1. ## gaussian elimination method

Am I doing this correctly for matices?
x has subscripts 1 and 2

-2x
+ 2x = 2
3x - 3x = -3

|-2, 2, 2 |
|3, -3, -3|

-2(-1\2) = 1 so

|1, -1, -1|
|3, -3, -3|

-3(1) + 3 = 0 so

|1, -1, -1|
|0, 0, 0 |

The unique solution is x subcript 1 = ? and x subscript 2 = ?

2. You only need to find the inverse of the coefficient matrix

$\displaystyle \left[
\begin{array}{ c c }
-2 & 2 \\
3 & -3
\end{array} \right]
$

But does the inverse for this matrix exist?

3. Originally Posted by hansfordmc

|1, -1, -1|
|0, 0, 0 |

The unique solution is x subcript 1 = ? and x subscript 2 = ?
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Whenever you get a only series of zeros in the final step of Gaussian Elimination, it implies $0.x_1+0.x_2=0$ which is obvious and holds for any values of $x_1,x_2$. In such case you do not get any unique solution, what you get is a family of solutions. In your problem, the family of solution is given by $\{(x_1,x_2)|x_2-x_1=1\}$

4. Hello, hansfordmc!

Am I doing this correctly for matices?

. . $\begin{array}{ccc}\text{-}2x + 2y &=& 2 \\ 3x - 3x &=& \text{-}3 \end{array}$

$\text}We have: }\;\left| \begin{array}{cc|c}\text{-}2 & 2 & 2 \\
3 & \text{-}3 & \text{-}3 \end{array}\right|$

. . . . $\begin{array}{c}\text{-}\frac{1}{2}R_1 \\ \\ \end{array}\left|\begin{array}{cc|c} 1 & \text{-}1 & \text{-}1 \\ 3 & \text{-}3 & \text{-}3 \end{array}\right|$

$\begin{array}{c} \\ R_2-3R_1 \\ \end{array}\left| \begin{array}{cc|c} 1 & \text{-}1 & \text{-}1 \\ 0 & 0 & 0 \end{array}\right|$ . This is correct!

$\text{The unique solution is: }\:\begin{Bmatrix}x\;=\;? \\ y \;=\;?\end{Bmatrix}$

As Sambit pointed out, a row of zeros indicates no unique solution.

5. So now, exactly what was the question? What you have shown so far is that both equations, $-2x_1+ 2x_2= 2$ and $3x_1- 3x_2= -3$ reduce to $x_1- x_2= -1$ or $x_1= x_2- 1$. If the question was "is there a unique solution", the answer is "no". If the problem was to find all possible solutions, then $(x_1, x_2)= (t- 1, t)$, is a solution for any number, t, and all solutions are of that form.