**Transpose**of a matrix/determinants from to is defined by:

The element in the i row j column is moved to j row i column.

For simplicity, you can imagine that the determinant is

Example:*flipped along the principal diagonal line*(which is the top-left to bottom-right one).Even if it isn't square matrix it still works:

In HKDSE, transpose is almost useless so I won't discuss its main properties, except one:

It may help when we can't solve a determinants, flipping it gives more clues. Take the HW as example:

Example: Factorize , hence evaluate .

Now notice that when , the original one is the transpose of X.

Therefore,

(the factorization has been demonstrated before) which gives the answer -210.

Note that transpose has given the symmetricities between rows and columns.

**Uniqueness of degree in determinants**

Sometimes the factorization does not give a unique degree in the polynomial expression which may cause difficulties in factorization.

Recall that determinants is the sum of product of elements from distinct rows and columns. Therefore

**it each row or column has unique degree, the factorization result has a unique degree.**

For example,

In such case you can treat 1 as a variable in order to equate the degree. The determinant becomes:

Note that

**this determinant is NOT cyclic**because each row are not in cyclic relationship.

Cyclic group of (a,b,c): (a,b,c) -> (b,c,a) -> (c,a,b) -> (a,b,c)

Cyclic group of (b,a,c): (b,a,c) -> (a,c,b) -> (c,b,a) -> (b,a,c)

They are two different group, and in this example, (c,a,b) can't reach (a,c,b) by cyclic relationship.

However we see that

**(a,b) is symmetrical**because

Therefore if is a factor of the det., is also a factor of the det.

Notice that when , , (the value of determinant = 0), so (c-a) is a factor of . As we have proven, (c-b) is also a factor of .

Though it's not cyclic, each row still contain a,b,c, which tell us (a+b+c) is a factor of it (this is an usual and important practice in factorizing determinants).

Equating degree, , so we have factorized out all factor of , we now assume , where k is a constant. By equating the coefficient, we get k = 1, and putting c = 1, we have

**Case for x=y=z**

When two equal variables "kill" the determinant, we say (a-b) is a factor. How to determine the factor when the condition "x=y=z" kill the determinant?

**If x=y=z is the condition to make determinant zero, then**

**is the factor of the determinant.**

Prove: . It's zero only if x=y=z, and therefore by factor theorem it's a factor of the determinants.

This is a simplified version of

*AM-GM inequalities or Rearrangement Inequalities*, HKALE players should be familiar with it.

Example:

Factorize , hence show that

Step 0: Notice that the determinant is

**cyclic**.

Step 1: Notice that the determinant vanishes when x=y=z, so is a factor of the determinant.

Step 2: Equating degree, the degree of determinant is 3 while the current factor is 2, that remains one more factor with degree 1, then that must be (a+b+c).

Step 3: Equating coefficient,

Step 4: The second part of the question:

Therefore .

**For ugly result which can't be factorized**

In this case we can't do much other than directly expansion.

Example: express in polynomial.

Notice that

1) This is symmetrical (as we have proven)

2) Each term of the polynomal does not contain a term with a square or a cube of a certain variable like .

Then we can conclude that , where are elemental functions for 3 variables.

By equating the coefficient, we have .

Example: express in polynomial.

Notice that: , therefore the

*sign of variable does change the value of determinant*. Since the degree is under 3, we deduce that where are constants. By equating the coefficient, we have .

Exercise:

1)Show that for distinct a,b,c, , then abc=1.

2)Factorize given that the product is four linear and one quadratic factors. [Sheffield past paper]

3)Factorize .

4)Show that (a,c) is symmetrical for , hence express it in polynomial.

5)Show that is symmetrical, hence express it in polynomial.

6)Why is not a factor of ?

Extra question:

For a square matrix where and are constants. The rest of the entries are 1. Show by induction, or otherwise, . (AL level; 6~7M)

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