1$

$=\begin{pmatrix}1&0&0&\frac{-ab-2a+2b+9}{-a+7}\\ 0&1&0&\frac{\left(b+1\right)\left(-a+4\right)}{-a+7}\\ 0&0&1&\frac{b+1}{-a+7}\end{pmatrix}$

solution is

$x=\frac{-ab-2a+2b+9}{7-a},\:\:\:\:y=\frac{\left(b+1\right)\left(-a+4\right)}{7-a},\:\:\:\:\:z=\frac{b+1}{7-a}$

put back $a=\alpha , \:\:\:\:\:\:\:\:\:\:b=\beta$

$x=\frac{-\alpha \beta -2\alpha +2\beta +9}{7-\alpha },\:\:\:\:y=\frac{\left(\beta +1\right)\left(-\alpha +4\right)}{7-\alpha },\:\:\:\:\:z=\frac{\beta +1}{7-\alpha }$

${\color{Red} x=\frac{-\alpha \beta -2\alpha +2\beta +9}{7-\alpha },\:\:\:\:y=\frac{\left(\beta +1\right)\left(4-\alpha \right)}{7-\alpha },\:\:\:\:\:z=\frac{\beta +1}{7-\alpha }}$

here ${\color{Red} \alpha \neq 7}$