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linear combinations on vectors.help

Mathematics
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write the matrix\[E= \left[\begin{matrix}3 & 1 \\ 1 & -1\end{matrix}\right]\]
as a linear combination of the matrices \[A=\left[\begin{matrix}1 & 1 \\ 1& 0\end{matrix}\right]\]
\[B=\left[\begin{matrix}0 & 0 \\ 1 & 1\end{matrix}\right]\]

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Other answers:

\[C=\left[\begin{matrix}0 & 2 \\ 0 & -1\end{matrix}\right]\]
are there just three?
basically ,you write the matrix E as a linear combination of the matrices A,B, and C
@richyw ,yes there only three
ok so the concepts you need here are matrix addition and scalar multiplication.
you can solve this one in different ways but it's simple enough to do by inspection
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?
ok i got that part
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.
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 \[3A-C=\left[\begin{matrix}3&1\\3&1\end{matrix}\right]\]
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
are you following?
yeah
alright cool well then you have the answer!
so it is 3A-2B-C=E?
yes!
thanx a lot!!!!!!!

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