Notation: $PA=LU$ where $P$ is the permutation matrix. Efficiently solve $A^Tx=b$

Question

$U$ is upper triangular so $U^T=U$ and $L$ is lower triangular.

I'm told to derive an expression for $A^{-T}$ which I don't know the meaning of? Is it $(A^T)^{-1}$?

If it's the second, $A^T=UL{P^{-1}}^T$ hence ${A^T}^{-1}=P^TL^{-1}U^{-1}$

Then I simply multiply it by $b$ to solve for $x$?

Answer 1

$U$ is an upper triangular so $U^T=U$ is not a true statement.

$$A^{-T}=(A^T)^{-1}$$

$$PA=LU$$

$$A^TP^T=U^TL^T$$

$$A^T=U^TL^TP^{-T}$$

$$(A^T)^{-1}=P^TL^{-T}U^{-T}$$

Interpret these results as forward and backward substitutions rather than general matrix multiplication.