Matrix Algebra - Matrix Manipulation

[trough]

My matrix algebra and order of operations has gotten rusty. I want to solve for a matrix in terms of other matrices given a set of matrix equations as follows:

{A} = [D] {C}

{A} = [T] {R} and {C} = [T] {S}

{R} = [Q} {S}

solve for [Q] in terms of [D] and [T]

I ended up with {R} = inv[T] [D] [T] {S} such that [Q]=inv[T] [D] [T] but I don't believe this is correct. Could you please advise?

Thanks.

 

[topsquark]

A question about your notation. Do [A] and {A} both represent the matrix A? Capital letters will suffice.

I will assume all necessary inverses matrices exist, otherwise the whole process is hosed.

[imath]A = DC, ~ A = TR, ~ C = TS, ~ R = QS[/imath]

So [imath]Q = R S^{-1}[/imath] and [imath]R = T^{-1} A[/imath] and [imath]S = T^{-1} C[/imath], so

[imath]Q = (T^{-1} A) (T^{-1} C)^{-1} = (T^{-1} (DC)) (T^{-1} C)^{-1} = T^{-1} D C (T^{-1}C)^{-1}[/imath].

Since [imath](T^{-1} C)^{-1} = C^{-1} T[/imath] we get that [imath]Q = T^{-1} D C C^{-1} T = T^{-1} D T[/imath], so I agree with you.

-Dan

 

[trough]

A question about your notation. Do [A] and {A} both represent the matrix A? Capital letters will suffice.

I will assume all necessary inverses matrices exist, otherwise the whole process is hosed.

[imath]A = DC, ~ A = TR, ~ C = TS, ~ R = QS[/imath]

So [imath]Q = R S^{-1}[/imath] and [imath]R = T^{-1} A[/imath] and [imath]S = T^{-1} C[/imath], so

[imath]Q = (T^{-1} A) (T^{-1} C)^{-1} = (T^{-1} (DC)) (T^{-1} C)^{-1} = T^{-1} D C (T^{-1}C)^{-1}[/imath].

Since [imath](T^{-1} C)^{-1} = C^{-1} T[/imath] we get that [imath]Q = T^{-1} D C C^{-1} T = T^{-1} D T[/imath], so I agree with you.

-Dan

Yes, [A] and {A} would both represent a matrix. technically speaking the {} notation represents a 3x1 vector and the [] notation is a 3x3 matrix. So you got same answer as me but when checking using real world numbers I don't think T^{-1} D T gives the right answer. That's what I'm stuck on. I believe the right answer is T^{-1} D transpose(T^{-1}) but I don't know how you get there.

 

[Steven G]

I don't think T^{-1} D T gives the right answer.

Please think again.
Can you show us your worked out example that contradicts topsquark's solution for Q?

 

[Dr.Peterson]

Yes, [A] and {A} would both represent a matrix. technically speaking the {} notation represents a 3x1 vector and the [] notation is a 3x3 matrix.

Is this notation something you were taught, or did you invent it?

I'll use capital letters for 3x3 matrices and lower case for column vectors.

So it appears that what you wrote as

{A} = [D] {C}

{A} = [T] {R} and {C} = [T] {S}

{R} = [Q} {S}

solve for [Q] in terms of [D] and [T]

is

a = Dc, a = Tr, c = Ts, r = Qs; solve for Q in terms of D and T.​

Since topsquark assumed everything, including C (which you say is a vector) has an inverse, his work has to be modified.

Do you see how? Can you show us your work (both to get your answer, and to check it)?

 

[Steven G]

Ahh!! The inverse of a 3x1 matrix!!!!

 

[topsquark]

Yes, [A] and {A} would both represent a matrix. technically speaking the {} notation represents a 3x1 vector and the [] notation is a 3x3 matrix. So you got same answer as me but when checking using real world numbers I don't think T^{-1} D T gives the right answer. That's what I'm stuck on. I believe the right answer is T^{-1} D transpose(T^{-1}) but I don't know how you get there.

Thanks for the correction. I should have thought of that.

Anyway, I'm still getting [imath]Q = T^{-1} D T[/imath], though by a different process. I'll echo Dr.Peterson's question: Why do you think this is wrong?

-Dan