In an LU decomposition \(A=LU\), which factor is lower triangular and which is upper triangular?

Under the course convention, what entries appear on the diagonal of \(L\) in an LU decomposition?

If \(A=LU\) and \(A\mathbf{x}=\mathbf b\), what two triangular systems are solved?

Why is LU decomposition useful when many systems have the same coefficient matrix \(A\)?

Solve \(L\mathbf u=\mathbf b\) for \(L=\begin{pmatrix}1&0\\3&1\end{pmatrix}\) and \(\mathbf b=\begin{pmatrix}2\\11\end{pmatrix}\).

Solve \(U\mathbf{x}=\mathbf u\) for \(U=\begin{pmatrix}2&1\\0&3\end{pmatrix}\) and \(\mathbf u=\begin{pmatrix}7\\9\end{pmatrix}\).

Given \(L=\begin{pmatrix}1&0\\2&1\end{pmatrix}\) and \(U=\begin{pmatrix}3&4\\0&5\end{pmatrix}\), compute \(A=LU\).

For \(A=LU\), solve using \(L=\begin{pmatrix}1&0\\4&1\end{pmatrix}\), \(U=\begin{pmatrix}2&1\\0&3\end{pmatrix}\), and \(\mathbf b=\begin{pmatrix}5\\23\end{pmatrix}\).

Solve \(A\mathbf{x}=\mathbf b\) for \(A=LU\), where \(L=\begin{pmatrix}1&0&0\\2&1&0\\-1&3&1\end{pmatrix}\), \(U=\begin{pmatrix}2&1&0\\0&1&4\\0&0&3\end{pmatrix}\), and \(\mathbf b=\begin{pmatrix}2\\5\\10\end{pmatrix}\).

Find an LU decomposition with unit diagonal \(L\) for \(A=\begin{pmatrix}2&1\\6&5\end{pmatrix}\).

Explain why forward substitution is used before backward substitution in an LU solve.

A student solves \(U\mathbf{x}=\mathbf b\) first when \(A=LU\). Explain the mistake.

Find a unit-diagonal LU decomposition for \(A=\begin{pmatrix}1&2&0\\3&7&1\\2&5&4\end{pmatrix}\).

Use the LU factors from \(A=\begin{pmatrix}1&2&0\\3&7&1\\2&5&4\end{pmatrix}\), \(L=\begin{pmatrix}1&0&0\\3&1&0\\2&1&1\end{pmatrix}\), \(U=\begin{pmatrix}1&2&0\\0&1&1\\0&0&3\end{pmatrix}\), to solve \(A\mathbf{x}=\begin{pmatrix}1\\5\\8\end{pmatrix}\).

For \(A=\begin{pmatrix}a&1\\2&3\end{pmatrix}\), identify when a no-row-swap LU decomposition using the first pivot fails.

For \(A=\begin{pmatrix}1&k\\2&4\end{pmatrix}\), find when the second pivot in no-row-swap LU is zero.

A single LU factorisation of \(A\) is used to solve two systems with right-hand sides \(\mathbf b_1\) and \(\mathbf b_2\). Describe exactly what work is reused and what work must be repeated.

Diagnose this error: "Since \(A=LU\), we may set \(\mathbf u=L\mathbf{x}\) and solve \(U\mathbf u=\mathbf b\)."

Explain why zero pivots matter in LU decomposition even if the original system might still be solvable.

Show why \(A=LU\) with triangular factors makes solving easier than multiplying out \(A\) and row-reducing from scratch each time.