# Thread: Linear Algebra

1. ## Linear Algebra

Is this is row echelon form?
Solve the linear system...
Im not sure how to do it because of the zeros on the bottom row, or is that the problem and therefore the system cannot be solved?

$
\begin{bmatrix}
1 & -3 & 8 \\0 & 1 & 0 \\0 & 0 & 1
\end{bmatrix}
$

Many thanks

2. Originally Posted by brd_7
Is this is row echelon form?
Solve the linear system...
Im not sure how to do it because of the zeros on the bottom row, or is that the problem and therefore the system cannot be solved?

$
\begin{bmatrix}
1 & -3 & 8 \\0 & 1 & 0 \\0 & 0 & 1
\end{bmatrix}
$

Many thanks
yes, All nonzero rows are above any rows of all zeroes, and The leading coefficient of a row is always strictly to the right of the leading coefficient of the row above it.
So, you must have another column. You can read off the bottom 2 rows for the answer, but the top row i syour problem. You waat 0's in the 2nd and 3rd columns. This is actually pretty simple. First, how would we get the 8 to be a zero?Well, the bottom row has a 1 below the 8, so the bottom row should help with this. What about -8R3+R1. That would take care of 1 zero. $
\begin{bmatrix}
1 & -3 & 0\\0 & 1 & 0 \\0 & 0 & 1
\end{bmatrix}
$
So, do the same thing for the -3, but using the 2nd row this time.

3. Im sorry, i should have said that the matrix above is augmented and so the further right column are actually answers to the bits before, eg. x -3y = 8.. So i dont think you've solved the linear system, but instead just got thr identity matrix.. i think .. Anyone?

4. Anyone?