Linear Algebra and Its Applications, exercise 1.3.7

Exercise 1.3.7. (a) Given a system of equations A, with the first two rows the same, at what point in elimination will it become clear that A is singular? Show a 3×3 example.

(b) Repeat (a), but instead assume that the first two columns are the same.

Answer: (a) If the first two rows of A are the same then the first elimination step will produce zeros in the second row. The second step will do a row exchange in an attempt to get a non-zero pivot in the second row. Assuming that this is successful elimination will continue and can proceed as usual, doing row exchanges when necessary to avoid the problem of finding a zero in the pivot position due to the row of zeros. However at the point where the zero row is the last row row exchange can no longer correct the problem, and elimination will fail because there will not be a non-zero pivot.

The following example shows the process. We start with the system of equations

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}u&+&v&+&w&=&b_1 \\ u&+&v&+&w&=&b_2 \\ 2u&+&v&+&w&=&b_3 \end{array}$

which in matrix form becomes

$\left[ \begin{array}{rrrl} 1&1&1&b_1 \\ 1&1&1&b_2 \\ 2&1&1&b_3 \end{array} \right]$

The first elimination step produces the following matrix

$\left[ \begin{array}{rrrc} 1&1&1&b_1 \\ 0&0&0&b_2 - b_1 \\ 0&-1&-1&b_3 - 2b_1 \end{array} \right]$

after which we can do a row exchange to produce

$\left[ \begin{array}{rrrc} 1&1&1&b_1 \\ 0&-1&-1&b_3 - 2b_1 \\ 0&0&0&b_2 - b_1 \end{array} \right]$

However at this point we cannot find a nonzero pivot, elimination fails, and it is clear that the system of equations is singular.

(b) If the first two columns of A are identical, then the first elimination step will produce zeros not only in the first column of the second and subsequent rows, but zeros in the second column of the second and subsequent rows as well. There will therefore be no way that row exchange could produce a nonzero pivot in the second column of the second row, and at that point we will know that elimination has failed and the system of equations in singular.

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}u&+&v&+&w&=&b_1 \\ u&+&v&+&2w&=&b_2 \\ u&+&v&+&3w&=&b_3 \end{array}$

which in matrix form becomes

$\left[ \begin{array}{rrrl} 1&1&1&b_1 \\ 1&1&2&b_2 \\ 1&1&3&b_3 \end{array} \right]$

The first elimination step produces the matrix

$\left[ \begin{array}{rrrc} 1&1&1&b_1 \\ 0&0&1&b_2 - b_1 \\ 0&0&2&b_3 - b_1 \end{array} \right]$

which has no nonzero pivot in the second column, even allowing for row exchange.

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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