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

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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