Question

Does the following matrix have an LU factorization? See Exercises 2.4.90 and 2.4.93.

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

Short answer

Yes, the given matrix does have an LU factorization.

Step-by-step solution

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 “m by n” matrix, written “$m\times n$.”

Given. 

Given matrix,

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

To find the given matrix is LU factorization or not. 

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

We have,

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

To solve,

$$\begin{gathered} det\left(A^{\left(1\right)}\right)=7\ne 0 \\ det\left(A^{\left(2\right)}\right)=1\ne 0 \\ det\left(A^{\left(3\right)}\right)=1\ne 0 \end{gathered}$$

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

Thus, $A$does have an LU factorization.