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Previous Problem Problem List Next Problem (1 point) Solve the system using row operations. \begin{cases} w + x &= -1 \\ x + y &= -5 \\ y + z &= 1 \\ w + z &= 5 \end{cases} \begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \\\ \\\ \\\ \end{bmatrix} + s \begin{bmatrix} \\\ \\\ \\\ \end{bmatrix} Preview My Answers Submit Answers You have attempted this problem 0 times.

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00:01 We're going to set the following problem up as a matrix.

00:04 Just be careful with the last equation.

00:06 There's only two of the three variables listed.

00:08 So make sure you remember that you technically have a 0 -z there that you're going to have to add into your matrix.

00:17 So what we have for our matrix is 1 -9 -3 -1, with an answer of 1, 2 -1 -1 -4 -0, 1 -9 -3 -0, 1 -9 -3 -3 -3.

00:30 2 -1, 1 -2 -0 -05.

00:36 Alright, so now we can go about solving our matrix.

00:40 And i'm going to do a bunch of steps combined here, just to kind of make our life, well, make it a little bit of a shorter process.

00:47 So i'm going to take row two, and i'm going to subtract two row ones from that.

00:53 I'm going to take row three and subtract a row one.

00:56 I'm going to take a row four and subtract a row one.

00:59 Pretty much all i'm doing is i'm taking that first row which has a one in the left position and i'm getting rid of all of the numbers underneath it.

01:08 So my first row is not changing, just the rest of my rows are.

01:15 So first step, double that row subtract it from the other one.

01:19 So 2 minus 2 is 0.

01:21 Negative 1 minus a negative 6 is a positive 5.

01:26 Negative 4 minus 2 is negative 6.

01:29 1 minus 2 is negative 2.

01:32 So row 2 is negative 2.

01:33 So row 2, is done.

01:35 Then we're going to do that subtracting from row 3.

01:38 So 1 minus 1 is 0.

01:41 Negative 3 minus a negative 3 is 0.

01:46 2 minus 1 is 1.

01:48 1 minus 1 is 0.

01:52 And then from our last row 1 minus 1 is 0.

01:55 0, negative 2 minus a negative 3 is 1, 0 minus 1 is negative 1, and 5 minus 1 is 4...

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