Question

linear combination question... help please?

Answer 1

write B as a linear combi of other matrices, if possible.

\[B = \left[\begin{matrix}6 & -2 & 5\\ -2 & 8 & 6\\ 5 & 6 & 6\end{matrix}\right]\]

Answer 2

\[A1 = \left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix}\right]\]

Answer 3

\[A2 = \left[\begin{matrix}1 & 1 & 1 \\ 1 & 0 & 1\\ 1 & 1 & 1 \end{matrix}\right]\]

Answer 4

\[A3 = \left[\begin{matrix}1 & 0 & 1 \\ 0 & 1 & 0\\ 1 & 0 & 1 \end{matrix}\right]\]

Answer 5

\[A4 = \left[\begin{matrix}0 & -1 & 0 \\ -1 & 1 & 1\\ 0 & 1 & 0 \end{matrix}\right]\]

Answer 6

What I tried so far is to say that there exists c1A1 + c2A2 + c3A3 + c4A4 = B. And then I try to elmentary row operations to be in row echelon form (augmented matrix) [B|0] I don't know if this's right, and got lost on the way... please help

Answer 7

well maybe, a bit confusing of my writing above. In other words, for that particular questions, how do you solve it, guys?? please help :(

Answer 8

i think you original set up is a good start; but im not sure if trying to go thru echelon is needed

Answer 9

c1A1 =

c1 0 0
0 c1 0
0 0 c1

c2A2 =

c2 c2 c2
c2 0 c2
c2 c2 c2

c3A3 =

c3 0 c3
0 c3 0
c3 0 c3

c4A4 =

0 -c4 c4
-c4 c4 c4
0 c4 0

since adding matrixes is just adding like parts ...

c1+c2+c3+0 = 6
0+c2 +0 -c4 = -2

etc

Answer 10

yeah, I got that as well.
c1 + c2 + c3 =6
c2 - c4 = -2
c2 + c3 = 5
c1 + c3 + c4 = 8
c2 + c4 = 6

and then? how to conclude that B is a lin combi or not?

Answer 11

you end up with 9 equations with 4 unknowns which should give you plenty of ammo to play with

if anything, you might be able to augment the setup into rref form as well

Answer 12

1 1 1 0 6
0 1 0 -1 -2
0 1 1 0 5
0 1 0 -1 -2
1 0 1 1 8
0 1 0 1 6
0 1 1 0 5
0 1 0 1 6
1 1 1 0 6


1 1 1 0 6
1 1 1 0 6
1 0 1 1 8
0 1 0 -1 -2
0 1 0 -1 -2
0 1 1 0 5
0 1 1 0 5
0 1 0 1 6
0 1 0 1 6


1 1 1 0 6
0 0 0 0 0
0 1 0 -1 -2
0 1 0 -1 -2
0 1 0 -1 -2
0 1 1 0 5
0 1 1 0 5
0 1 0 1 6
0 1 0 1 6

1 1 1 0 6
0 0 0 0 0
0 1 0 -1 -2
0 0 0 0 0
0 0 0 0 0
0 1 1 0 5
0 0 0 0 0
0 1 0 1 6
0 0 0 0 0




1 1 1 0 6
0 1 0 -1 -2
0 0 1 1 7
0 0 0 2 8


1 1 1 0 6
0 1 0 -1 -2
0 0 1 1 7
0 0 0 1 4

Answer 13

1 1 1 0 6
0 1 0 -1 -2
0 0 1 1 7
0 0 0 1 4


1 0 0 0 1
0 1 0 0 2
0 0 1 0 3
0 0 0 1 4


try

c1=1
c2=2
c3=3
c4=4

Answer 14

is what i did make sense?

Answer 15

yeah I got it now, a bit problem with simplify to rref form things, but I'll be fine. Thanks so much :)

Answer 16

youre welcome, and good luck :)