Texts given matrix a 4 4 4 3 3 1 1 0 9 2 5 0 5 4 3 4 7 1 9 6...

Name: texts given matrix a 4 4 4 3 3 1 1 0 9 2 5 0 5 4 3 4 7 1 9 6 6 find a basis for the column space of a find all solutions of the system of linear equations ax 3 2 0 3 6 0 and write them param 08626
Uploaded: 2022-09-16T11:00:42-08:00
Duration: 5 min 5 s
Channel: Ben Blakesley
Description: texts given matrix a 4 4 4 3 3 1 1 0 9 2 5 0 5 4 3 4 7 1 9 6 6 find a basis for the column space of a find all solutions of the system of linear equations ax 3 2 0 3 6 0 and write them param 08626

Texts: Given matrix A = -4 4 4 3 3 1 -1 0 -9 2 -5 0 -5 -4 -3 4 7 1 9 6 -6 Find a basis for the column space of A. Find all solutions of the system of linear equations Ax = 3 2 0 -3 -6 0 and write them parametrically as a vector or explain why there are none.

Transcript

00:01 Okay, so we have this matrix a, and we want to find a basis for the solution space of a times a general vector x, y, z, w equals zero.

00:11 So we're going to create the augmented matrix, a, with the zero vector, one, two, three, zero, one, two, five, two, two, four, seven, one, minus two, zero, three, augmented with the right -hand side of this equation, which is just all zeros.

00:31 And then we're going to perform elementary row operations to reduce this to a simpler matrix and then try and find the solutions.

00:40 So firstly, let's look at row four.

00:44 So let's write this.

00:46 So we can leave the right hand side alone because we're always going to get zeros.

00:50 Row four, i'm going to add, let's say, row one to row four.

00:56 So the first components cancel.

00:58 We have minus one plus one, which is zero.

01:01 Here we have minus 2 plus 2, which is 0.

01:07 Here we have 0 plus 3, which is 3.

01:10 And then we have 3 plus 0, which is 3, which is 3, sorry.

01:19 Row 1, i'm going to leave alone 1, 2, 3, 0.

01:24 Row 2, i'm going to subtract row 1.

01:27 So we get 1 minus 1 is 0.

01:30 2 minus 2 is 0 5 minus 3 is 2 and 2 minus 0 is 2 and then row 3 i'm going to subtract 2 times row 1 so we get 2 minus 2 which is 0 we get 4 minus 4 which is 0 we get 7 minus 6 which is 1, and we get 1 minus 0, which is 1.

02:08 So we see now that these 3 look very similar.

02:12 So the next operations i'm going to do are the following.

02:16 I'm going to again leave row 1 alone, 1, 2, 3, 0, 0, and then this is also going to be all 0 0 here, because nothing changes.

02:25 I'm going to divide row 2 by 2, so this gives us 0 -0 -1 -1, 0 -0 .0...

Question

Texts: Given matrix A = -4 4 4 3 3 1 -1 0 -9 2 -5 0 -5 -4 -3 4 7 1 9 6 -6 Find a basis for the column space of A. Find all solutions of the system of linear equations Ax = 3 2 0 -3 -6 0 and write them parametrically as a vector or explain why there are none.

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