Linear Algebra and Its Applications, Review Exercise 1.15

Review exercise 1.15. For the following $n$ by $n$ matrix $A$ and its inverse

$A = \begin{bmatrix} n&-1&\cdots&-1 \\ -1&n&\cdots&-1 \\ \vdots&\vdots&\ddots&\vdots \\ -1&-1&\cdots&n \end{bmatrix} \quad A^{-1} = \frac{1}{n+1} \begin{bmatrix} c&1&\cdots&1 \\ 1&c&\cdots&1 \\ \vdots&\vdots&\ddots&\vdots \\ 1&1&\cdots&c \end{bmatrix}$

what is the value of $c$ ?

Answer: Since $AA^{-1} = I$ we must have

$1 = \frac{1}{n+1} (nc + \sum_{k=2}^{n} (-1) \cdot 1)$

$= \frac{1}{n+1} (nc - \sum_{k=2}^{n} 1)$

$= \frac{1}{n+1} (nc - (n -1)) = \frac{1}{n+1} (c-1)n + 1)$

Multiplying both sides by $n+1$ we obtain $n+1 = (c-1)n + 1$ or $n = (c-1)n$ so that $c = 2$ .

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 Cancel reply

Enter your comment here...

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

Email (required) (Address never made public)

Name (required)

Website

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

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

Cancel

Connecting to %s