quick question

[sadzade]

 

[ksdhart2]

Seems to me like the most straightforward way to go is to break out the pencil and paper and crank it through. Start with the most generic matrix possible:

\(\displaystyle X = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}\)

Then do matrix multiplication per the usual rules to end up with two matrices in terms of these unknown parameters, on either side of the equation:

\(\displaystyle \begin{bmatrix} -2 & 2 \\ -6 & 5 \end{bmatrix}X = \begin{bmatrix} \text{??} & \text{??} \\ \text{??} & \text{??} \end{bmatrix}\)

\(\displaystyle \begin{bmatrix} -8 & 4 \\ -6 & -5 \end{bmatrix}X = \begin{bmatrix} \text{???} & \text{???} \\ \text{???} & \text{???} \end{bmatrix}\)

And finally perform the matrix addition per the usual rules:

\(\displaystyle \begin{bmatrix} \text{??} & \text{??} \\ \text{??} & \text{??} \end{bmatrix} + \begin{bmatrix} 6 & -8 \\ 7 & -6 \end{bmatrix} = \begin{bmatrix} \text{???} & \text{???} \\ \text{???} & \text{???} \end{bmatrix}\)

...to create four equations in four unknowns (the parameters \(\displaystyle x_1 \:, x_2, \: x_3, \: x_4\))

 

[Steven G]

AX + B = CX
CX-AX = B
(C-A)X = B
X = (C-A)^-1B

So basically find (C-A) and then find its inverse

 

[sadzade]

Thanks For the replies. Have a nice day.