Matrix arithmetic: Let A, C be 3x3 matrices. For any A, find C when CA = 2C.

[TNA]

Hello all,

trying to solve this matrix problem, follows like that:

two matrix 3x3 A and C, for any A find C when CA=2C,

I tried to apply an identity like A^{-1}*A=I, however I can't solve it, looks simply however I'm missing something,

I appreciate any thoughts, Thanks!

 

[Ishuda]

Hello all,

trying to solve this matrix problem, follows like that:

two matrix 3x3 A and C, for any A find C when CA=2C,

I tried to apply an identity like A^{-1}*A=I, however I can't solve it, looks simply however I'm missing something,

I appreciate any thoughts, Thanks!

Suppose you assumed C was invertible and applied C^-1*C= I, i.e. Left multiplied by C^-1

 

[TNA]

Suppose you assumed C was invertible and applied C^-1*C= I, i.e. Left multiplied by C^-1

Yes I did this way, however i think its leads to nowhere,
-> A*I= 2*I -> A=2;
-> C2=2C -> C=C ; :-(

 

[ksdhart2]

Well, for one thing, you should know right away that answer can't be right because it doesn't make sense. A is a matrix, and 2 is a scalar. It is therefore impossible to have A = 2. In addition, you appear to have forgotten that matrices do not obey the commutative property of multiplication. That is to say, for matrices A and B, \(\displaystyle AB \ne BA\). Because order of the terms matters, you have to be careful when multiplying.

The original equation was \(\displaystyle CA = 2C\). For now we're assuming that C is invertible. If we do that, we can multiply both sides of the equation by C^-1. But, here's the important part: Are we left-multiplying or right-multiplying? I'll outline what would happen if you did both, and let you pick which one gives the desired result. But be sure to note that if you left-multiply on one side of the equation, you must also left-multiply on the other side. You absolutely cannot mix and match.

If we left-multiply, we'll get: \(\displaystyle C^{-1}CA = C2C\). If we right-multiply, we'll get: \(\displaystyle CAC^{-1}=2CC\).