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Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 6: Q55E (page 276)

Does the following matrix have an LU factorization? See Exercises 2.4.90 and 2.4.93.
$A = [\begin{matrix} 7 & 4 & 2 \\ 5 & 3 & 1 \\ 3 & 1 & 4 \end{matrix}]$

Short Answer

Expert verified

Yes, the given matrix does have an LU factorization.

Step by step solution

01

Matrix Definition.

Matrix is aset of numbers arranged in rows and columns so as to form a rectangulararray.

The numbers are called the elements, or entries, of the matrix.

If there are $m$ rows and $n$ columns, the matrix is said to be an 鈥�m by n鈥� matrix, written 鈥� $m 脳 n$ .鈥�

02

Given.

Given matrix,

$A = [\begin{matrix} 7 & 4 & 2 \\ 5 & 3 & 1 \\ 3 & 1 & 4 \end{matrix}]$

03

To find the given matrix is LU factorization or not.

As seen from Exercise 2.4.93.c, an $n 脳 n$ matrix has an LU factorization if all of its principal sub matrices are invertible.

We have,

$A^{(1)} = [7] A^{(2)} = [\begin{matrix} 7 & 4 \\ 5 & 3 \end{matrix}], A^{(3)} = [\begin{matrix} 7 & 4 & 2 \\ 5 & 3 & 1 \\ 3 & 1 & 4 \end{matrix}]$

To solve,

$d e t (A^{(1)}) = 7 鈮� 0 d e t (A^{(2)}) = 1 鈮� 0 d e t (A^{(3)}) = 1 鈮� 0$

Since,all of its principal sub matrices $A$ are invertible.

Thus, $A$ does have an LU factorization.

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Most popular questions from this chapter

An $8 脳 8$ matrix fails to be invertible if (and only if) its determinant is nonzero.

For an invertiblenxnmatrix A, what is adj(adj A)?

Let $P_{n}$ be the $n 脳 n$ matrix whose entries are all ones, except for zeros directly below the main diagonal; for example,

role="math" localid="1659508976827" $P_{5} [\begin{matrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \end{matrix}]$

Find the determinant of $P_{n}$ .

Use Gaussian elimination to find the determinant of the matrices A in Exercises 1 through 10.

4. $[\begin{matrix} 1 & - 1 & 2 & - 2 \\ - 1 & 2 & 1 & 6 \\ 2 & 1 & 14 & 10 \\ - 2 & 6 & 10 & 33 \end{matrix}]$

$d e t [\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{matrix}] = 1$

See all solutions

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