## Linear Algebra and Its Applications, Exercise 1.6.20

Exercise 1.6.20. Suppose $A$ is a 3 by 3 matrix for which the third row is the sum of the first and second rows. Show that there is no solution to the equation $Ax = \begin{bmatrix} 1&2&4 \end{bmatrix}^T$, and determine whether $A$ has an inverse or not.

Answer: For $x = \begin{bmatrix} u \\ v \\ w \end{bmatrix}$ we have the following system of three equations: $Ax = \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} \rightarrow \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$ $\rightarrow \setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}a_{11}u&+&a_{12}v&+&a_{13}w&=&1 \\ a_{21}u&+&a_{22}v&+&a_{23}w&=&2 \\ a_{31}u&+&a_{32}v&+&a_{33}w&=&4 \end{array}$

If we subtract the sum of the first two equations from the third equation, we obtain the following: $[a_{31} - (a_{11}+a_{21})]u + [a_{32} - (a_{12}+a_{22})]v + [a_{33} - (a_{13}+a_{23})]w = 4 - (1+2)$ $\rightarrow 0\cdot u + 0 \cdot v + 0 \cdot w = 1 \rightarrow 0 = 1$

Since we have reached a contradiction we conclude that the system of equations has no solutions. The corresponding matrix $A$ is singular and therefore has no inverse.

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.

### 2 Responses to Linear Algebra and Its Applications, Exercise 1.6.20

1. Swagatika Panda says:

Hi. Your posts are really helpful. I have a question from exercise 1.6. (4th edition). It is: Show that A^2 = 0 is possible but A’A = 0 is not possible (unless A= zero matrix). Your response will he helpful to me. Thanks.

• hecker says:

I’m glad you like the blog and find it useful. I don’t known exactly which exercise you are talking about. Are you referring to multiplying the transpose of A times A? If so, I will give you a hint: When you multiply A-transpose times A you are multiplying entries in each row of A-transpose by entries in each column of A. But by definition the rows of A-transpose are the columns of A. So, for example, when you compute the 1, 1 entry of the product matrix you are multiplying the entries of the first row of A-transpose by the entries of the first column of A, which means you are multiplying the entries of the first column of A by themselves–and then you take the sum, per the definition of matrix multiplication. So the question for you is: under what conditions could that sum be zero?