## Linear Algebra and Its Applications, Review Exercise 1.5

Review exercise 1.5. For each of the systems of equations in review exercise 1.4, factor the corresponding matrices into the forms $A = LU$ or $PA = LU$.

Answer: For the first system

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}u&&&+&w&=&4 \\ u&+&v&&&=&3 \\ u&+&v&+&w&=&6 \end{array}$

the corresponding matrix is

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

and the final matrix after elimination is

$U = \begin{bmatrix} 1&0&1 \\ 0&1&-1 \\ 0&0&1 \end{bmatrix}$

The eliminations steps were as follows:

1. Subtract the first equation from the second equation ($l_{21} = 1$).
2. Subtract the first equation from the third equation ($l_{31} = 1$).
3. Subtract the second equation from the third equation ($l_{32} = 1$).

The matrix of multipliers is thus

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

and we have the following factorization:

$A = \begin{bmatrix} 1&0&1 \\ 1&1&0 \\ 1&1&1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 1&1&0 \\ 1&1&1 \end{bmatrix} \begin{bmatrix} 1&0&1 \\ 0&1&-1 \\ 0&0&1 \end{bmatrix} = LU$

For the second system

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}&&v&+&w&=&0 \\ u&&&+&w&=&0 \\ u&+&v&&&=&6 \end{array}$

the corresponding matrix is

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

We did an initial exchange of the first and third rows, corresponding to multiplying by the permutation matrix

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

The final matrix after elimination was

$U = \begin{bmatrix} 1&1&0 \\ 0&-1&1 \\ 0&0&2 \end{bmatrix}$

The elimination steps were as follows:

1. Subtract the first equation from the second ($l_{21} = 1$).
2. Add the second equation to the third ($l_{32} = -1$).

The matrix of multipliers is thus

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

and we have the following factorization:

$PA = \begin{bmatrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix} \begin{bmatrix} 0&1&1 \\ 1&0&1 \\ 1&1&0 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 1&1&0 \\ 0&-1&1 \end{bmatrix} \begin{bmatrix} 1&1&0 \\ 0&-1&1 \\ 0&0&2 \end{bmatrix} = LU$

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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