## 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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