The Timpson Lyceum · Linear Algebra (2026)

2.5 Exercises

Problem sheet · 20 points · 4 extra credit

Matrix Factorizations

1. 1. (3 pts) Find an LU factorization $A = LU$ (with $L$ unit lower triangular and $U$ upper triangular) for

   $A = \begin{bmatrix} 2 & 1 & 1 \\ 4 & 3 & 3 \\ 8 & 7 & 9 \end{bmatrix},$

   recording the row-reduction multipliers in $L$.

2. 2. (3 pts) Using the factorization from Problem 1, solve $A\mathbf{x} = \mathbf{b}$ for $\mathbf{b} = \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}$ by forward substitution ($L\mathbf{y} = \mathbf{b}$) followed by back substitution ($U\mathbf{x} = \mathbf{y}$).

3. 3. (3 pts) Find an LU factorization of

   $A = \begin{bmatrix} 3 & -1 & 1 \\ 9 & -1 & 4 \\ -6 & 4 & 1 \end{bmatrix}.$

4. 4. (2 pts) Suppose you must solve $A\mathbf{x} = \mathbf{b}$ for the same $n\times n$ matrix $A$ but many different right-hand sides $\mathbf{b}$. Explain why computing an LU factorization once, then reusing it, is more efficient than row reducing $[\,A\mid\mathbf{b}\,]$ from scratch each time.

5. 5. (3 pts) State whether each is true or false, and justify.

   (a) Every square matrix has an LU factorization.

   (b) In an LU factorization with $L$ unit lower triangular, where each elimination step has the form $R_i \to R_i - m\,R_j$, the entries of $L$ below the diagonal are the negatives of the multipliers $m$.

   (c) If $A = LU$ with $U$ having no zero on its diagonal, then $A$ is invertible.

6. 6. (3 pts) Explain why, when $A$ can be reduced to echelon form using only row-replacement operations that add a multiple of a higher row to a lower row (no interchanges, no scaling), the resulting factorization $A = LU$ has $L$ unit lower triangular.

7. 7. (3 pts) The matrix $A = \begin{bmatrix} 0 & 1 \\ 2 & 3 \end{bmatrix}$ has no LU factorization with $L$ unit lower triangular. Explain why, and give a permutation matrix $P$ so that $PA$ does have such a factorization; exhibit $PA = LU$.

8. 8. (4 pts ••) Prove that an LU factorization is unique in the following sense: if $A = L_1 U_1 = L_2 U_2$, where $L_1, L_2$ are unit lower triangular and $U_1, U_2$ are invertible upper triangular, then $L_1 = L_2$ and $U_1 = U_2$.