# Thread: [SOLVED] Solve linear systen with given matrix

1. ## [SOLVED] Solve linear systen with given matrix

Solve the linear system with the given augmented matrix (the last column should have the dots before it but I'm not sure how to enter those)

$\displaystyle \begin{pmatrix}4 & 2 & -1 & 5\\3 & 3 & 6 & 1\\5 & 1 & -8 & 8\end{pmatrix}$
First I subtracted the 2nd row from the 1st to get:
$\displaystyle \begin{pmatrix}1 & -1 & -7 & 4\\3 & 3 & 6 & 1\\5 & 1 & -8 & 8\end{pmatrix}$
Then I subtracted 3 x row 1 from row 2 and 5 x row 1 from row 3 to get:
$\displaystyle \begin{pmatrix}1 & -1 & -7 & 4\\0 & 6 & 27 & -11\\0 & 6 & 27 & -12\end{pmatrix}$
I subtracted row 3 from row 2 to get:
$\displaystyle \begin{pmatrix}1 & -1 & -7 & 4\\0 & 0 & 0 & 1\\0 & 6 & 27 & -12\end{pmatrix}$
Since I have a row of 0 0 0 1, I believe there is no solution. Am I doing this wrong? or maybe missed a step?

2. What you have done is correct it shows inconsistency meaning there are no solutions.

3. Hello, Jennifer!

Solve the linear system with the given augmented matrix.

. . . $\displaystyle \left(\begin{array}{ccc|c}4 & 2 & \text{-}1 & 5\\3 & 3 & 6 & 1\\5 & 1 & \text{-}8 & 8\end{array}\right)$

My work . . .

$\displaystyle \begin{array}{c}R_1-R_2 \\ \\ \end{array} \left(\begin{array}{ccc|c} 1 & \text{-}1 & \text{-}7 & 4\\3 & 3 & 6 & 1\\5 & 1 & \text{-}8 & 8\end{array}\right)$

$\displaystyle \begin{array}{c}\\ R_2-3R_1 \\R_3-5R_1 \end{array}\left(\begin{array}{ccc|c}1 & \text{-}1 & \text{-}7 & 4\\0 & 6 & 27 & \text{-}11\\0 & 6 & 27 & \text{-}12\end{array}\right)$

$\displaystyle \begin{array}{c}\\ R_2-R_3 \\ \\ \end{array} \left(\begin{array}{ccc|c} 1 & \text{-}1 & \text{-}7 & 4\\0 & 0 & 0 & 1\\0 & 6 & 27 & \text{-}12\end{array}\right)$

Since I have a row of $\displaystyle 0\;0\;0\;1$, I believe there is no solution. . . . . Right!
Am I doing this wrong . . . or maybe missed a step? . . . . no

Ya done real good!