Linear Algebra and Its Applications, exercise 1.4.10

Exercise 1.4.10. Given a matrix A with entries a_{ij}, what are the following entries?

  1. first pivot
  2. the multiplier l_{i1} that is used to multiply the first row and subtract it from the ith row
  3. the value that replaces a_{ij} once the above operation occurs
  4. second pivot

Answer: The first pivot is a_{11}, the entry in the first row and first column. (This assumes that that entry is non-zero and thus no row exchange is needed.)

For row i we need to eliminate the entry a_{i1}, the first entry in that row. Since l_{i1} is used to multiply the first row, we must have

a_{i1} - l_{i1}a_{11} = 0 \quad \Rightarrow \quad l_{i1}a_{11} = a_{i1} \quad \Rightarrow \quad l_{i1} = a_{i1}/a_{11}

Since l_{i1} is used to multiply all the entries in the first row and subtract the resulting value from corresponding entries in the ith row, the result of applying this operation to the entry a_{ij} is therefore

a_{ij} - l_{i1}a_{1j}

This operation is done on all rows, including the second row and the entry in the second column in that row, a_{22}. The resulting value

a_{22} - l_{21}a_{12}

then becomes the second pivot.

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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