Why determinant | a ^ - 1 | = | a | ^ - 1, how to prove it? Thank you

Why determinant | a ^ - 1 | = | a | ^ - 1, how to prove it? Thank you

Because AA ^ - 1 = E
So | AA ^ - 1 | = | e | = 1
So | a | a ^ - 1 | = 1
So | a ^ - 1 | = | a ^ - 1

A determinant proof problem How to prove this determinant, the first column is 11 1, the second column is a ^ 2 B ^ 2 C ^ 2, and the third column is a ^ 3 B ^ 3 C ^ 3, which is equal to (AB + BC + Ca) × the first column 11 1 1 the second column a a a A and the third column a ^ 2 B ^ 2 C ^ 2 Sorry for typing the second column of the second determinant, which is a B C

Have you ever studied Vandermonde determinant? If you have, I can help you to prove it. If you haven't learned it, please look at the textbook and ask again
How can the second row of your second determinant be AAA yes? If it is zero

Determinant proof: | a ^ 2 A * B B ^ 2 | 2 * a a + B 2 * B | = (a-b) ^ 2 | 11 1 | how to prove it!

D =
c1-2c2+c3
(a-b)^2 ab b^2
0 a+b 2b
0 1 1
c2-c3
= (a-b)^2 *
(a-b)^2 ab-b^2 b^2
0 a-b 2b
0 0 1
= (a-b)^3

Using determinant to prove that three points are collinear It is said in the book that a X1 Y1 B x2 Y2 C X3 Y3 is collinear if and only if x1 y1 1 x2 y2 1 =0 x3 y3 1 But for example, a (1,1) B (3213433) C (1,1) They have a determinant of 0 but are not collinear Is it necessary or not

Collinear means that they are on the same line. In your example, a (1,1), B (3213433), and C (1,1) are indeed collinear. They are on the straight line determined by a and B. in addition, what you mentioned is a necessary and sufficient condition

Line generation, how to find determinant? 4 1 2 4 1 2 0 2 10 5 2 0 0 1 1 7

4 1 2 4 1 2 0 210 5 2 0 0 1 1 7r3-2r1-2r2, r1-4r2 0 -7 2 -4 1 2 0 2 0 -1 -2 -12 0 1 1 7r1r2 1 2 0 2 0 -7 2 -4 0 -1 -2 -12 0 1 1 7r2-7r3,r4+r3 1 2 0 2 0 0 16 80 0 -1 -2 -...

Line determinant If the equation is: 1 2 3 4 1 3-x^2 3 4 3 4 1 2 3 4 1 5-x ^ 2 the determinant equation = 0, find X; find the detailed explanation process tat

0

0

Extract common factor in the first line
The second line extracts the common factor
.
The common factor is extracted in line n
The common factor was extracted in line n + 1
Then the determinant is changed into Vandermonde determinant
By using Vandermonde determinant formula and sorting out

Prove the following determinant ｜a∧2+1/a∧2 a 1/a 1｜ The following format is the same = 0 Replace a with BCD ｜ ｜ It is known that ABCD = 1

D=D1+D2
D1=
| a² a 1/a 1 |
| b² b 1/b 1 |
| c² c 1/c 1 |
| d² d 1/d 1 |
=
| a 1 1/a² 1/a |
(abcd)*| b 1 1/b² 1/b |
| c 1 1/c² 1/c |
| d 1 1/a² 1/d |
=
| a 1 1/a² 1/a |
| b 1 1/b² 1/b |
| c 1 1/c² 1/c |
| d 1 1/d² 1/d |
D2=
| 1/a² a 1/a 1 |
| 1/b² b 1/b 1 |
| 1/c² c 1/c 1 |
| 1/d² d 1/d 1 |
=
| a 1 1/a² 1/a |
(-1)³ | a 1 1/b² 1/b |
| a 1 1/c² 1/c |
| a 1 1/d² 1/d |
∴D=D1+D2=0

Can all n-order determinants be transformed into upper (or lower) triangular determinants? How to prove (or not) this proposition? What's more, when calculating the determinant value, why is it so easy to make mistakes? It's always full of ambition to calculate, and finally find out that the result value is wrong. It's tragic! Is it because everyone has the experience of this initial stage or my own problem? I hope under the guidance of experts!

It can be proved by induction
Consider column 1 in D first
If the elements in column 1 are all 0, then the determinant is equal to 0
Otherwise, swap a nonzero element to the upper left corner and use it to make the rest of the elements in column 1 to 0
At this point, the first row and the first column of d do not need to be moved
The determinant is reduced by one order
Treat column 2 in the same way
In this way, the determinant can be changed into an upper triangular determinant

Prove the following determinant 0 a a-b a-c a-d -a 0 b b-c b-d b-a -b 0 c c-d c-a c-b -c 0 d d-a d-b d-c -d 0 This determinant is equal to 0

The odd order antisymmetric determinant must be zero. The following figure is the proof of order 5. The economic mathematics team will help you solve the problem, please adopt it in time. Thank you!