## 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 .

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