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I don't know how to do the following problem.

Suppose that $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$, and $x_{5}$ are roots for the equation $x^{5}-2x^{4}+x^{3}+1=0$. Let $a_{ij}=1+x_{i}x_{j}$ for $i=j$, and $a_{ij}=x_{i}x_{j}$ for $i\neq j$, where $1\leq i\leq 5$ and $1\leq j\leq 5$. Evaluate \begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} & a_{15}\\ a_{21} & a_{22} & a_{23} & a_{24} & a_{25}\\ a_{31} & a_{32} & a_{33} & a_{34} & a_{35}\\ a_{41} & a_{42} & a_{43} & a_{44} & a_{45}\\ a_{51} & a_{52} & a_{53} & a_{54} & a_{55} \end{vmatrix}

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You should get comfortable with the definition of a determinant. – Asaf Karagila Mar 8 '13 at 3:10
@AsafKaragila Actually, this does not require to know more than the determinant of a diagonal(izable) matrix. – 1015 Mar 8 '13 at 3:13
@julien: True. Regardless to that, being comfortable with the definition of a determinant allows one to sit and unravel the matrix and to preform the needed calculations, even if that would be the longer and bruteforce attempt at solving the problem. There is a huge difference between posting "I don't know how to do it" and "I know the answer is $X$, but I did that by a bruteforce method. Is there a shorter way to derive it?" – Asaf Karagila Mar 8 '13 at 3:15

First observe that all the $x_j$'s are nonzero since their product is $-1$, by looking at the constant term of the polynomial.

Now note that $$ A=I_5+B\quad\mbox{with}\quad B=(x_ix_j). $$ It is clear that $B=xx^T$ is of rank $1$, so $0$ is an eigenvalue of $B$ with multiplicity $4$.

Now $$ \mbox{tr}\;B=x_1^2+\ldots+x_5^2. $$

Using Newton's identities for the power sums $s_k=\sum_{j=1}^5 x_j^k$ of the roots of the polynomial $$ t^5-2t^4+t^3+1=t^5-a_1t^4+a_2t^3-a_3t^2+a_4t-a_5, $$ we have $$ s_2=a_1s_1-2a_2=2\cdot 2-2\cdot 1=2. $$

So the trace yields a distinct eigenvalue. Hence $B$ is diagonalizable with diagonal $(0,0,0,0,2)$.

It follows that $A=I_5+B$ is diagonalizable with diagonal $(1,1,1,1,3)$.

So $$ \det A=3. $$

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