# Solve for A

• Oct 7th 2010, 07:43 PM
l flipboi l
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!
• Oct 7th 2010, 08:02 PM
tonio
Quote:

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!

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

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

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

Tonio
• Oct 7th 2010, 08:08 PM
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?
• Oct 7th 2010, 09:38 PM
tonio
Quote:

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