# Thread: Linear system solutions using Gauss-Jordan elimination

1. ## Linear system solutions using Gauss-Jordan elimination

$
\left\{
\begin{matrix}
ax - 3y + z = a\\
-x + 6y = -2\\
x + 5z = 4\\
\end{matrix}\right.
$

system matrix:
$
\[
A =
\left[ \begin{array}{ccc|c}
a & -3 & 1 & a\\
-1 & 6 & 0 & -2\\
1 & 0 & 5 & 4\end{array}\right]

$

I'm not sure where to start. Should I switch row 3 and row 1?

2. Originally Posted by qaz
$
\left\{
\begin{matrix}
ax - 3y + z = a\\
-x + 6y = -2\\
x + 5z = 4\\
\end{matrix}\right.
$

system matrix:
$
\[
A =
\left[ \begin{array}{ccc|c}
a & -3 & 1 & a\\
-1 & 6 & 0 & -2\\
1 & 0 & 5 & 4\end{array}\right]

$

I'm not sure where to start. Should I switch the 3rd line with the 1st?
What's the question? Are you supposed to solve for x, y, z in terms of a or find a value of a for which the system has a unique solution or ...?

If you're required to solve for x, y, z in terms of a, you can indeed start by switching row 1 and row 3.

3. Your first objective, in row reduction, is to get a "1" in the first column, first row. Yes, you can do that by swapping the first and third rows (which is a row operation). In general, as long as the number already is "first column, first row" is not 0, you can get a "1" there by dividing the entire row by that number (here, "a") but for this problem swapping rows is simpler.

4. Originally Posted by qaz
$
\left\{
\begin{matrix}
ax - 3y + z = a\\
-x + 6y = -2\\
x + 5z = 4\\
\end{matrix}\right.
$

system matrix:
$
\[
A =
\left[ \begin{array}{ccc|c}
a & -3 & 1 & a\\
-1 & 6 & 0 & -2\\
1 & 0 & 5 & 4\end{array}\right]

$

I'm not sure where to start. Should I switch the 3rd line with the 1st?
I would switch the 1st and 3rd, then add row 1 to 2, divide row 2 by 6, .....

You keep going until you have an upper triangular matrix or you if you are feeling ambitious you can do the rref i.e. until you have the identity matrix.

5. Hello,
So I did start by switching row 1 and row 3...

$
\[
\left[ \begin{array}{ccc|c}
a & -3 & 1 & a\\
-1 & 6 & 0 & -2\\
1 & 0 & 5 & 4\end{array}\right]
$

$\begin{array}{c}\\ R_3<->R_1\end{array} \left[\begin{array}{ccc|c}1&0&5&4 \\ -1&6&0&-2 \\ a&-3&1&a \end{array}\right]$

$\begin{array}{c}\\ R_2+R_1\\R_3-aR_1\end{array} \left[\begin{array}{ccc|c}1&0&5&4 \\ 0&6&5&2 \\ 0&-3&1-5a&-3a \end{array}\right]$

$\begin{array}{c}\\ R_3+\frac{1}{2}R_2 \end{array} \left[\begin{array}{ccc|c}1&0&5&4 \\ 0&6&5&2 \\ 0&0&-5a+7/2&-3a+1 \end{array}\right]$

$\begin{array}{c}\\ \frac{1}{-5a+7/2}R_3 \end{array} \left[\begin{array}{ccc|c}1&0&5&4 \\ 0&6&5&2 \\ 0&0&1&3/5a+2/7 \end{array}\right]$

$\begin{array}{c}\\ \frac{1}{5}R_1-R_3\\\frac{1}{5}R_2-R_3 \end{array} \left[\begin{array}{ccc|c}1/5&0&0&3/5a-18/35 \\ 0&6/5&0&3/5a-4/35 \\ 0&0&1&3/5a+2/7 \end{array}\right]$

$\begin{array}{c}\\ \frac{1}{5}R_1\\\frac{6}{5}R_2 \end{array} \left[\begin{array}{ccc|c}1&0&0&3/25a-18/175 \\ 0&1&0&18/25a-24/175 \\ 0&0&1&3/5a+2/7 \end{array}\right]$

if $a \ne \frac{5}{3}*-\frac{2}{7}$ than rank(A) = 3 = rank([A|B]) = 3 -> Possible,

if $a=\frac{5}{3}*-\frac{2}{7}$ than rank(A) = 3 > rank([A|B]) = 2 -> ??

6. Originally Posted by qaz

$\begin{array}{c}\\ R_3+\frac{1}{2}R_2 \end{array} \left[\begin{array}{ccc|c}1&0&5&4 \\ 0&6&5&2 \\ 0&0&-5a+7/2&-3a+1 \end{array}\right]$
Assuming everything is correct. If a = 7/10, your matrix is inconsistent.

If a = 1/2, then your solution is (x,y,z).