##### Practice Midterm #6 (Second Part)
tanishaharlalka
For the second part, when it asks is { (1,1,2,0) , (1,0,3,1) , (2,1,4,3), (1,-2,4,5) } is a basis for R4, did you guys get that it is NOT a basis because the dimension of the image is not equal to 4? In reduced row echelon form, I got a (0,0,0,0) for the last row, thus showing that it is not a basis for R4.
jared_brock: May 4, 2015, 10:16 p.m.
You may want to double check (or perhaps I should), for I got that all components equal zero, i.e. I was able to get rre form.
TLindberg: May 4, 2015, 10:27 p.m.
When I row reduced the matrix \begin{pmatrix} 1 & 1 & 2 & 0 \\ 1 & 0 & 3 & 1 \\ 2 & 1 & 4 & 3 \\ 1 & -2 & 4 & 5 \end{pmatrix} I found that the last row was reduced to (0,0,0,0). Therefore the set of vectors would not be a basis for R4 since the they are not linearly independent. (No idea if this matrix will display correctly by the way, so forgive me for that.)
valerie: May 4, 2015, 10:30 p.m.
I also got (0,0,0,0) for the last row, showing it is not a basis.
jared_brock: May 4, 2015, 10:33 p.m.
I believe that may be the incorrect matrix. Note that the first term (first four digits) are the "a" term, and etc. for the 2nd, 3rd, and 4th term. So, think a(1,1,2,0), b(1,0,3,1), c(2.1.4.3), and d(1,-2,4,5) leading to equations a+b+2c+d=0, a+c-2d=0, 2a+3b+4c+3d=0, and b+3c+5d=0. Therefore, the matrix should be: first column 1120, second column 1031, third column 2143, and 4th column 1-235. Basically yours rows and columns but switched.
jared_brock: May 4, 2015, 10:37 p.m.
Row reducing this matrix results in a reduced row echelon form, I'm just not 100 percent sure if my logic for creating the matrix is correct, but if it is I hope that is helpful!
danielkim: May 4, 2015, 10:46 p.m.
B = { (1,1,2,0), (1,0,3,1) , (2,1,4,3), (1,-2,4,5) } is not a basis for R4 because when you rref the matrix, the right most column vector (1,-2,4,5)^T (T is the transpose) is linearly dependent on the previous three column vectors. In other words, it is a linear combination of the previous three column vectors. Therefore it is not B is not a basis for R4
jared_brock: May 4, 2015, 10:52 p.m.
Ah, my mistake, I made an error composing the matrix and I did get all zeros for one of the rows, therefore it's not a basis. Sorry about that!
ConnorCDougherty: May 5, 2015, 6:57 p.m.
Jared displayed the matrix correctly I believe, and yes I found the last row to be all 0s. This means that B is not a basis in R4.