Here is my attempt :  If order of $A$ is $2$, then $$A=         \begin{pmatrix}         2 &amp; 2  \\         2 &amp; 5  \\         \end{pmatrix}  \Rightarrow \det A=6.$$ Also the eigenvalues in this case are $1,6$.  If order $A$ is $3$, then $$A=         \begin{pmatrix}         2 &amp; 2 &amp; 3 \\         2 &amp; 5 &amp; 6 \\         3 &amp; 6 &amp; 10 \\         \end{pmatrix} \Rightarrow \det A=15$$ Also the eigenvalues in these case are $1,1,15.$  If order of $A$ is $4$, then $$         \begin{pmatrix}         2 &amp; 2 &amp; 3 &amp; 4 \\         2 &amp; 5 &amp; 6 &amp; 8\\         3 &amp; 6 &amp; 10 &amp; 12\\         4 &amp; 8 &amp; 12 &amp; 17\\ \end{pmatrix} \Rightarrow \det A=-21.$$ I found that $1$ is an eigenvalue here by directly putting $\lambda=1$ in the characteristic determinant. But $-21$ is not the eigenvalue.  I couldn't use information about eigenvalues so I looked at the pattern of determinants :  $6,15,-21,...?$ Observed that they are $3\times 2,3 \times 5,-3 \times 7$. From this pattern it might be possible that next term is $-3 \times 11$ since $2,5,7$ are primes and next is $11$. Also the negative sign can be in the pattern $+,+,-,-,+,+,-,-,+,+...$ As it seems to me I will have to calculate order $5$ too to get some more concrete idea with respect to this way. But isn't there some elegant method?  I see that every such order $n$ matrix is "engraved to upper-left inside" order $n+1$ such matrix and every $A$ is symmetric. Thanks.  PS : I wrongly calculated $\det A_{4 \times 4}=-21$. It should be $31$.
This question aims to create an &quot;&quot;  of numerous questions that ask about determinants of specific matrices (I may have missed a few):              The general question of this type is  Let $A$ be a square matrix of rank$~1$, let $I$ the identity matrix of the same size, and $\lambda$ a scalar. What is the determinant of $A+\lambda I$?  A clearly very closely related question is  What is the characteristic polynomial of a matrix $A$ of rank$~1$?
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