The question asks if the det of the following 3*3 matrix is divisible by $$ x^4,x^3,x^2,x $$ $$ \begin{bmatrix} a^2+x^2 & ab & ac \\ ab & b^2+x^2 & bc \\ ac & bc & c^2+x^2 \\ \end{bmatrix} $$ I just shovelled through using the standard method and got $$ x^4(b^2+c^2+a^2+x^2) $$ So the answer is "Yes" but is there a simpler way like splitting the det as a product of two dets? If not that is there anything else?
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The matrix can be written as $$A = x^2I_3 + \begin{bmatrix}a\\b\\c\end{bmatrix} \begin{bmatrix}a & b & c\end{bmatrix}$$ Now by Sylvester determinant theorem, we have \begin{align} \det(A) & = x^6 \det\left(I_3 + \dfrac1{x^2}\begin{bmatrix}a\\b\\c\end{bmatrix} \begin{bmatrix}a & b & c\end{bmatrix}\right) = x^6 \left(1+\dfrac1{x^2}\begin{bmatrix}a & b & c\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}\right)\\ & = x^4(x^2+a^2+b^2+c^2) \end{align}
Adhvaitha
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1In this case, this is essentially the same computation as given by the http://en.wikipedia.org/wiki/Matrix_determinant_lemma , which gives a formula for the determinant of a rank-$1$ modification of a matrix. – Travis Willse May 02 '15 at 04:24
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What if I don't know about Sylvester theorem? – GrandAlpha May 02 '15 at 04:39
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1@GrandAlpha Then know it. – Adhvaitha May 02 '15 at 16:43
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There is this theorem that if on substituting $x = a$ if n rows of the determinant become identical or proportional then $(x-a)^{n-1}$ is a factor of the determinant. In this case $3$ rows become proportional so both $x$ and $x^2$ are factors
Sumit
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