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### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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

Yes, the given matrix does have an LU factorization.

See the step by step solution

## Step 1: Matrix Definition.

Matrix is a set of numbers arranged in rows and columns so as to form a rectangular array.

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 “m by n” matrix, written “$m×n$.”

## Step 2: Given.

Given matrix,

$A=\left[\begin{array}{ccc}7& 4& 2\\ 5& 3& 1\\ 3& 1& 4\end{array}\right]$

## Step 3: To find the given matrix is LU factorization or not.

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

We have,

${A}^{\left(1\right)}=\left[7\right]\phantom{\rule{0ex}{0ex}}{A}^{\left(2\right)}=\left[\begin{array}{cc}7& 4\\ 5& 3\end{array}\right],\phantom{\rule{0ex}{0ex}}{A}^{\left(3\right)}=\left[\begin{array}{ccc}7& 4& 2\\ 5& 3& 1\\ 3& 1& 4\end{array}\right]$

To solve,

$det\left({A}^{\left(1\right)}\right)=7\ne 0\phantom{\rule{0ex}{0ex}}det\left({A}^{\left(2\right)}\right)=1\ne 0\phantom{\rule{0ex}{0ex}}det\left({A}^{\left(3\right)}\right)=1\ne 0$

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

Thus, $A$ does have an LU factorization.