I was wondering if you have a matrix (AB)t
does matrix A and B both become transposed?
like A^t and B^t?
thank you
An amount of money invested for 1 year in a savings account will earn $1,500. That same amount of money invested in a mini-mall development will earn $4,500 interest in a year, because the interest paid is 10% more than that paid by the savings account. Find the rate of interest paid by each investment.
(AB)t = (Bt)(At)
Yes, when you take the transpose of the product of two matrices A and B, denoted as (AB)^T, both matrices A and B do get transposed individually.
To explain this in more detail, let's consider two matrices A and B.
Matrix A:
A = [a₁₁ a₁₂ a₁₃
a₂₁ a₂₂ a₂₃
a₃₁ a₃₂ a₃₃]
Matrix B:
B = [b₁₁ b₁₂ b₁₃
b₂₁ b₂₂ b₂₃
b₃₁ b₃₂ b₃₃]
To find the transpose of the product (AB)^T, we first need to calculate the product AB.
Matrix product (AB):
AB = [a₁₁b₁₁ + a₁₂b₂₁ + a₁₃b₃₁ a₁₁b₁₂ + a₁₂b₂₂ + a₁₃b₃₂ a₁₁b₁₃ + a₁₂b₂₃ + a₁₃b₃₃
a₂₁b₁₁ + a₂₂b₂₁ + a₂₃b₃₁ a₂₁b₁₂ + a₂₂b₂₂ + a₂₃b₃₂ a₂₁b₁₃ + a₂₂b₂₃ + a₂₃b₃₃
a₃₁b₁₁ + a₃₂b₂₁ + a₃₃b₃₁ a₃₁b₁₂ + a₃₂b₂₂ + a₃₃b₃₂ a₃₁b₁₃ + a₃₂b₂₃ + a₃₃b₃₃]
Now, to find the transpose of (AB)^T, we need to take the transpose of the result we obtained in the previous step.
((AB)^T):
(AB)^T = [(a₁₁b₁₁ + a₁₂b₂₁ + a₁₃b₃₁) (a₁₁b₁₂ + a₁₂b₂₂ + a₁₃b₃₂) (a₁₁b₁₃ + a₁₂b₂₃ + a₁₃b₃₃)
(a₂₁b₁₁ + a₂₂b₂₁ + a₂₃b₃₁) (a₂₁b₁₂ + a₂₂b₂₂ + a₂₃b₃₂) (a₂₁b₁₃ + a₂₂b₂₃ + a₂₃b₃₃)
(a₃₁b₁₁ + a₃₂b₂₁ + a₃₃b₃₁) (a₃₁b₁₂ + a₃₂b₂₂ + a₃₃b₃₂) (a₃₁b₁₃ + a₃₂b₂₃ + a₃₃b₃₃)]
So, when you take the transpose of the product (AB)^T, each element of the resulting matrix retains the same order, but the rows become columns and the columns become rows. Matrix A and B remain unchanged individually.