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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)

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- Sep 24th 2011, 06:03 AMkmjtCalculate the solution using X = A^-1B
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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 AMPlatoRe: Calculate the solution using X = A^-1B
Go to this web page.

Scroll about half way down.

You will find material of inverses of a $\displaystyle 2\times 2$ matrix. - Sep 24th 2011, 11:20 AMHallsofIvyRe: 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 PMDevenoRe: 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.