# Factorizing determinants and rules to simplify them

1. Feb 23, 2015

### PcumP_Ravenclaw

1. The problem statement, all variables and given/known data
2. Evaluate the determinants

$\begin{vmatrix} 1 & 1 & 1\\ x & a & b \\ x^2 & a^2 & b^2 \\ \end{vmatrix}$

$\begin{vmatrix} x & a & b \\ x^2 & a^2 & b^2 \\ x^3 & a^3 & b^3 \\ \end{vmatrix}$

2. Relevant equations

Rules of determinants are given in the attachment.

3. The attempt at a solution

Using the scalar triple product definition of the determinant I calculated the first determinant to be

$ab^2 - ba^2 + bx^2 - xb^2 + ax^2 - xa^2$

The second determinant is $a^2b^3 - b^2a^3 + b^2x^3 - x^2b^3 + a^3x^2 - x^2a^3$

How do I factorize this? Are the two determinant values same. If so can you please show the rules are applied in the two matrices above. From linear combination I only understand that a scalar multiple of a row/column can be added to another row/column. How do they multiply different elements in each row. e.g.

The first row of first matrix becomes x a b.I thought it could only be x x x or a a a or b b b.

danke...

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2. Feb 23, 2015

### Stephen Tashi

3. Feb 23, 2015

### BvU

$a^2b^3 - b^2a^3 + b^2x^3 - x^2b^3 + a^3x^2 - x^2a^3$ is definitely wrong in all terms. The first row doesn't contain ones, but x, a, b !

(and instead of $a^3x^2 - x^2a^3$ you probably meant $a^3x^2 - x^3a^2$ ? )

4. Feb 23, 2015

### LCKurtz

Hint: Note that the first determinant is a second degree polynomial in $x$, call it $P(x)$. What are $P(a)$ and $P(b)$?

Hint2: What happens to the second determinant if you factor out an $x$? An $a$? A $b$?

5. Feb 24, 2015

### HallsofIvy

One way of evaluating determinants is to use "row operations" to reduce the matrix to an upper triangular matrix. The determinant of such a matrix is just the product of the diagonal numbers.

There are 3 "row operations":
1) Multiply a row by a number. This multiplies the entire determinant by that number.
2) Swap two rows. This changes the sign of the determinant.
3) Add any multiple of a row to another row. This does not change the determinant.

For example, starting with the matrix
$$\begin{bmatrix}1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{bmatrix}$$
we can get 0s in the first column, second and third rows, by adding -x times the first row to the second and $-x^2$ times the first row to the third:
$$\begin{bmatrix}1 & 1 & 1 \\ 0 & y- x & z- x \\ 0 & y^2- x^2 & z^2- x^2 \end{bmatrix}$$
This does not change the determinant.

We can get 0 in the second column, second row by adding -(y+ x) times the second row from the third row:
$$\begin{bmatrix}1 & 1 & 1 \\ 0 & y- x & z- x \\ 0 & 0 & z^2- y^2 \end{bmatrix}$$
This does not change the determinant so this determinant, and the original determinant, are just the product of the numbers on the diagonal, $(1)(y- x)(z^2- y^2)$.

6. Feb 24, 2015

### epenguin

Maybe not the best, not completeLy what they are looking for, but much faster than your laborious expansion would be the "subtract the columns" rule. Row operationsbare also quite easy. If you do this nevertheless also note Kurz's comments.

This is also a form of determinant that comes up quite often in e.g. theory of equations and elimination known as the Vandernonde determinant.