Question

linear combinations on vectors.help

Answer 1

write the matrix\[E= \left[\begin{matrix}3 & 1 \\ 1 & -1\end{matrix}\right]\]

Answer 2

as a linear combination of the matrices
\[A=\left[\begin{matrix}1 & 1 \\ 1& 0\end{matrix}\right]\]

Answer 3

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

Answer 4

\[C=\left[\begin{matrix}0 & 2 \\ 0 & -1\end{matrix}\right]\]

Answer 5

are there just three?

Answer 6

basically ,you write the matrix E as a linear combination of the matrices A,B, and C

Answer 7

@richyw ,yes there only three

Answer 8

ok so the concepts you need here are matrix addition and scalar multiplication.

Answer 9

you can solve this one in different ways but it's simple enough to do by inspection

Answer 10

so looking at E, the first thing I notice is that \(e_{11}=3\) now looking at matrices A, B, and C. The only one that has anything but 0 in that position is matrix A, so we know already that it must be 3 times matrix A. which gives. \[\left[\begin{matrix}3 & 3 \\ 3 & 0\end{matrix}\right]\] right?

Answer 11

ok i got that part

Answer 12

alright so now look at matrix E and notice that position \(e_{12}\) is 2, so we need to subtract a certain amount from matrix A to produce a 2 there.

Answer 13

well, matrix B has a zero in that position, so it is no good, so subtract one times matrix C from matrix A and we get \[3A-C=\left[\begin{matrix}3&1\\3&1\end{matrix}\right]\]

Answer 14

sorry I should have said "subtract one times matrix C from three times matrix A". Now I hope you can see what multiple of matrix B you bust subtract to get E

Answer 15

are you following?

Answer 16

yeah

Answer 17

alright cool well then you have the answer!

Answer 18

so it is
3A-2B-C=E?

Answer 19

yes!

Answer 20

thanx a lot!!!!!!!