## Linear Algebra and Its Applications, Review Exercise 1.26

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.

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