## 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.

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

1. 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?