Math problems! Linear algebra determinant! Thank you gratefully!

The first question is to master a method of separating determinants by rows or columns.

Left = Separate by first column.

by ay+bz az+bx| +ax ay+bz az+bx|bz az+bx ax+by| |ay az+bx ax+by|bx ax+by ay+bz| |az ax+by ay+bz|As long as you understand this step, you're good to go.

The above two items are then separated in this way by separating the second and third columns, the one on the left.

by bz az+bx|+|by ay az+bx|(The latter one is 0, don't say you don't know), bz bx ax+by| |bz az ax+by|

bx by ax+bz| |bx ax ay+bz|The right item is also separated in the same format, you can also get a determinant and a 0 Okay to go to the third step, and then divide it, you can get 4 more determinants, two of which are 0 The final result is.

by bz bx| +ax ay az|

bz bz by| |ay az ax|

bx by bz| |az ax bz|

If you can't prove it at this point, then ask me again.

I know the first question, and the second question must be able to do it, it's not easy to type, so I'll give you points.

1|ax+by ay+bz az+bx|=[(ax+by)(az+bx)(ay+bz)]^3-(az+bx)^3-(ax+by)^3-

ay+bz az+bx ax+by|(ay+bz) 3 results and the comparison below.

az+bx ax+by ay+bz|

x y z|Set of 3x3 determinant formulas.

y z x|=3xzy-z 3-x 3-y 3, this result *(a+b)=3axyz+3bxyz-(a+b)*(z 3-x 3-

z x y|y 3) is the same as the above simplified, so it is equal.

2.First transpose the determinant (a->at) or symmetry transformation, and become.

a b c d |

a+1 b+1 c+1 d+1|

a+2 b+2 c+2 d+2|

a+3 b+3 c+3 d+3|

The transpose transformation does not change the value of the determinant, and then subtracts the first line from the second, third, and fourth lines respectively (this elementary transformation does not change the value of the determinant).

a b c d|After the Gaussian elimination is found, the two rows are 0 0 0 0, then the value of this determinant is 0

This is because any determinant has as long as one line is all zero (or has two identical rows), the determinant value is 0

The first determinant change is a column change, the second is a column change, and the third is a row change.

4) An arrow goes past the fourth column of the second row, and a -1 is missing. Not 0.

Finding the determinant is a polynomial about x, and this polynomial is equal to 0 is the equation.

This is a strip row copy.

formula, in accordance with the attack.

The first column expands and gets.

Two determinants, one of which is the n-1 order determinant dn-1 and the other determinant, according to the first line, gives the n-2 order determinant dn-2, that is, dn=

2adn-1

a²dn-2

Then dn-adn-1 = a(dn-1-adn-2) satisfies the proportional property and is therefore obtained recursively.

a²(dn-2-adn-3)

aⁿ⁻²d2-ad1)

aⁿ⁻²3a²-2a²)

a May also be written:

dn-adn-1 =aⁿ

a(dn-1-adn-2)=aⁿ

a²(dn-2-adn-3)=aⁿ

.aⁿ⁻²d2-ad1)=aⁿ

Add the above n-1 equations to obtain.

dn-aⁿ⁻¹d1 = (n-1)aⁿ

Then DNA = A D1 + N-1)A = 2A + N-1)A

n+1)a choose a

Choose A, and the recursive method can be obtained.

One. (1) Because the row mark is a natural order arrangement, the column mark is the inverse ordinal number of the 532416.

t(532416) = 4+2+1+1+0+0 = 8 is an even number, so this item has a positive sign.

2) In the same way, because t(162435) = 0+4+0+1+0+0 = 5 is an odd number, this term has a negative sign.

3) a21a53a16a42a65a34 = a16a21a34a42a53a65

t(614235) = 5+0+2+0+0+0 = 7, so it's a minus.

4) a51a32a13a44a65a26 = a13a26a32a44a51a65

t(362415) = 2+4+1+1+0+0 = 8, so with a positive sign.

5) a61a52a43a34a25a16 = a16a25a34a43a52a61

t(654321) = 5+4+3+2+1+0 = 15, so with a negative sign.

Note: 1It is also possible to arrange the columns in natural order and calculate the inverse ordinal number of the row labels.

2.It is also possible to calculate the sum of the inverse ordinal numbers arranged by rows and columns.

Two. Columns 1 and 5 are missingTake k=1, l=5

t(31425) = 2+0+1+0+0 = 3.So with a negative sign at this time, it is what you want.

Note: If a positive sign is required, after calculating t(31425) = 3, the position of 1 and 5 is even.

When calculating the determinant, there should be a formula, 丌(-1) (i+j)aij, and then sum this thing.

These two questions of yours are asking the power of -1.

So you take the ij of each aij and sum it, even numbers are positive, odd numbers are negative, and then you multiply the product of the 6 terms of 1, and finally the symbol of 1 is the second problem, which is also the same principle.

Find the k and l that let the last 1 be a minus

Well, I can't remember the formula clearly, it shouldn't be as simple as i+j.

Each line is added to the first row =(n+1)(-1) (n+2) a1a2......an

Only row transformations are made. Lines are added and subtracted, the common factor of the lines is proposed, and finally the first three lines are exchanged. Finish.

Look at the process experience.

Satisfied, please promptly. Thank you!

Solution: The transpose of a is:

a -b -c -d

b a d -c

c -d a b

d c -b a

then aa'= (a 2+b 2+c 2+d 2)e, so |a|^2 = |aa'| = (a^2+b^2+c^2+d^2)^4.

Considering |a|In A 4 with a positive sign, so there is |a| = (a^2+b^2+c^2+d^2)^2.