Activity 5.1.4

Sage will create $LU$ factorizations; once we have a matrix A, we write P, L, U = A.LU() to obtain the matrices $P$, $L$, and $U$ such that $PA = LU$.

1. In this example, we found the $LU$ factorization $$A = \begin{bmatrix} 2 & -3 & 1 \\ -4 & 5 & 0 \\ 2 & -2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -3 & 1 \\ 0 & -1 & 2 \\ 0 & 0 & 3 \end{bmatrix}=LU.$$ Using Sage, define the matrix $A$, and then ask Sage for the $LU$ factorization. What are the matrices $P$, $L$, and $U$?

   Notice that Sage finds a different $LU$ factorization than we found in the previous example because Sage uses partial pivoting, as described in the previous section, when it performs Gaussian elimination.

2. Define the vector $\bvec = \threevec8{-13}{8}$ in Sage and compute $P\bvec$.

3. Use the matrices L and U to solve $L\cvec = P\bvec$ and $U\xvec = \cvec$. You should find the same solution $\xvec$ that we found in the previous example.

4. Use the factorization to solve the equation $A\xvec = \threevec9{-16}{10}$.

5. How does the factorization show us that $A$ is invertible and that, therefore, every equation $A\xvec=\bvec$ has a unique solution?

6. Suppose that we have the matrix $$B = \left[\begin{array}{rrr} 3 & -1 & 2 \\ 2 & -1 & 1 \\ 2 & 1 & 3 \\ \end{array}\right]$$. Use Sage to find the $LU$ factorization. Explain how the factorization shows that $B$ is not invertible.

7. Consider the matrix $$C = \left[\begin{array}{rrrr} -2 & 1 & 2 & -1 \\ 1 & -1 & 0 & 2 \\ 3 & 2 & -1 & 0 \\ \end{array}\right]$$ and find its $LU$ factorization. Explain why $C$ and $U$ have the same null space and use this observation to find a basis for $\nul(A)$.