hello,

Can someone please help me with this problem?

(I+2A)^-1 = [ -1 2 ]

[ 4 5 ]

not sure how to make a matrix...

it's suppose to be a 2 x 2 matrix with row 1 = -1, 2 and row 2 = 4, 5

Thanks!

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- Oct 7th 2010, 07:43 PMl flipboi lSolve for A
hello,

Can someone please help me with this problem?

(I+2A)^-1 = [ -1 2 ]

[ 4 5 ]

not sure how to make a matrix...

it's suppose to be a 2 x 2 matrix with row 1 = -1, 2 and row 2 = 4, 5

Thanks! - Oct 7th 2010, 08:02 PMtonio
$\displaystyle (I+2A)^{-1}=\begin{pmatrix}\!\!-1&2\\4&5\end{pmatrix}\Longleftrightarrow I+2A=\begin{pmatrix}\!\!-1&2\\4&5\end{pmatrix}^{-1}=\begin{pmatrix}\!\!-5/13&2/13\\4/13&1/13\end{pmatrix}\Longleftrightarrow$

$\displaystyle 2A=\begin{pmatrix}\!\!-18/13&2/13\\4/13&\!\!-12/13\end{pmatrix}$ ...and now divide by 2 and find your matrix.

Note that at the end youcheck that $\displaystyle I+2A$ is actually invertible, otherwise the problem's no solution.__must__

Tonio - Oct 7th 2010, 08:08 PMl flipboi l
Thanks!

What i'm stumped with is I, what do you have to do with it?

from the looks of it, since the matrix for the identity is row 1 = [1 0] row 2 = [0 1]

just subtract 1? when you move the I to the right side? - Oct 7th 2010, 09:38 PMtonio