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

In summary, the given conversation is about proving that there are no possible values for matrices A and B (2x2 and real number matrices) that satisfy the equation AB-BA=I2. The individual attempting to solve the problem asks for guidance and writes out the full equation, which results in a contradiction. The expert confirms this and suggests moving on to the next question.
  • #1
annoymage
362
0

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?
 
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  • #2
Write down the full equation

[tex]
\left(
\begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}
\right)\left(
\begin{array}{cc}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{array}
\right)-\left(
\begin{array}{cc}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{array}
\right)\left(
\begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}
\right)=\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right)[/tex]

After you do that it will be obvious
 
  • #3
i did, but then i get,

[tex]\begin{pmatrix} bg-fc & af+bh+be+df \\ ce+dg-aq-ch & cf-bg \end{pmatrix}[/tex]

assume that my

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

then? solve the equation right?
 
  • #4
owh wait,

bg-fc=1

cf-bg=1

its contradiction..

right?
 
  • #5
Yeah that's the idea I think
 
  • #6
hoho, that's proof alright,, thank you very much

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

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

How do you prove a linear system?

To prove a linear system, you must show that all the equations in the system can be satisfied by the same set of values for the variables. This can be done by using various methods such as substitution, elimination, or graphing. If a solution exists for all equations in the system, then it is proven to be linear.

What is the difference between a consistent and inconsistent linear system?

In a consistent linear system, there exists at least one solution that satisfies all the equations in the system. In an inconsistent system, there is no solution that satisfies all the equations. This means that the equations are contradictory and cannot be solved simultaneously.

Can a linear system have more than one solution?

Yes, a linear system can have one, infinite, or no solutions. If a system has one solution, it is called a consistent system with a unique solution. If it has infinite solutions, it is called a consistent system with infinitely many solutions. If there are no solutions, it is called an inconsistent system.

How can you use matrices to prove a linear system?

Matrices can be used to represent a linear system in a compact and organized manner. By performing row operations on the matrix, such as row reduction, you can determine the solution(s) to the system. If you end up with a reduced row echelon form of the matrix, you can easily read off the solution(s) to the system.

What is the importance of proving a linear system?

Proving a linear system is important because it allows us to determine whether the system has a solution or not. This can be useful in various fields such as engineering, physics, and economics where linear systems are commonly used to model real-world situations. Knowing the properties of a linear system also helps us in solving and analyzing more complex systems.

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