1

This is a question from Gilbert Strang's "Intro to Linear Algebra"

section header

question from book

solution from book

$A$ has 6 permutations but only two of them have non-zero terms, and both these permutations are one row-exchange away from $I$, making the terms negative. Therefore, $\text{det} (A)=-1-1=-2$.

My thought was that the $6\times6$ matrix, given all the zeros, would only have 4 permutations (out of the 720) that had non-zero terms.

$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 0\\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 0\\ \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ \end{bmatrix} $$

Since all four involve two row-exchanges to get to $I$, they would each give a term 0f $+1$, making the determinant equal to $4$. However, the book says the answer is $-1$ and I'm having trouble seeing where this comes from. Did I miss some permutations?

SteveV
  • 71
  • 2
    Indeed, the determinant of a block matrix with blocks $A$ and $A$ is simply $\det(A)^2$, which is $(-2)^2=4$. See this duplicate, for example. So there is a typo in the book, or in the exercise. Can you give a link, so that one can verify it? – Dietrich Burde Jan 28 '25 at 19:17
  • Thanks, I added images from the book to the post. It is from the 4th edition. Thanks, also, for confirming that the determinant of D is 0 – SteveV Jan 28 '25 at 19:30
  • Indeed, $\det(D)=\det(A)\det(B)=0$, because the telephone matrix $B$ has rank $2$ (add the first and third row). And the book is wrong with $\det(C)$, so you have found a "real" typo. – Dietrich Burde Jan 28 '25 at 19:50
  • I suppose its different because the book does not teach the various formulas discussed in the other post before posing the question. The question asks one to solve it using the "big formula" related to permutations. Is it preferable that I delete it since it related to a typo, or is it a nice addition, since it might help other people stuck on the same typo? – SteveV Jan 28 '25 at 20:16
  • No, the determinant of two blocks $A$ is immediately $\det(A)^2$, even without the formulas (just definition of a determinant and Laplace development). I don't think this typo is interesting. We would have millions of such posts then. – Dietrich Burde Jan 28 '25 at 20:18
  • I'm not at all suggesting your answer is incorrect. I am just suggesting that perhaps Strang thought it useful to have students find an answer without relying on the fact that two blocks of $A=$det$(A)^2$ or Laplace development. These are not discussed in the book before the question and I rather enjoyed thinking through just how many permutations were non-zero in this case. I am happy to delete...could someone else offer an opinion. This site has plenty of questions from this book that I've found useful. I wish there were more. – SteveV Jan 28 '25 at 21:23
  • These are not discussed in the book before the question - this is not true. The image says it you have linked: use the big formula with $n!$ terms, also known as Laplace formula. It immediately implies that the determinant of $C$ equals $\det(A)^2$. – Dietrich Burde Jan 28 '25 at 21:39

0 Answers0