enter image description here

I'm not sure how to start on these types of problems. I know that the dot product of the columns have to be zero between each pair.

Do I set up a system of linear equations for this? If so, what would it even look like?

  • 2
    $\begingroup$ This is a great problem. It gives me ideas for my own course. $\endgroup$ – James S. Cook Mar 30 at 17:39
  • 1
    $\begingroup$ Remember that two columns are orthogonal if and only if their dot product is zero. For matrix A what you need to do is find a, b, c, d so that the dot product is zero pairwise for all columns. And something similar for matrix B. $\endgroup$ – ErotemeObelus Mar 30 at 18:19

Your matrix $A$ is$$A=\begin{bmatrix}-1&1&-1&1\\1&1&-1&a\\0&1&b&c\\0&0&1&d\end{bmatrix}.$$The first to columns are clearly orthogonal. Also, the first and the third columns are orthogonal too, no matter what $B$ is. On the other hand, the second and the second columns are orthogonal if and only if $1\times(-1)+1\times(-1)+1\times b+0\times1=0$, which means that $b=2$. Let us then assume that $b=2$

Now, the first and the fourth columns are orthogonal if and only if $a=1$. So let us assume that $a=1$. Then the second and the fourth columns are orthogonal if and only if $c=-2$. Now, complete the process in order to determine the only possible value for $d$. Can you deal with the other matrix now?

  • $\begingroup$ for the second matrix the first column's missing number is -1 and the third column is 1 ,-1 -,1, is there a step by step method to solve these types of problems $\endgroup$ – Samurai Bale Mar 30 at 18:02
  • $\begingroup$ The method that I described is step-by-step. $\endgroup$ – José Carlos Santos Mar 30 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.