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I'm having some trouble with the following exercice:

Using only the properties of the determinant, prove that: $$\det \left( \begin{matrix} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^3 & b^3 & c^3 & d^3 \end{matrix} \right) = (d − a)(d − c)(d − b)(c − b)(c − a)(b − a)$$

I know how to solve this using Laplace formula but I don't see how I can use the properties of the determinant to do this. I tried to put the matrix in it's reduced form but I'm having some trouble doing it. How can I solve this problem?

Edit: this is the list of properties that I'm allowed to use:

  • If a matrix $A$ has two identical lines then $\det (A) = 0$
  • $\det(I_n) = 1$
  • Let:

$$A =\left( \begin{matrix} a_{11} & a_{12} & ... & a_{1n} \\ \vdots & \vdots & & \vdots \\ \alpha a'_{i1} + \beta a''_{i1} & \alpha a'_{i2} + \beta a''_{i2} & ... &\alpha a'_{in} + \beta a''_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & ... & a_{nn} \end{matrix} \right)$$

If $$A' = \left( \begin{matrix} a_{11} & a_{12} & ... & a_{1n} \\ \vdots & \vdots & & \vdots \\ a'_{i1}& a'_{i2}& ... & a'_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & ... & a_{nn} \end{matrix} \right)$$

and $$A'' = \left( \begin{matrix} a_{11} & a_{12} & ... & a_{1n} \\ \vdots & \vdots & & \vdots \\ a''_{i1}& a''_{i2}& ... & a''_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & ... & a_{nn} \end{matrix} \right)$$

Then $\det(A) = \alpha \det(A') + \beta \det(A'')$

These are the properties that I'm allowed to use.

Rodrigo de Azevedo
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656475
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    I'm not sure if it's the intended solution, but a standard way of evaluating this "Vandermonde determinant" is to treat $d$ as a variable, so that the determinant is a polynomial of degree $3$ in $d$, and then gain information about that polynomial by plugging in $d=a$ and $d=b$ and $d=c$. – Greg Martin Feb 16 '21 at 18:03
  • I don't think that's what the teacher wanted us to do, but that's very nice. I'll try that anyway, Thanks @GregMartin – 656475 Feb 16 '21 at 18:04

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