Find all matrices [imath]B[/imath] = [imath]\begin{bmatrix} a & b \\ c & d \end{bmatrix}[/imath] such that [imath]AB=BA[/imath].

[Integrate]

Let [imath]A[/imath] = [imath]\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}[/imath]
Find all matrices [imath]B[/imath] = [imath]\begin{bmatrix} a & b \\ c & d \end{bmatrix}[/imath] such that [imath]AB=BA[/imath]. Then use the results to exhibit (2 × 2) matrices B and C such that AB = BA and AC =/ CA

[imath]\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}[/imath][imath]\begin{bmatrix} a & b \\ c & d \end{bmatrix}[/imath]=[imath]\begin{bmatrix} a & b \\ c & d \end{bmatrix}[/imath][imath]\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}[/imath]

[imath]\begin{bmatrix} a+2c & b+2d \\ 3a+4c & 3b+4d \end{bmatrix}[/imath]=[imath]\begin{bmatrix} a+3b & 2a+4b \\ c+3d & 2c+4b \end{bmatrix}[/imath]

[imath]a+2c = a+3b[/imath] , [imath]b+2d = 2a+4b[/imath]
[imath]3a+4c=c+3d [/imath] , [imath]3b+3d = 2c+4b[/imath]

[imath]2c-3b=0[/imath]
[imath]-2a-3b-d=0[/imath]
[imath]3a+3c-3d=0[/imath]
[imath]3b-2c=0[/imath]

[imath]\begin{bmatrix} 1 & 3/2 & 0 & -1 & 0\\ 1 & 0 & 1 & 1 & -1\\ 0 & 1 & -2/3 & 0 & 0\\ 0 & -3/2 & 1 & 0 & 0 \end{bmatrix}[/imath] [imath]\longrightarrow[/imath] [imath]\begin{bmatrix} 1 & 0 & 1 & -1 & 0\\ 0 & 1 & -2/3 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}[/imath] [imath]\longrightarrow[/imath] [imath]b=2/3c[/imath] and [imath]a=-c+d[/imath]



So I understand that I am supposed to plug these in to get [imath]\begin{bmatrix} -c+d & 2/3c \\ c & d\end{bmatrix}[/imath].


I understood that this is what I needed to do, but I am not sure why it works.


For the second part of this question I don't even know what it wants. Just pick a random matrix for C?

 

[Dr.Peterson]

I understood that this is what I needed to do, but I am not sure why it works.

Can you say what it means to "work"? I'm not sure what is missing.

Have you checked that any matrix of this form commutes with A, by doing the multiplication?

For the second part of this question I don't even know what it wants. Just pick a random matrix for C?

Since you now know what it takes to make B commute with A, you are to pick a pair b, c and create the corresponding matrix B, and then make a matrix C that doesn't fit this pattern (e.g. change one entry from your B so it doesn't satisfy the equations). Then calculate AB, BA, AC, and CA, and show that [imath]AB=BA[/imath] but [imath]AC\ne CA[/imath].

 

[zahid0987]

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