Compute this determinant only using determinant properties

Question

Compute $$\begin{vmatrix} 1 & x & x^2 & x^3 \\ 1 & y & y^2 & y^3 \\ 1 & z & z^2 & z^3 \\ 1 & x+y+z & xy+yz+zx & xyz \end{vmatrix}$$,where $x,y,z \in \mathbb{R}$.
I tried to create zeroes on the first line,but this doesn't seem to work.

Answer 1

That determinant is $0$, because the fourth column is equal to the sum of:

• the first column times $xyz$;
• the second column times $-xy-xz-yz$;
• the third column times $x+y+z$.

Answer 2

Here is a solution whose spirit is more "matricial" than "determinantal", but that explains the why and how of this exercice, and gives the result (determinant = 0).

Let us recall first that "the" third degree polynomial with roots $x,y,z$ can be written :

$$(t-x)(t-y)(t-z)=-xyz+(xy+yz+zx)t-(x+y+z)t^2+t^3$$

(we have chosen to use the article "the" because we comply to the classical convention with dominant coefficient is $1$).

In particular, taking $t=x$, the LHS is $0$, thus we have

$$-xyz+(xy+yz+zx)x-(x+y+z)x^2+x^3=0 \tag{1}$$

Similar identities for $y$ and $z$ :

$$-xyz+(xy+yz+zx)y-(x+y+z)y^2+y^3=0 \tag{2}$$

$$-xyz+(xy+yz+zx)z-(x+y+z)z^2+z^3=0 \tag{3}$$

Now consider the matrix product :

$$\begin{pmatrix}1 & x & x^2 & x^3 \\ 1 & y & y^2 & y^3 \\ 1 & z & z^2 & z^3 \\ 1 & x+y+z & xy+yz+zx & xyz\end{pmatrix}\begin{pmatrix}-xyz\\xy+yz+zx\\-(x+y+z)\\1\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix} \tag{4}$$

(the three first zeros are due to identities (1), (2), (3) ; the last zero being due to terms cancellations).

What conclusion drawn from (4) ? That the image of a non zero vector is the zero vector ; as the zero vector is also sent onto the zero vector, the transformation associated with the matrix is not a bijection ; thus its determinant is zero.

Remark 1 : This is in fact very close to the (direct) solution given by @José Carlos Santos, but for the fact that the coefficients survene in a natural way.

Remark 2 : A similar result ($\det M = 0$) can be generalized to any odd number of parameters (odd in order than the cancellations can be done for obtaining a zero last coefficient). In the even case with letters $x,y,z,t$, we get the following result for the analoguous determinant

$$\det(M)=(t - x)(t - y)(t - z)(y - x)(z - x)(z - y)(t^2x^2 + t^2y^2 + t^2z^2 + x^2y^2 + x^2z^2 + y^2z^2)$$

which is rather simple and a kind of generalization of the Vandermonde determinant.

Answer 3

It seems worth giving a complete solution for the general case. Obviously I am indebted to @Jean Marie.

So let $x_1,\dots, x_n$ be variables, and let $a_0,a_1, \dots, a_n$ be the elementary symmetric functions in these, so that

$$ p(X):=\prod_{i}(X-x_i)=\sum_{j=0}^{n} (-1)^{n-j}a_{n-j}X^{j}, $$ and $$ a_s=\sum_{i_1<i_2<\dots<i_s} x_{i_{1}}x_{i_{2}}\dots x_{i_{s}}. $$

Now consider the determinant of the $(n+1)\times (n+1)$ matrix $$ \begin{bmatrix} 1 & 1 & \dots & 1 & a_0\\ x_1 & x_2 & \dots & x_n & a_1\\ \vdots &\vdots&\ddots &\vdots &\vdots\\ x_1^n & x_2^n & \dots & x_n^n & a_n\\ \end{bmatrix}, $$ whose rows we label for convenience from $0$ to $n$.

As $a_0=1$ the value of the determinant will be unchanged after the following row-operation: $$ \text{replace } R_{n} \text{ by }\sum_{j=0}^{n} (-1)^{n-j}a_{n-j} R_{j}. $$

This means we can consider the determinant of this matrix: $$ \begin{bmatrix} 1 & 1 & \dots & 1 & a_0\\ x_1 & x_2 & \dots & x_n & a_1\\ \vdots &\vdots&\ddots &\vdots &\vdots\\ x_1^{n-1} & x_2^{n-1} & \dots & x_n^{n-1} & a_{n-1}\\ p(x_1) & p(x_2) & \dots & p(x_n) & \sum_{j=0}^{n} (-1)^{n-j}a_{n-j}a_j\\ \end{bmatrix}, $$ which simplifies to $$ \begin{bmatrix} 1 & 1 & \dots & 1 & a_0\\ x_1 & x_2 & \dots & x_n & a_1\\ \vdots &\vdots&\ddots &\vdots &\vdots\\ x_1^{n-1} & x_2^{n-1} & \dots & x_n^{n-1} & a_{n-1}\\ 0 & 0 & \dots & 0 & \sum_{j=0}^{n} (-1)^{n-j}a_{n-j}a_j\\ \end{bmatrix}. $$

The original determinant is therefore a product of the Vandermonde determinant $\prod_{i<j}(x_i-x_j)$ and the symmetric function $\alpha:=\sum_{j=0}^{n} (-1)^{n-j}a_{n-j}a_j$. It is the latter that we must simplify.

In the case when $n$ is odd $\alpha$ simplifies at once to $0$; the equal terms $a_{n-j}a_j$ and $a_j a_{n-j}$ have opposite sign and so cancel out.

In the case when $n=2k$ is even things are a little more complicated, but in this case (as we show below) $$ \alpha=\sum_{i_1<i_2<\dots<i_k} x_{i_{1}}^2 x_{i_{2}}^2 \dots x_{i_{k}}^2, $$ the $k$-th elementary symmetric function in the squares of the original variables.

To see this, note first that $$ (-1)^n p(-X)=\prod_{i}(X+x_i)=\sum_{j=0}^{n} a_{n-j}X^{j}, $$ so that $$ (\sum_{j=0}^{n} (-1)^{n-j}a_{n-j}X^{j})(\sum_{j=0}^{n} a_{n-j}X^{j})=\prod_{i}(X-x_i) \prod_{i}(X+x_i)=\prod_{i}(X^2 - x_i^2). $$

If we now pick out the coefficient of $X^{2k}$ on each side we have the promised $ \alpha=\sum_{i_1<i_2<\dots<i_k} x_{i_{1}}^2 x_{i_{2}}^2 \dots x_{i_{k}}^2. $