Proving Linear System: No Solution for AB-BA=I2 (2x2 Real Number Matrices)

[annoymage]

Homework Statement

Show that there are no A,B (2x2 and real number matrices)

such that AB-BA=I₂

Homework Equations

N/A

The Attempt at a Solution

can anyone give me clue, how to prove this?

 

[Gregg]

Write down the full equation

<br /> \left(<br /> \begin{array}{cc}<br /> a_{11} &amp; a_{12} \\<br /> a_{21} &amp; a_{22}<br /> \end{array}<br /> \right)\left(<br /> \begin{array}{cc}<br /> b_{11} &amp; b_{12} \\<br /> b_{21} &amp; b_{22}<br /> \end{array}<br /> \right)-\left(<br /> \begin{array}{cc}<br /> b_{11} &amp; b_{12} \\<br /> b_{21} &amp; b_{22}<br /> \end{array}<br /> \right)\left(<br /> \begin{array}{cc}<br /> a_{11} &amp; a_{12} \\<br /> a_{21} &amp; a_{22}<br /> \end{array}<br /> \right)=\left(<br /> \begin{array}{cc}<br /> 1 &amp; 0 \\<br /> 0 &amp; 1<br /> \end{array}<br /> \right)

After you do that it will be obvious

 

[annoymage]

i did, but then i get,

\begin{pmatrix} bg-fc &amp; af+bh+be+df \\ ce+dg-aq-ch &amp; cf-bg \end{pmatrix}

assume that my

a11=a
a12=b
.
.
.
b21=g
b22=h

then? solve the equation right?

 

[annoymage]

owh wait,

bg-fc=1

cf-bg=1

its contradiction..

right?

 

[Gregg]

Yeah that's the idea I think

 

[annoymage]

hoho, that's proof alright,, thank you very much

next, can you please check my next question i posted.. ^^