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

**burdonjo** · November 9th 2009, 07:58 AM

{ 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.

**kjchauhan** · November 9th 2009, 08:57 AM

The given matrix is a singular matrix, so solution does not exist.

**Soroban** · November 9th 2009, 09:02 AM

Hello, burdonjo!

This system has an infinite number of solutions.

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

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

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

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

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

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

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

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

**burdonjo** · November 9th 2009, 10:35 AM

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

**HallsofIvy** · November 10th 2009, 05:15 AM

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.

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