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Chapter 8: Problem 37

Make an \(L U\) decomposition of the matrix $$ A=\left(\begin{array}{ccc} 3 & 6 & 9 \\ 1 & 0 & 5 \\ 2 & -2 & 16 \end{array}\right) $$ and hence solve \(A x=b\), where (i) \(b=\left(\begin{array}{lll}21 & 9 & 28\end{array}\right)^{\mathrm{T}}\), (ii) \(b=\left(\begin{array}{lll}21 & 7 & 22\end{array}\right)^{\mathrm{T}}\).

Short Answer

Expert verified
Part (i): \(x = \left( \begin{array}{c} 4.75 \ 0.75 \ 0.25 \end{array} \right)\)Part (ii): \(x = \left( \begin{array}{c} -1 \ 1 \ 2 \end{array} \right)\)

Step by step solution

01

- Understand the task

We need to decompose matrix A into a product of a lower triangular matrix L and an upper triangular matrix U. Then, we need to solve the equation system for two different vectors b using this decomposition.
02

- Write matrix A

Matrix A is given as: \[ A = \left( \begin{array}{ccc} 3 & 6 & 9 \ 1 & 0 & 5 \ 2 & -2 & 16 \end{array} \right) \]
03

- Initialize L and U

Start by setting U equal to A and L as an identity matrix of the same size: \[ U = A = \left( \begin{array}{ccc} 3 & 6 & 9 \ 1 & 0 & 5 \ 2 & -2 & 16 \end{array} \right), \quad L = \left( \begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array} \right) \]
04

- Eliminate element U[2,1] with row operations

To make the element U[2,1] zero, we can subtract \(\frac{1}{3} \cdot row 1\) from row 2. This gives us an updated U and updates L accordingly: \[ L = \left( \begin{array}{ccc} 1 & 0 & 0 \ \frac{1}{3} & 1 & 0 \ 0 & 0 & 1 \end{array} \right), \quad U = \left( \begin{array}{ccc} 3 & 6 & 9 \ 0 & -2 & 2 \ 2 & -2 & 16 \end{array} \right) \]
05

- Eliminate element U[3,1] with row operations

Next, to make U[3,1] zero, we can subtract \(\frac{2}{3} \cdot row 1\) from row 3. Update U and L accordingly: \[ L = \left( \begin{array}{ccc} 1 & 0 & 0 \ \frac{1}{3} & 1 & 0 \ \frac{2}{3} & 0 & 1 \end{array} \right), \quad U = \left( \begin{array}{ccc} 3 & 6 & 9 \ 0 & -2 & 2 \ 0 & -6 & 10 \end{array} \right) \]
06

- Eliminate element U[3,2] with row operations

Finally, make U[3,2] zero by subtracting 3 \(\cdot row 2\) from row 3 and update matrices L and U: \[ L = \left( \begin{array}{ccc} 1 & 0 & 0 \ \frac{1}{3} & 1 & 0 \ \frac{2}{3} & 3 & 1 \end{array} \right), \quad U = \left( \begin{array}{ccc} 3 & 6 & 9 \ 0 & -2 & 2 \ 0 & 0 & 4 \end{array} \right) \]
07

- Check if LU decomposition is correct

Verify that \(LU = A\): \[ L = \left( \begin{array}{ccc} 1 & 0 & 0 \ \frac{1}{3} & 1 & 0 \ \frac{2}{3} & 3 & 1 \end{array} \right), \quad U = \left( \begin{array}{ccc} 3 & 6 & 9 \ 0 & -2 & 2 \ 0 & 0 & 4 \end{array} \right) \]
08

- Solving Ax=b using L and U (part i)

For \(b = \left( \begin{array}{c} 21 \ 9 \ 28 \end{array} \right)\), first solve \(Ly=b\) by forward substitution: 1. \(y_1 = 21\) 2. \(\frac{1}{3}y_1 + y_2 = 9\) which implies \( y_2 = 2\) 3. \(\frac{2}{3}y_1 + 3y_2 + y_3 = 28\) which implies \( y_3 = 1 \).Thus, \(y = \left( \begin{array}{c} 21 \ 2 \ 1 \end{array} \right)\). Now solve \(Ux=y\) using back substitution: 1. \(4x_3 = 1\), thus \(x_3 = \frac{1}{4}\) 2. \(-2x_2 + 2x_3 = 2\), thus \(x_2 = 0.75\) 3. \(3x_1 + 6x_2 + 9x_3 = 21 \), thus \(x_1 = 4.75\). So, \(x = \left( \begin{array}{c} 4.75 \ 0.75 \ 0.25 \end{array} \right)\)
09

- Solving Ax=b using L and U (part ii)

For \(b = \left( \begin{array}{c} 21 \ 7 \ 22 \end{array} \right)\), first solve \(Ly=b\) by forward substitution: 1. \(y_1 = 21\) 2. \(\frac{1}{3}y_1 + y_2 = 7\) which implies \( y_2 = 0\) 3. \(\frac{2}{3}y_1 + 3y_2 + y_3 = 22\) which implies \( y_3 = 8 \). Thus, \(y = \left( \begin{array}{c} 21 \ 0 \ 8 \end{array} \right)\). Now solve \(Ux=y\) using back substitution: 1. \(4x_3 = 8\), thus \(x_3 = 2\) 2. \(-2x_2 + 2x_3 = 0\), thus \(x_2 = 1\) 3. \(3x_1 + 6x_2 + 9x_3 = 21\), thus \(x_1 = -1\). So, \(x = \left( \begin{array}{c} -1 \ 1 \ 2 \end{array} \right)\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Decomposition
Matrix decomposition is a crucial concept in linear algebra. It involves breaking down a matrix into simpler, easily handled components. One common method is the LU decomposition.
LU decomposition factors a matrix A into two matrices: L (lower triangular) and U (upper triangular). This simplifies solving linear equations and other operations.
The method is practical in numerical analysis for matrix inversion and determinant calculation. It also helps in solving systems of linear equations by allowing easier processes like forward and back substitution.
This technique is pivotal in computational mathematics, offering efficiency and accuracy.
Linear Algebra
Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It's foundational for understanding complex structures and solutions.
Matrix operations, determinants, vector spaces, and eigenvalues are core subjects. These elements are essential in various fields such as physics, engineering, and computer science.
LU decomposition is a method within linear algebra that demonstrates the breakdown of matrices for simplified computation. It optimizes solving systems of equations, making linear algebra highly applicable in real-world problems and technologies.
Gaussian Elimination
Gaussian elimination is a method for solving linear systems. It transforms a matrix into a row echelon form using row operations which simplifies solving the system.
The process works by:
  • Swapping rows
  • Multiplying rows by a scalar
  • Adding or subtracting rows

In LU decomposition, Gaussian elimination facilitates the creation of matrix U while matrix L tracks the transformations. This highlights its importance in matrix factorization and simplifying the resolution of multiple equations.
Forward Substitution
Forward substitution is used to solve lower triangular systems. After performing LU decomposition, you’re left with Ly = b, where L is a lower triangular matrix and y is an intermediate vector.
You start solving from the top of the matrix (first row), moving downwards:
  • Solve for the first variable (simple as there's no earlier variable)
  • Progressively substitute known values to solve subsequent variables

This step ensures all intermediate variables are found, leading into back substitution.
Back Substitution
Back substitution is utilized after forward substitution. Here, we solve the upper triangular system Ux = y, where U is the upper triangular matrix formed during LU decomposition.
It solves systems from the bottom up:
  • Start with the last variable (easiest as it's directly isolated)
  • Use the value of the last variable to find the second to last, and so forth, progressively substituting backwards

This method ensures all original variables are determined, completing the solution of the original linear system.

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Most popular questions from this chapter

The four matrices \(\mathrm{S}_{x}, \mathrm{~S}_{y}, \mathrm{~S}_{z}\) and \(\mathrm{I}\) are defined by $$ \begin{array}{ll} \mathrm{S}_{x}=\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right), & \mathrm{S}_{y}=\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right) \\ \mathrm{S}_{z} & =\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right), & \mathrm{I}=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) \end{array} $$ where \(i^{2}=-1\). Show that \(\mathrm{S}_{x}^{2}=\mathrm{I}\) and \(\mathrm{S}_{x} \mathrm{~S}_{y}=i \mathrm{~S}_{z}\), and obtain similar results by permutting \(x, y\) and \(z\). Given that \(\mathbf{v}\) is a vector with Cartesian components \(\left(v_{x}, v_{y}, v_{z}\right)\), the matrix \(\mathrm{S}(\mathbf{v})\) is defined as $$ \mathrm{S}(\mathbf{v})=v_{x} \mathrm{~S}_{x}+v_{y} \mathrm{~S}_{y}+v_{z} \mathrm{~S}_{z} $$ Prove that, for general non-zero vectors a and b, $$ \mathrm{S}(\mathbf{a}) \mathrm{S}(\mathbf{b})=\mathbf{a} \cdot \mathbf{b} \mid+i \mathrm{~S}(\mathbf{a} \times \mathbf{b}) $$ Without further calculation, deduce that \(\mathrm{S}(\mathbf{a})\) and \(\mathrm{S}(\mathbf{b})\) commute if and only if a and \(\mathbf{b}\) are parallel vectors.

One method of determining the nullity (and hence the rank) of an \(M \times N\) matrix A is as follows. \- Write down an augmented transpose of \(A\), by adding on the right an \(N \times N\) unit matrix and thus producing an \(N \times(M+N)\) array \(\mathrm{B}\). \- Subtract a suitable multiple of the first row of \(B\) from each of the other lower rows so as to make \(B_{i 1}=0\) for \(i>1\) \- Subtract a suitable multiple of the second row (or the uppermost row that does not start with \(M\) zero values) from each of the other lower rows so as to make \(B_{i 2}=0\) for \(i>2\) \- Continue in this way until all remaining rows have zeroes in the first \(M\) places. The number of such rows is equal to the nullity of \(A\) and the \(N\) rightmost entries of these rows are the components of vectors that span the null space. They can be made orthogonal if they are not so already. Use this method to show that the nullity of $$ A=\left(\begin{array}{cccc} -1 & 3 & 2 & 7 \\ 3 & 10 & -6 & 17 \\ -1 & -2 & 2 & -3 \\ 2 & 3 & -4 & 4 \\ 4 & 0 & -8 & -4 \end{array}\right) $$ is 2 and that an orthogonal base for the null space of \(A\) is provided by any two column matrices of the form \(\left(\begin{array}{lll}2+\alpha_{i} & -2 \alpha_{i} & 1 & \alpha_{i}\end{array}\right)^{\mathrm{T}}\) for which the \(\alpha_{i}(i=1,2)\) are real and satisfy \(6 \alpha_{1} \alpha_{2}+2\left(\alpha_{1}+\alpha_{2}\right)+5=0\).

The commutator [X, Y] of two matrices is defined by the equation $$ [X, Y]=X Y-Y X $$ Two anti-commuting matrices \(A\) and \(B\) satisfy $$ \mathrm{A}^{2}=\mathrm{I}, \quad \mathrm{B}^{2}=\mathrm{I}, \quad[\mathrm{A}, \mathrm{B}]=2 i \mathrm{C} $$ (a) Prove that \(\mathrm{C}^{2}=\mathrm{I}\) and that \([\mathrm{B}, \mathrm{C}]=2 i \mathrm{~A}\). (b) Evaluate \([[[A, B],[B, C]],[A, B]]\).

Show that the following equations have solutions only if \(\eta=1\) or 2 , and find them in these cases: $$ \begin{aligned} x+y+z &=1 \\ x+2 y+4 z &=\eta \\ x+4 y+10 z &=\eta^{2} \end{aligned} $$

(a) Show that if \(A\) is Hermitian and \(U\) is unitary then \(U^{-1} \mathrm{AU}\) is Hermitian. (b) Show that if \(A\) is anti-Hermitian then \(i A\) is Hermitian. (c) Prove that the product of two Hermitian matrices \(A\) and \(B\) is Hermitian if and only if \(A\) and \(B\) commute. (d) Prove that if \(\mathrm{S}\) is a real antisymmetric matrix then \(\mathrm{A}=(\mathrm{I}-\mathrm{S})(\mathrm{I}+\mathrm{S})^{-1}\) is orthogonal. If \(A\) is given by $$ A=\left(\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right) $$ then find the matrix \(\mathrm{S}\) that is needed to express \(\mathrm{A}\) in the above form. (e) If \(\mathrm{K}\) is skew-hermitian, i.e. \(\mathrm{K}^{\dagger}=-\mathrm{K}\), prove that \(\mathrm{V}=(\mathrm{I}+\mathrm{K})(\mathrm{I}-\mathrm{K})^{-1}\) is unitary.
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