Linear Algebra and Its Applications, Review Exercise 1.26

Posted on August 7, 2011 by

Review exercise 1.26. (a) Given a 3 by 3 matrix A what vector x would make the product Ax have 1 times column 1 of A plus 2 times column 3?

(b) Construct a matrix A for which the sum of column 1 and 2 times column 3 is zero, and show that A is singular. Why?

Answer: (a) If we choose

x = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}

then Ax will consist of column 1 of A plus 2 times column 3. In essence if

A = \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{bmatrix}

then

Ax = \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} = 1 \cdot \begin{bmatrix} a_{11} \\ a_{21} \\ a_{31} \end{bmatrix} + 0 \cdot \begin{bmatrix} a_{12} \\ a_{22} \\ a_{32} \end{bmatrix} + 2 \cdot \begin{bmatrix} a_{13} \\ a_{23} \\ a_{33} \end{bmatrix}

(b) The following matrix has column 1 plus 2 times column 3 equal to 0.

A = \begin{bmatrix} 2&1&-1 \\ -4&2&2 \\ 6&0&-3 \end{bmatrix}

Elimination of this matrix proceeds as follows:

\begin{bmatrix} 2&1&-1 \\ -4&2&2 \\ 6&0&-3 \end{bmatrix} \rightarrow \begin{bmatrix} 2&1&-3 \\ 0&4&0 \\ 0&-3&0 \end{bmatrix} \rightarrow \begin{bmatrix} 2&1&-3 \\ 0&4&0 \\ 0&0&0 \end{bmatrix}

Since there is no pivot in row 3 the matrix is singular. The problem is that since column 3 is a multiple of column 1 (being equal to -2 times column 1) the elimination steps that produce zeros in the (2, 1) and (3, 1) position will also produce zeros in the (2, 3) and (3, 3) positions, so that there is no possible pivot in column 3.

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.

This entry was posted in linear algebra. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

Gravatar
WordPress.com Logo

You are commenting using your WordPress.com account. ( Log Out /  Change )

Google photo

You are commenting using your Google account. ( Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. ( Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. ( Log Out /  Change )

Cancel

Connecting to %s