# Determinant of symmetric matrix

Given the following matrix, is there a way to compute the determinant other than using laplace till there're $3\times3$ determinants?

\begin{pmatrix} 2 & 1 &1 &1&1 \\ 1 & 2 & 1& 1 &1\\ 1& 1 & 2 & 1 &1\\ 1&1 &1 &2&1\\ 1&1&1&1&-2 \end{pmatrix}

• A calculator is certainly another way... :P But I'll try to think about one that you're actually looking for. :) – apnorton Jun 12 '13 at 12:53

You can substract the first row from every other rows and get matrix of form: $$\begin{pmatrix} 2 & 1 &1 &1&1 \\ -1 & 1 & 0& 0 &0\\ -1& 0 & 1 & 0 &0\\ -1&0 &0 &1&0\\ -1&0&0&0&-3 \end{pmatrix}.$$ Computing the determinant is now much easier.

• would you mind to elaborate that? I'm not really sure what you did there – deemel Jun 12 '13 at 12:56
• I.e perform the row operation $R_i - R_1$ for $i \geq 2$. – Kaish Jun 12 '13 at 12:57
• Bartek subtracted the top row from each of the other rows. Are you unfamiliar with the effects of elementary row and column operations on the determinant? – Gerry Myerson Jun 12 '13 at 12:57
• Adding any row multiplied by non-zero scalar (in this case scalar is $-1$) to another row doesn't change the determinant of matrix. – Bartek Pawlik Jun 12 '13 at 13:02

Look at the matrix $$A = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & -3 \end{pmatrix}.$$ It has rank two and its nonzero eigenvalues have sum $1$ (the trace) and product $$\det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} + \det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} + \det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} + \det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} = -16$$

So the characteristic polynomial is $t^3 (t^2 - t - 16)$. Evaluate this at $t = -1$ and you get $$\det(-I - A) = (-1)^5 \det \begin{pmatrix} 2 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 & 1 \\ 1 & 1 & 2 & 1 & 1 \\ 1 & 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 1 & -2 \end{pmatrix}.$$

• Also, another comment: since it appears that you are writing your characteristic polynomial as $\det(t I - A)$, I think evaluating at $t = -1$ will give you minus the determinant asked for. – Willie Wong Jun 12 '13 at 13:06
• @WillieWong Oops again, didn't think it out far enough, writing it down should make it clear. – user82098 Jun 12 '13 at 13:11

Call your matrix $A$ and let $J_n$ denotes the all-one matrix of size $n$. By Laplace expansion along the last row, we have \begin{align*} \det(A) &=\left(\sum_{j=1}^\color{red}{4}(-1)^{5+j}a_{5j}M_{5j} + 2M_{55}\right) - 4M_{55}\\ &=\det(I_5+J_5)-4\det(I_4+J_4). \end{align*} As $\det(aI_n+bJ_n)=a^{n-1}(a+nb)$, we get $\det(A)=6-4(5)=-14$.