# Matrix equation

• Oct 19th 2012, 07:33 AM
dumbledore
Matrix equation
Given the matrix equation

$\displaystyle X^{-1}A^{T} + B^{-1}C = 0$

Where $\displaystyle A,B,C,$ and $\displaystyle X$ are 2 x 2 invertible matrices. Solve for $\displaystyle X$ in terms of $\displaystyle A,B,C$.

ii) Find X when $\displaystyle A = \begin{matrix} [1 & 1] \\ [0 & 2] \end{matrix}, B= \begin{matrix} [1 & 2] \\ [1 & 1] \end{matrix}, C= \begin{matrix} [1 & 2] \\ [1 & 4] \end{matrix}$
• Oct 19th 2012, 08:54 AM
MaxJasper
Re: Matrix equation
$\displaystyle x=\left(\begin{array}{cc} 1 & 3 \\ 1 & 2\end{array}\right)$(Shake)
• Oct 20th 2012, 06:33 AM
dumbledore
Re: Matrix equation
For solving for X (part i) of the question , is this correct?

$\displaystyle B^{-1} C = - X^{-1} A^{T}$

left multiply by X

$\displaystyle X B^{-1} C = - (X X^{-1}) A^{T}$

$\displaystyle X B^{-1} C = - I A^{T}$

right multiply by $\displaystyle C^{-1} B$

$\displaystyle X B^{-1} C (C^{-1} B) = - A^{T} (C^{-1} B)$

$\displaystyle X = - A^{T} C^{-1} B$

as

$\displaystyle B^{-1} C (C^{-1} B) = B^{-1} (C C^{-1}) B =$

$\displaystyle = B^{-1} I B = B^{-1} B = I$
• Oct 20th 2012, 08:55 AM
HallsofIvy
Re: Matrix equation
Yes, of course that is correct.
• Oct 20th 2012, 09:43 AM
dumbledore
Re: Matrix equation
Thanks HallsofIvy. The problem now is that when I plug in my 2x2 matrices to the equation $\displaystyle X = -A^{T} C^{-1} B$ , I do not get the matrix that MaxJasper provided above.
• Oct 20th 2012, 12:55 PM
MaxJasper
Re: Matrix equation
Typo: Sorry, minus sign was dropped, so correct x is:

$\displaystyle \text{x}=-\left(\begin{array}{cc} 1 & 3 \\ 1 & 2\end{array}\right)$(Nod)
• Oct 20th 2012, 01:03 PM
dumbledore
Re: Matrix equation
Got it!