## Linear Algebra and Its Applications, Exercise 1.5.2

Exercise 1.5.2. Assume we have a matrix A with

$A = \begin{bmatrix} 1&0&0 \\ 2&1&0 \\ 1&4&1 \end{bmatrix} \begin{bmatrix} 5&7&8 \\ 0&2&3 \\ 0&0&6 \end{bmatrix}$

What multiple of row 2 of A will elimination subtract from row 3? What will be the pivots? Will a row exchange be required?

Answer: We have A = LU where L is a lower triangular matrix and U an upper triangular matrix:

$L = \begin{bmatrix} 1&0&0 \\ 2&1&0 \\ 1&4&1 \end{bmatrix} \quad U = \begin{bmatrix} 5&7&8 \\ 0&2&3 \\ 0&0&6 \end{bmatrix}$

The multipliers $l_{ij}$ are the entries below the diagonal of L, with $l_{ij}$ being the multiplier that multiplies the pivot row when it is subtracted from row i, and that produces a zero in the i, j position. Since in the exercise we are subtracting from row 3 we have i = 3, and since we are multiplying row 2 we have j = 2. The multiplier is therefore $l_{32}$ or 4.

The pivots are the diagonal entries of U, or 5, 2, and 6.

There is no permutation matrix P present, so no row exchanges are required.

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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