Question

Given a matrix that has been partially solved \[\left[\begin{matrix}1 & -5 & 4 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 2& 0\end{matrix}\right]\]

"continue the appropriate row operations and describe the solution set of the original system"

Answer 1

sorry that should be AUGMENTED matrix!

Answer 2

So, since the last column is all 0's, not matter what operations I do, it will stay 0's right? And since there are 4 rows, this would imply 4 variables correct? So my solution is (0,0,0,0) right?

Answer 3

that is correct :)

Answer 4

thanks!

Answer 5

\[\left[\begin{array}{cccc|c}1 & -5 & 4 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 2& 0\end{array}\right]\]\[\sim\left[\begin{array}{cccc|c}1 & -5 & 4 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1& 0\end{array}\right]\qquad \begin{array}{c}R_3\to R_3/3\\R_4\to R_2/2\end{array}\]\[\sim\left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1& 0\end{array}\right]\qquad \begin{array}{c}R_1\to R_1+5R_2-4R_3\\R_2\to R_2-R_4\end{array}\]_____________________________________
\[\left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right]=\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]\]