A faster way of calculating this determinant?

Question

I'm doing a problem involving Cramer's rule, and one of the determinants I have to work with is as follows:

\begin{vmatrix} 1&1&1\\ a&b&c\\ a^3&b^3&c^3 \end{vmatrix}

So I started off by getting the matrix to a triangular matrix so I can just take the product of the diagonal cells by doing this:

\begin{equation} R_1 \times-c + R_2 \end{equation}

\begin{equation} R_3 \times -\frac{1}{c^3} + R_1 \end{equation}

\begin{equation} R_2 \times(\frac{1-\frac{b^3}{c^3}}{b-c})+R_1 \end{equation}

I then got this matrix

\begin{vmatrix} ((1-\frac{a^3}{c^3})-(a-c)(\frac{1-\frac{b^3}{c^3}}{b-c}))&0&0\\ a-c&b-c&0\\ a^3&b^3&c^3\\ \end{vmatrix}

By summing the diagonal cells, I got this as a final answer:

\begin{equation} (c-a)(a-b)(b-c)(a+b+c) \end{equation}

However, it was a lengthy process, and I can't help but thinking this is not the type of calculations I can afford the time in a written exam, especially if I have to repeat this four times solving three linear equations with Cramer's rule.

I'd really appreciate it if you have any thoughts as to speed this process up.

Just as an added detail, the original question was (translated from Chinese):

With regards to the following set of equations

\begin{equation} x + y + z = 1\\ ax + by + cz = d\\ a^3x + b^3y + c^3z = d^3 \end{equation}

1. Under what conditions can Cramer's rule be used?

2. Please solve the set of equations with Cramer's rule.

For the first question, the immediate thought that pops up is that

\begin{equation} det(A) ≠ 0 \end{equation}

For the second question, it's simply

\begin{equation} \frac{\begin{vmatrix}A_1\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}}, \frac{\begin{vmatrix}A_2\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}}, \frac{\begin{vmatrix}A_3\end{vmatrix}}{\begin{vmatrix}A\end{vmatrix}} \end{equation}

which is the part I'm having trouble solving quickly.

Again, appreciate any hints or thoughts on this. Thanks.

Answer 1

Determinants are not changed by adding multiples of a column to another. So $$ \begin{vmatrix}1&1&1\\ a&b&c\\ a^3&b^3&c^3\end{vmatrix}=\begin{vmatrix}1&0&0\\ a&b-a&c-a\\ a^3&b^3-a^3&c^3-a^3\end{vmatrix}=(b-a)(c^3-a^3)-(b^3-a^3)(c-a) $$

Edit: since in the comments to the other answer you were asking how to factor, here it is: you have the identity $x^3-y^3=(x-y)(x^2+xy+y^2)$. Then \begin{align} (b-a)(c^3-a^3)-(b^3-a^3)(c-a)&=(b-a)(c-a)(c^2+ac+a^2)-(b-a)(c-a)(b^2+ab+a^2)\\ &=(b-a)(c-a)(c^2+ac+a^2-b^2-ab-a^2)\\ &=(b-a)(c-a)(c^2+ac-b^2-ab)\\ &=(b-a)(c-a)[(c^2-b^2)+a(c-b)]\\ &=(b-a)(c-a)(c-b)(c+b+a). \end{align}

Answer 2

You can compute the determinant of a generic $3\times 3$ matrix using a neat trick, if we have:

$$\mathbf{A}=\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$$

Then we have the sum of the diagonals (highlighted in green) minus the sum of the antidiagonals (highlighted in red) as follows:

$$\det(\mathbf{A}) = \underbrace{aei}_{\begin{pmatrix}{\color{green} a} & b & c \\ d & {\color{green} e} & f \\ g & h & {\color{green} i}\end{pmatrix}} + \underbrace{bfg}_{\begin{pmatrix}a & {\color{green}b} & c \\ d & e & {\color{green} f} \\ {\color{green} g} & h & i\end{pmatrix}} + \underbrace{cdh}_{\begin{pmatrix}a & b & {\color{green} c} \\ {\color{green} d} & e & f \\ g & {\color{green} h} & i\end{pmatrix}} - \underbrace{afh}_{\begin{pmatrix}{\color{red} a} & b & c \\ d & e & {\color{red} f} \\ g & \color{red} h & i\end{pmatrix}} - \underbrace{bdi}_{\begin{pmatrix}a & \color{red} b & c \\ \color{red} d & e & f \\ g & h & \color{red} i\end{pmatrix}} - \underbrace{ceg}_{\begin{pmatrix}a & b & \color{red}c \\ d & \color{red} e & f \\ \color{red} g & h & i\end{pmatrix}}$$

Thus if we have your matrix:

$$\mathbf{M}=\begin{pmatrix}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{pmatrix}$$

Then:

$$\det(\mathbf{M})=bc^{3}+ca^{3}+ab^{3}-cb^{3}-ac^{3}-ba^{3}$$

Answer 3

By cofactor expansion along the fourth column, if you take the determinant $$\begin{vmatrix} 1&1&1&1\\ a&b&c&t\\ a^2&b^2&c^2&t^2\\ a^3&b^3&c^3&t^3 \end{vmatrix}$$

and find the coefficient of $t^2$, it will be the negative of the determinant you're after.

This makes the problem a matter of understanding the Vandermonde Matrix, which is any matrix of the form $$\begin{bmatrix} 1 & 1 & \dots & 1\\ \alpha_1 & \alpha_2 & \dots & \alpha_m\\ \alpha_1^2 & \alpha_2^2 & \dots & \alpha_m^2\\ \vdots & \vdots & \ddots &\vdots \\ \alpha_1^{n-1} & \alpha_2^{n-1} & \dots & \alpha_m^{n-1} \end{bmatrix}.$$ In the case where $n=m$, its determinant can be expressed as $\prod_{1\le i<j\le n} (\alpha_j-\alpha_i)$, helping achieve the factored form you are after.

Further explanation: If $n=m=4$ then $$\prod_{1\le i<j\le 4} (\alpha_j-\alpha_i)=\left(\prod_{1\le i<j\le 3}(\alpha_j-\alpha_i)\right)\cdot\left(\prod_{1\le i<j=4}(\alpha_j-\alpha_i)\right)$$ Letting $\alpha_1=a,\alpha_2=b,\alpha_3=c,\alpha_4=t$, we have $$\left((b-a)(c-b)(c-a)\right)\cdot\left((a-t)(b-t)(c-t)\right)$$ where the only way to get a $t^2$ term when expanding the right product is to multiply by one of $a$, $b$, or $c$. You could also say it's a matter of total degree.

Answer 4

Let $P(a,b,c)$ be the determinant in question. Then the general properties of determinants imply that:

• $P$ is a homogeneous polynomial of degree $4$.
• $P$ is alternating: that is, $P(a,b,c)=-P(b,a,c)=-P(a,c,b)=-P(c,b,a)$.

In particular, $P$ vanishes whenever any two of $a,b,c$ are equal, and so it must have the linear homogeneous polynomials $(b-a)$, $(c-b)$, and $(a-c)$ as factors.

Now, consider the polynomial $Q(a,b,c)=\frac{P(a,b,c)}{(b-a)(c-b)(a-c)}$. This is homogeneous of degree $1$. Moreover, since $P$ and $(b-a)(c-b)(a-c)$ are both alternating, $Q$ is symmetric: that is, $Q(a,b,c)=Q(b,a,c)=Q(a,c,d)=Q(c,b,a)$.

But the only symmetric polynomials in three variables which are homogeneous of degree $1$ are multiples of $a+b+c$. So we must have $Q(a,b,c)=K \cdot (a+b+c)$ for some constant $K$, and thus

$$P(a,b,c)=K(a+b+c)(b-a)(c-b)(a-c)$$

This is actually good enough to do the original Chinese question (once you convince yourself that $K$ must be nonzero), since all the $K$s will cancel out when you apply Cramer's rule. If you want to know the actual value of the determinant, you could compute $K$ by finding the value of $P$ at a single point where it's nonzero.