Matrix Problem: solve for X where AX = XB

[enott312]

I need to solve for X where AX = XB

Unfortunately I don't know how to code the matrix on this forum

Matrix A
3 2
-1 0

Matrix B
0 1
2 -1

Having just started to learn how to use matrices I am not sure what to do.
My instincts say to bring XB to same side as AX and have AX-XB = 0 then factorise but I know that matrices are not communicative under multiplication.
Any help would be most welcome.

 

[HallsofIvy]

I need to solve for X where AX = XB

Unfortunately I don't know how to code the matrix on this forum

Using "Latex" you would write your matrix A as
"\begin{bmatrix}3 & 2 \\ 1 & 0 \end{bmatrix}"
with "[t e x]" at the beginning and "[/t e x]" (without the spaces) at the end.
That gives \(\displaystyle \begin{bmatrix}3 & 2 \\ 1 & 0 \end{bmatrix}\).
(Using "pmatrix" rather than "bmatrix" gives \(\displaystyle \begin{pmatrix}3 & 2 \\ 1 & 0 \end{pmatrix}\).)
Check this: https://www.latex-tutorial.com/tutorials/

Matrix A
3 2
-1 0

Matrix B
0 1
2 -1

Having just started to learn how to use matrices I am not sure what to do.
My instincts say to bring XB to same side as AX and have AX-XB = 0 then factorise but I know that matrices are not communicative under multiplication.
Any help would be most welcome.

First, in order that both those multiplications exist, X must be a 2 by 2 matrix. The simplest thing to do is to write \(\displaystyle X= \begin{bmatrix}w & x \\ y & z\end{bmatrix}\) and actually do the multiplications:
\(\displaystyle AX= \begin{bmatrix}3 & 2 \\ 1 & 0 \end{bmatrix}\begin{bmatrix}w & x \\ y & z\end{bmatrix}= \begin{bmatrix}3w+ 2y & 3x+ 2z \\ w & x\end{bmatrix}\)
and
\(\displaystyle XY= \begin{bmatrix}w & x \\ y & z\end{bmatrix}\begin{bmatrix}0 & 1 \\ 2 & -1 \end{bmatrix}= \begin{bmatrix}2x & w- x \\ 2z & y- z\end{bmatrix}\)

So AX= XY is the matrix equation \(\displaystyle \begin{bmatrix}3w+ 2y & 3x+ 2z \\ w & x\end{bmatrix}=\begin{bmatrix}2x & w- x \\ 2z & y- z\end{bmatrix}\)

and, since two matrices are equal if and only if "corresponding terms" are equal, that gives the four numerical equations
3w+ 2y= 2x
3x+ 2z= w- x
w= 2z and
x= y- z to solve for the four unknowns, w, x, y, and z.

 

[enott312]

Using "Latex" you would write your matrix A as
"\begin{bmatrix}3 & 2 \\ 1 & 0 \end{bmatrix}"
with "[t e x]" at the beginning and "[/t e x]" (without the spaces) at the end.
That gives \(\displaystyle \begin{bmatrix}3 & 2 \\ 1 & 0 \end{bmatrix}\).
(Using "pmatrix" rather than "bmatrix" gives \(\displaystyle \begin{pmatrix}3 & 2 \\ 1 & 0 \end{pmatrix}\).)
Check this: https://www.latex-tutorial.com/tutorials/


First, in order that both those multiplications exist, X must be a 2 by 2 matrix. The simplest thing to do is to write \(\displaystyle X= \begin{bmatrix}w & x \\ y & z\end{bmatrix}\) and actually do the multiplications:
\(\displaystyle AX= \begin{bmatrix}3 & 2 \\ 1 & 0 \end{bmatrix}\begin{bmatrix}w & x \\ y & z\end{bmatrix}= \begin{bmatrix}3w+ 2y & 3x+ 2z \\ w & x\end{bmatrix}\)
and
\(\displaystyle XY= \begin{bmatrix}w & x \\ y & z\end{bmatrix}\begin{bmatrix}0 & 1 \\ 2 & -1 \end{bmatrix}= \begin{bmatrix}2x & w- x \\ 2z & y- z\end{bmatrix}\)

So AX= XY is the matrix equation \(\displaystyle \begin{bmatrix}3w+ 2y & 3x+ 2z \\ w & x\end{bmatrix}=\begin{bmatrix}2x & w- x \\ 2z & y- z\end{bmatrix}\)

and, since two matrices are equal if and only if "corresponding terms" are equal, that gives the four numerical equations
3w+ 2y= 2x
3x+ 2z= w- x
w= 2z and
x= y- z to solve for the four unknowns, w, x, y, and z.

Thank you for that, on both counts. I was trying to find a numerical answer until it finally twigged that I needed an algebraic on. My solution comes out to \(\displaystyle \begin{bmatrix}2x & x \\ -2x & -x\end{bmatrix}\)