# Calculate the solution using X = A^-1B

• Sep 24th 2011, 06:03 AM
kmjt
Calculate the solution using X = A^-1B
http://img.photobucket.com/albums/v1...at100024AM.png

I wrote the system of equations in the form AX = B, however how would I go about calculating the solution using the equation X = A^-1B? (Headbang)
• Sep 24th 2011, 06:17 AM
Plato
Re: Calculate the solution using X = A^-1B
Go to this web page.
You will find material of inverses of a $\displaystyle 2\times 2$ matrix.
• Sep 24th 2011, 11:20 AM
HallsofIvy
Re: Calculate the solution using X = A^-1B
You would find the inverse of the matrix and multiply by it! That is exactly what $\displaystyle A^{-1}B$ means.

The inverse of the two by two matrix [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix} is $\displaystyle \frac{1}{\Delta}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$ where $\displaystyle \Delta= ad- bc$, the determinant of the matrix. Be sure to do the multiplication in the order shown.
• Sep 24th 2011, 04:52 PM
Deveno
Re: Calculate the solution using X = A^-1B
if $\displaystyle AX = B$

then $\displaystyle A^{-1}(AX) = A^{-1}B$ so

$\displaystyle (A^{-1}A)X = A^{-1}B$, that is:

$\displaystyle IX = X = A^{-1}B$.

so if A is indeed invertible, then calculating $\displaystyle A^{-1}B$ tells you which vector "X" is.