Risolvi {ccc}{a^2+2a}&{2a+1}&{1}{2a+1}&{a+2}&{1}{3}&{3}&{1}=(a-1)^3

\left| \begin{array} { c c c } { a ^ { 2 } + 2 a } & { 2 a + 1 } & { 1 } \\ { 2 a + 1 } & { a + 2 } & { 1 } \\ { 3 } & { 3 } & { 1 } \end{array} \right| = ( a - 1 ) ^ { 3 }

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vero

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5 problemi simili a:

\left| \begin{array} { c c c } { a ^ { 2 } + 2 a } & { 2 a + 1 } & { 1 } \\ { 2 a + 1 } & { a + 2 } & { 1 } \\ { 3 } & { 3 } & { 1 } \end{array} \right| = ( a - 1 ) ^ { 3 }

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Prove that the paraboloids: \displaystyle\frac{{x}^{{2}}}{{{a}_{{1}}^{{2}}}}+\frac{{y}^{{2}}}{{{b}_{{1}}^{{2}}}}=\frac{{{2}{z}}}{{c}_{{1}}} ; \displaystyle\frac{{x}^{{2}}}{{{a}_{{2}}^{{2}}}}+\frac{{y}^{{2}}}{{{b}_{{2}}^{{2}}}}=\frac{{{2}{z}}}{{c}_{{2}}} ...

https://socratic.org/questions/prove-that-the-paraboloids-x-2-a-1-2-y-2-b-1-2-2z-c-1-x-2-a-2-2-y-2-b-2-2-2z-c-2-1

Cesareo R. May 12, 2017 See https://socratic.org/questions/prove-that-the-paraboloids-x-2-a-1-2-y-2-b-1-2-2z-c-1-x-2-a-2-2-y-2-b-2-2-2z-c-2

Calculating the Jacobian for a linear transformation (in Kadane's 'Principles of Uncertainty')

https://math.stackexchange.com/questions/1744252/calculating-the-jacobian-for-a-linear-transformation-in-kadanes-principles-of

Let us take the dimension n=2. Transformation: Y=BXB^T is \begin{bmatrix}y_1&y_3\\y_2&y_4\end{bmatrix}=\begin{bmatrix}a&c\\b&d\end{bmatrix}\begin{bmatrix}x_1&x_3\\x_2&x_4\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix} ...

determinant linear factorisation

https://math.stackexchange.com/questions/333579/determinant-linear-factorisation

Playing around, and beginning pretty much as you began (except carrying out the row operations correctly): D = \begin{vmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c & 1 \end{vmatrix}\\ = \begin{vmatrix} (a^3-b^3)+(a^2-b^2) & a-b & 0 \\ (b^3-c^3)+(b^2-c^2) & b-c & 0 \\ c^3+c^2 & c & 1 \end{vmatrix}\\ = \begin{vmatrix} (a^3-b^3)+(a^2-b^2) & a-b \\ (b^3-c^3)+(b^2-c^2) & b-c \end{vmatrix}\\ = (a-b)(b-c) \begin{vmatrix} a^2+ab+b^2+a+b & 1 \\ b^2+bc+c^2+b+c & 1 \end{vmatrix}\\ = (a-b)(b-c)(a^2+ab+b^2+a+b-b^2-bc-c^2-b-c)\\ = (a-b)(b-c)(a^2+ab+a-bc-c^2-c)\\ = (a-b)(b-c)(a^2+ab+a+ac-ac-bc-c^2-c)\\ = (a-b)(b-c)(a-c)(a+b+c+1), ...

To obtain the condition for vanishing of the given determinant

https://math.stackexchange.com/questions/1617875/to-obtain-the-condition-for-vanishing-of-the-given-determinant

This is NOT a one line solution you are expecting but it is an idea which is very useful in solving such determinants which may remind one of Vandermonde matrix. I'm replacing a by x (for the sake ...

Determinant of a Block Matrix times Inverse

https://math.stackexchange.com/questions/992686/determinant-of-a-block-matrix-times-inverse

You can use another decomposition. \begin{pmatrix} 1 & -\alpha A^{-1} \\ 0 & A^{-1} \end{pmatrix} \begin{pmatrix} a & \alpha \\ \beta & A \end{pmatrix} = \begin{pmatrix} a - \alpha A^{-1} \beta & 0 \\ A^{-1} \beta & I \end{pmatrix} ...

A determinant problem

https://math.stackexchange.com/questions/941616/a-determinant-problem

Let g(x) = (x-1)(x-\alpha)(x-\beta), a monic polynomial with the roots 1, \alpha, \beta. Then the Vandermonde matrix of roots to g is M = \left(\begin{array}{ccc}1 & 1 & 1\\1 & \alpha & \alpha^2\\ 1 & \beta & \beta^2\end{array}\right) ...

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