# I can't solve this matrix using the Gauss-Jordan method, i need help!!

• Nov 9th 2009, 07:58 AM
burdonjo
I can't solve this matrix using the Gauss-Jordan method, i need help!!
{ 2 -2 -4 | -2
-2 3 1 | 7
3 -3 -6 | -3}

I CANT FIGURE THIS ONE OUT! I can only do it on a calculator, obviously without showing work, but, i need to figure this problem out showing all steps. I would really appreciate it if someone would be able to help me out, thanks.
• Nov 9th 2009, 08:57 AM
kjchauhan
matrix
The given matrix is a singular matrix, so solution does not exist.
• Nov 9th 2009, 09:02 AM
Soroban
Hello, burdonjo!

This system has an infinite number of solutions.

Quote:

Solve: .$\displaystyle \left|\begin{array}{ccc|c}2 & \text{-}2& \text{-}4 & \text{-}2 \\ \text{-}2 & 3 & 1 & 7 \\ 3 & \text{-}3 & \text{-}6 & \text{-}3\end{array}\right|$

. . . $\displaystyle \begin{array}{c}\frac{1}{2}R_1 \\ \\ \frac{1}{3}R_3\end{array} \left|\begin{array}{ccc|c}1 & \text{-}1 & \text{-}2 & \text{-}1 \\ \text{-}2 & 3 & 1 & 7 \\ 1 & \text{-}1 & \text{-}2 & \text{-}1 \end{array}\right|$

$\displaystyle \begin{array}{c}\\ R_2 + 2R_1 \\ R_3 - R_1\end{array} \left|\begin{array}{ccc|c}1 & \text{-}1 & \text{-}2 & \text{-}1 \\ 0 & 1 & \text{-}3 & 5 \\ 0 & 0 & 0 & 0 \end{array}\right|$

. $\displaystyle \begin{array}{c}R_1+R_2 \\ \\ \\ \end{array} \left|\begin{array}{ccc|c}1 & 0 & \text{-}5 & 4 \\ 0 & 1 & \text{-}3 & 5 \\ 0 & 0 & 0 & 0 \end{array}\right|$

We have:. . $\displaystyle \begin{Bmatrix}x - 5z &=& 4 \\ y -3x &=& 5 \end{Bmatrix}\quad\Rightarrow\quad \begin{Bmatrix}x &=& 5z + 4 \\ y &=& 3z + 5 \\ z &=& z \end{Bmatrix}$

On the right, replace $\displaystyle z$ with a parameter $\displaystyle t.$

. . . . $\displaystyle \begin{Bmatrix}x &=& 5t + 4 \\ y &=& 3t + 5 \\ z &=& t \end{Bmatrix}$

This represents all the solutions for the system of equations,
. . one solution for each value of $\displaystyle t.$

• Nov 9th 2009, 10:35 AM
burdonjo
Bottom row
When the bottom row of the matrix is zeros all the way across, does that mean that the matrix has "no solution"? Sorry to bother you
• Nov 10th 2009, 05:15 AM
HallsofIvy
Quote:

Originally Posted by burdonjo
When the bottom row of the matrix is zeros all the way across, does that mean that the matrix has "no solution"? Sorry to bother you

Are you talking about the row reduced coefficient matrix or the row reduced augmented matrix (with the right side of the system of equations as the last column)? If any column of the row reduced augmented matrix has all "0"s except that the last column is not 0, then there exist no solution. If all rwos of the row reduced augmented matrix that have all "0" also have "0" in the last column, then there exist an infinite number of solutions.

In either case, there is not a "unique" solution.