## 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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