Examples

Basic LU factorization

Basic LU factorization

Problem: Show how to obtain the LU factorization:

$$A=\left(\begin{array}{rrrr}2& -3& 5& -8\\ 4& -5& 11& -16\\ 6& -4& 17& -26\\ 8& -6& 5& -45\end{array}\right)=\left(\begin{array}{rrrr}1& 0& 0& 0\\ 2& 1& 0& 0\\ 3& 5& 1& 0\\ 4& 6& 7& 1\end{array}\right)\left(\begin{array}{rrrr}2& -3& 5& -8\\ 0& 1& 1& 0\\ 0& 0& -3& -2\\ 0& 0& 0& 1\end{array}\right)=LU$$

Solution: Full steps for the product multiplication showing $A=LU$ as above can be found here. Full steps for the reverse, factoring $A$ into $LU$, can be found here. (Click “Details” to show the steps.) We explain the latter details below. You are encouraged to use MatrixCalc.org to help you master the process, but remember that on the exams you must be able to run the process by hand.

Since $A$ is a $4\times 4$ matrix, we are looking for $L_1,L_2,L_3$ such that $A=(L_1{L}_2{L}_3)U$ and each $L_i$ is a single-column row-add combo matrix. Since the factorization is given to us, we know the answers should be:

$$L_1=\left(\begin{array}{rrrr}1& 0& 0& 0\\ 2& 1& 0& 0\\ 3& 0& 1& 0\\ 4& 0& 0& 1\end{array}\right),L_2=\left(\begin{array}{rrrr}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 5& 1& 0\\ 0& 6& 0& 1\end{array}\right),L_3=\left(\begin{array}{rrrr}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 7& 1\end{array}\right).$$

Obtain these as follows. To eliminate the subdiagonal entries in the first column of $A$, we must add $-2$ times row $1$ into row $2$, then $-3$ times row $1$ into row $3$, and then $-4$ times row $1$ into row $4$.

To cook up $L_1$, we first place ones on the diagonal, zeros elsewhere (namely $I_4$). Then we arrange the first column so that it’s cleared by the same sequence of steps, except now we have $1$ in the first row whereas $A$ had $2$. Thus: $-2$ times $1$ added to $2$ gives $0$, then $-3$ times $1$ added to $3$ gives $0$, and $-4$ times $1$ added to $4$ gives $0$. The sequence $(1,2,3,4)$ is therefore appropriate for the first column in $L_2$, since it will be cleared by whatever row-adds clear the first column of $A$.

We have thus found $L_1$ as the matrix which is sent to $I_4$ by the action of the elimination process just described. This elimination process is represented by the action of the single-column row-add combo matrix:

$$L_1^{-1}=\left(\begin{array}{rrrr}1& 0& 0& 0\\ -2& 1& 0& 0\\ -3& 0& 1& 0\\ -4& 0& 0& 1\end{array}\right)$$

and therefore $L_1^{-1}$ just given is the inverse of $L_1$ that we sought: since $L_1^{-1}{L}_1=I_4$.

After applying the process to $A$ and finding $L_1$, we proceed to find $L_2$, creating its column $1,-5,-6$ in order to clear the $5,6$ from the second column of $L_1^{-1}A$:

$$L_1^{-1}A=\left(\begin{array}{rrrr}1& 0& 0& 0\\ -2& 1& 0& 0\\ -3& 0& 1& 0\\ -4& 0& 0& 1\end{array}\right)\left(\begin{array}{rrrr}2& -3& 5& -8\\ 4& -5& 11& -16\\ 6& -4& 17& -26\\ 8& -6& 5& -45\end{array}\right)=\left(\begin{array}{rrrr}2& -3& 5& -8\\ 0& 1& 1& 0\\ 0& 5& 2& -2\\ 0& 6& -15& -13\end{array}\right)$$ $$L_2^{-1}(L_1^{-1}A)=\left(\begin{array}{rrrr}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& -5& 1& 0\\ 0& -6& 0& 1\end{array}\right)\left(\begin{array}{rrrr}2& -3& 5& -8\\ 0& 1& 1& 0\\ 0& 5& 2& -2\\ 0& 6& -15& -13\end{array}\right)=\left(\begin{array}{rrrr}2& -3& 5& -8\\ 0& 1& 1& 0\\ 0& 0& -3& -2\\ 0& 0& -21& -13\end{array}\right).$$

Therefore

$$L_2=\left(\begin{array}{rrrr}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 5& 1& 0\\ 0& 6& 0& 1\end{array}\right)={\left(\begin{array}{rrrr}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& -5& 1& 0\\ 0& -6& 0& 1\end{array}\right)}^{-1}$$

is the right choice for $L_2$, and we are almost finished. Observe that $-21=-3\cdot 7$. Thus:

$$L_3^{-1}(L_2^{-1}{L}_1^{-1}A)=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& -7& 1\end{array}\right)\left(\begin{array}{cccc}2& -3& 5& -8\\ 0& 1& 1& 0\\ 0& 0& -3& -2\\ 0& 0& -21& -13\end{array}\right)=\left(\begin{array}{cccc}2& -3& 5& -8\\ 0& 1& 1& 0\\ 0& 0& -3& -2\\ 0& 0& 0& 1\end{array}\right)$$

and so

$$L_3=\left(\begin{array}{rrrr}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 7& 1\end{array}\right)={\left(\begin{array}{rrrr}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& -7& 1\end{array}\right)}^{-1}.$$