For each of the listed matrices A and vectors 𝐛, find a permuted L U factorization of the matrix

step 1 Okay, so part A here, we consider this linear system. We have 2x1 minus 3x2 plus 5x3 is equal to

For each of the listed matrices A and vectors 𝐛, find a...

For each of the listed matrices $A$ and vectors $\mathbf{b}$, find a permuted $L U$ factorization of the matrix, and use your factorization to solve the system $A \mathbf{x}=\mathbf{b}$. (a) $\left(\begin{array}{lr}0 & 1 \\ 2 & -1\end{array}\right),\left(\begin{array}{l}3 \\ 2\end{array}\right)$, (b) $\left(\begin{array}{rrr}0 & 0 & -4 \\ 1 & 2 & 3 \\ 0 & 1 & 7\end{array}\right),\left(\begin{array}{r}1 \\ 2 \\ -1\end{array}\right)$, (c) $\left(\begin{array}{rrr}0 & 1 & -3 \\ 0 & 2 & 3 \\ 1 & 0 & 2\end{array}\right),\left(\begin{array}{r}1 \\ 2 \\ -1\end{array}\right)$, (d) $\left(\begin{array}{rrrr}1 & 2 & -1 & 0 \\ 3 & 6 & 2 & -1 \\ 1 & 1 & -7 & 2 \\ 1 & -1 & 2 & 1\end{array}\right),\left(\begin{array}{l}1 \\ 0 \\ 0 \\ 3\end{array}\right)$, (e) $\left(\begin{array}{rrrr}0 & 1 & 0 & 0 \\ 2 & 3 & 1 & 0 \\ 1 & 4 & -1 & 2 \\ 7 & -1 & 2 & 3\end{array}\right),\left(\begin{array}{r}-1 \\ -4 \\ 0 \\ 5\end{array}\right)$, (f) $\left(\begin{array}{rrrrr}0 & 0 & 2 & 3 & 4 \\ 0 & 1 & -7 & 2 & 3 \\ 1 & 4 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 7 & 3\end{array}\right),\left(\begin{array}{r}-3 \\ -2 \\ 0 \\ 0 \\ -7\end{array}\right)$.

For each of the listed matrices $A$ and vectors $\mathbf{b}$, find a permuted $L U$ factorization of the matrix, and use your factorization to solve the system $A \mathbf{x}=\mathbf{b}$. (a) $\left(\begin{array}{lr}0 & 1 \\ 2 & -1\end{array}\right),\left(\begin{array}{l}3 \\ 2\end{array}\right)$,
(b) $\left(\begin{array}{rrr}0 & 0 & -4 \\ 1 & 2 & 3 \\ 0 & 1 & 7\end{array}\right),\left(\begin{array}{r}1 \\ 2 \\ -1\end{array}\right)$,
(c) $\left(\begin{array}{rrr}0 & 1 & -3 \\ 0 & 2 & 3 \\ 1 & 0 & 2\end{array}\right),\left(\begin{array}{r}1 \\ 2 \\ -1\end{array}\right)$,
(d) $\left(\begin{array}{rrrr}1 & 2 & -1 & 0 \\ 3 & 6 & 2 & -1 \\ 1 & 1 & -7 & 2 \\ 1 & -1 & 2 & 1\end{array}\right),\left(\begin{array}{l}1 \\ 0 \\ 0 \\ 3\end{array}\right)$,
(e) $\left(\begin{array}{rrrr}0 & 1 & 0 & 0 \\ 2 & 3 & 1 & 0 \\ 1 & 4 & -1 & 2 \\ 7 & -1 & 2 & 3\end{array}\right),\left(\begin{array}{r}-1 \\ -4 \\ 0 \\ 5\end{array}\right)$,
(f) $\left(\begin{array}{rrrrr}0 & 0 & 2 & 3 & 4 \\ 0 & 1 & -7 & 2 & 3 \\ 1 & 4 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 7 & 3\end{array}\right),\left(\begin{array}{r}-3 \\ -2 \\ 0 \\ 0 \\ -7\end{array}\right)$.

Key Concepts

LU Factorization

LU factorization is the process of decomposing a matrix A into the product of a lower triangular matrix L and an upper triangular matrix U. This decomposition is fundamental in numerical linear algebra because it allows for efficient solutions of systems of linear equations, computation of determinants, and matrix inversions. In many practical situations, modifications such as permutation are required to enable the factorization process, especially when encountering zero or small pivot elements.

Permutation Matrix and Partial Pivoting

A permutation matrix is used to rearrange the rows (or columns) of a matrix and is key in the context of LU factorization when the factorization without row interchanges fails or becomes numerically unstable. Partial pivoting involves reordering the rows via a permutation matrix to position the largest available pivot at the diagonal during the factorization process, thereby enhancing numerical stability and ensuring that the LU decomposition is valid even if the original matrix has zero or near-zero entries on its diagonal.

Forward Substitution

Forward substitution is the method used to solve a system of equations when given a lower triangular matrix L in the factorized system L*y = P*b. Because L has zeros above the diagonal, the solution proceeds step-by-step, starting from the first equation and substituting the already computed variables into subsequent equations. This process efficiently computes the intermediary vector y necessary for solving the complete system.

Backward Substitution

Backward substitution is the process applied after forward substitution to solve the system U*x = y, where U is an upper triangular matrix having zeros below the diagonal. Beginning from the last equation and progressing upward, variables are computed in reverse order, making use of the previously computed values. This step completes the solution of the original linear system.

Solving Linear Systems Using Factorizations

The overall strategy for solving a linear system A*x = b using LU factorization involves first finding the LU (or permuted LU) decomposition of A, then solving the system in two steps: first by using forward substitution on L*y = P*b (if a permutation is applied), and then by using backward substitution on U*x = y. This layered approach leverages the structure of triangular matrices to reduce computational complexity and increase numerical robustness.

Transcript

00:01 Okay, so part a here, we consider this linear system.

00:03 We have 2x1 minus 3x2 plus 5x3 is equal to 7, 9x1 minus x2 plus x3 is equal to negative 1, and x1 plus 5x2 plus 4x3 is equal to 0.

00:16 So we're trying to find matrices a, x, and b, such that the linear system can be expressed as a single matrix equation ax is equal to b.

00:28 Okay.

00:30 So we know that two matrices are equal if the corresponding entries are equal.

00:37 So we can replace the three equations in this system by, well, a single matrix equation.

00:44 We can have, so the 2x1 minus 3x2 plus 5x2, right, is equal to basically a column matrix on the right.

00:58 And while we can then factor out the x1 plus x2 plus x3, right, we can have x1 times the first column matrix.

01:16 So x1 times 291, the column matrix 291 plus x2 times the column matrix negative 3, negative 1, and so on, plus the x3 times the 514 column matrix.

01:30 And then, while we get our matrix here of 2, negative 3, 5, right, which is just basically all, just the coefficients in our equation.

01:44 So it's our matrix.

01:45 We have 9, negative 1, 1, 1, 5, 4.

01:53 So there's basically our matrix, and then times, while times the column matrix, x1, x2, x3.

02:01 Is going to get us back to our system of equations.

02:05 And then this is just equal to, well, to the b is just what we're equal to here.

02:10 The column matrix 7 -1 -0...

Please give Ace some feedback