In practice, one rarely finds the inverse of a matrix \(A\text{.}\) It requires considerable effort to compute, and we can solve any equation of the form \(A\xvec = \bvec\) using an \(LU\) factorization, which means that the inverse isn't necessary. In any case, the best way to compute an inverse is using an \(LU\) factorization, as this exericse demonstrates.

  1. Suppose that \(PA = LU\text{.}\) Explain why \(A^{-1} = U^{-1}L^{-1}P\text{.}\)

    Since \(L\) and \(U\) are triangular, finding their inverses is relatively efficient. That makes this an effective means of finding \(A^{-1}\text{.}\)

  2. Consider the matrix

    \begin{equation*} A = \left[\begin{array}{rrr} 3 \amp 4 \amp -1 \\ 2 \amp 4 \amp 1 \\ -3 \amp 1 \amp 4 \\ \end{array}\right] \text{.} \end{equation*}

    Find the \(LU\) factorization of \(A\) and use it to find \(A^{-1}\text{.}\)