## Linear Algebra and Its Applications, Exercise 1.5.4

Exercise 1.5.4. Given the matrices

$A = \begin{bmatrix} 2&1 \\ 8&7 \end{bmatrix} \quad A = \begin{bmatrix} 3&1&1 \\ 1&3&1 \\ 1&1&3 \end{bmatrix} \quad A = \begin{bmatrix} 1&1&1 \\ 1&4&4 \\ 1&4&8 \end{bmatrix}$

use elimination to find the factors L and U for each of the matrices.

Answer: For the first matrix we subtract 4 times the first row from the second to obtain U:

$\begin{bmatrix} 2&1 \\ 8&7 \end{bmatrix} \Rightarrow \begin{bmatrix} 2&1 \\ 0&3 \end{bmatrix} = U$

The other factor L then is the identity matrix with 4 in the 2,1 position:

$L = \begin{bmatrix} 1&0 \\ 4&1 \end{bmatrix}$

so that

$A = LU = \begin{bmatrix} 1&0 \\ 4&1 \end{bmatrix} \begin{bmatrix} 2&1 \\ 0&3 \end{bmatrix} = \begin{bmatrix} 2&1 \\ 8&7 \end{bmatrix}$

For the second matrix we multiply the first row by 1/3 and subtract it from the second row, and also multiply the first row by 1/3 and subtract it from the second row:

$\begin{bmatrix} 3&1&1 \\ 1&3&1 \\ 1&1&3 \end{bmatrix} \Rightarrow \begin{bmatrix} 3&1&1 \\ 0&\frac{8}{3}&\frac{2}{3} \\ 0&\frac{2}{3}&\frac{8}{3} \end{bmatrix}$

We can multiply the second row of the resulting matriix by 1/4 and subtract it from the third row to obtain the upper triangular matrix U:

$\begin{bmatrix} 3&1&1 \\ 0&\frac{8}{3}&\frac{2}{3} \\ 0&\frac{2}{3}&\frac{8}{3} \end{bmatrix} \Rightarrow \begin{bmatrix} 3&1&1 \\ 0&\frac{8}{3}&\frac{2}{3} \\ 0&0&\frac{5}{2} \end{bmatrix} = U$

The other factor L then is the identity matrix with 1/3 in the 2,1 and 3,1 positions and 1/4 in the 3,2 position:

$L = \begin{bmatrix} 1&0&0 \\ \frac{1}{3}&1&0 \\ \frac{1}{3}&\frac{1}{4}&1 \end{bmatrix}$

so that

$LU = \begin{bmatrix} 1&0&0 \\ \frac{1}{3}&1&0 \\ \frac{1}{3}&\frac{1}{4}&1 \end{bmatrix} \begin{bmatrix} 3&1&1 \\ 0&\frac{8}{3}&\frac{2}{3} \\ 0&0&\frac{5}{2} \end{bmatrix} = \begin{bmatrix} 3&1&1 \\ 1&3&1 \\ 1&1&3 \end{bmatrix} = A$

For the third matrix we multiply the first row by 1 and subtract it from the second and third rows:

$\begin{bmatrix} 1&1&1 \\ 1&4&4 \\ 1&4&8 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&1 \\ 0&3&3 \\ 0&3&7 \end{bmatrix}$

and then multiply the second row of the resulting matrix by 1 and subtract it from the third row to obtain the upper triangular matrix U:

$\begin{bmatrix} 1&1&1 \\ 0&3&3 \\ 0&3&7 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&1 \\ 0&3&3 \\ 0&0&4 \end{bmatrix} = U$

The other factor L then is the identity matrix with 1 in the 2,1 and 3,1 positions and 1 in the 3,2 position as well:

$L = \begin{bmatrix} 1&0&0 \\ 1&1&0 \\ 1&1&1 \end{bmatrix}$

so that

$LU = \begin{bmatrix} 1&0&0 \\ 1&1&0 \\ 1&1&1 \end{bmatrix} \begin{bmatrix} 1&1&1 \\ 0&3&3 \\ 0&0&4 \end{bmatrix} = \begin{bmatrix} 1&1&1 \\ 1&4&4 \\ 1&4&8 \end{bmatrix} = A$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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