1. ## Solve for A

hello,

(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!

2. Originally Posted by l flipboi l
hello,

(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!
$(I+2A)^{-1}=\begin{pmatrix}\!\!-1&2\\4&5\end{pmatrix}\Longleftrightarrow I+2A=\begin{pmatrix}\!\!-1&2\\4&5\end{pmatrix}^{-1}=\begin{pmatrix}\!\!-5/13&2/13\\4/13&1/13\end{pmatrix}\Longleftrightarrow$

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

Note that at the end you must check that $I+2A$ is actually invertible, otherwise the problem's no solution.

Tonio

3. 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?

4. Originally Posted by l 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?

You have to do is substraction of matrices...you should know this if your dealing with this problem.

Tonio