From the decomposition $PAQ=LU$ of a matrix $A$, explain how to solve:
- $Ax=b$
- $A^{-1}$
- $det(A)$
($P$ and $Q$ are permutation matrixes)
When working with decompositions $A=LU$ or $PA=LU$ , this question is quite simpler. I don't quite understand how to do this.
Since we have $PAQ=LU$ we can write $A$ as $A=P^{-1}LUQ^{-1}$
So $Ax=b \Leftrightarrow (P^{-1}LUQ^{-1})x=b$
We can solve this with a system where: $Lz=Pb \wedge Uy=z \wedge x=Qy$.
I think that part is correct.
For the rest, how can I solve $A^{-1}$ and $det(A)$ using the decomposition $PAQ=LU$?
EDIT:
If I wanted to determinate $A^{-1}$ with $PA=LU$, it would be like this:
$AA^{-1}=I \Leftrightarrow PAA^{-1}=PI \Leftrightarrow PA(x_1...x_n)=(P_{e_1}...P{e_n}) \Leftrightarrow LU(x_1...x_n)=(P_{e_1}...P{e_n}) \Leftrightarrow LUx_i=P_{e_i}$
So the solution is solving this system: $L_{y_i}=P_{e_i} \wedge U_{x_i}=y_i$
However I don't know how to do this 'process' with matrix $Q$ involved.
The quetion on the determinant is more or less answered in the comments. If you use the convention that $L$ has ones on the diagonal, then $\det(A)=\pm \det(U)$ where $\pm$ is due to the fact that $\det(P)\det(U)=\pm 1$. Moreover $\det(U)=u_{11}\cdot ...\cdot u_{nn}$. What this means is that you can calculate $|\det(A)|$ easily and also you can see if $A$ is singular by just looking at the diagonal elements of $U$.
One has $PAQ=LU$ so $PA=LUQ^{-1}$ hence $P=LUQ^{-1}A^{-1}$. Now let $x_i$ be the columns of $A^{-1}$ and $p_i$ the columns of $P$ then one needs to solve $p_i=LUQ^{-1}x_i$, for $i\in \{1,...,n\}$. This can be easily solved by setting $Q^{-1}x_i=y_i$ and solving $p_i=LUy_i$. Finally notice that both $Q$ and $P$ are orthogonal matrices, hence $Q^{-1}=Q^T$.