Linear Algebra: it is proved that if the element of determinant | a | of order n (n > = 2) is 1 or - 1, then | a | is even RT

Linear Algebra: it is proved that if the element of determinant | a | of order n (n > = 2) is 1 or - 1, then | a | is even RT

Expand | a | into n! Terms, each of which is 1 or - 1

Determinant problem of linear algebra, using determinant properties to solve, urgent, online, etc four thousand one hundred and twenty-four one thousand two hundred and two 10 520 0117

4 1 2 4
1 2 0 2
10 5 2 0
0 1 1 7
r3-2r1-2r2,r1-4r2
0 -7 2 -4
1 2 0 2
0 -1 -2 -12
0 1 1 7
r1r2
1 2 0 2
0 -7 2 -4
0 -1 -2 -12
0 1 1 7
r2-7r3,r4+r3
1 2 0 2
0 0 16 80
0 -1 -2 -12
0 0 -1 -5
r2+16r4
1 2 0 2
0 0 0 0
0 -1 -2 -12
0 0 -1 -5
Determinant = 0

Properties of determinant of linear algebraic matrix: please prove the property of determinant of square matrix: if a and B are matrices, then determinant of AB product is equal to determinant of a and B Can we prove that d = [a O] is a block matrix [-E B] det(D)=detAdetB After primary transformation D[A AB] The process of [- E O] transformation is to construct AB from the original position of O det(D)=det(AB) So det (AB) = detadetb Can a sub block matrix be treated as a matrix element with the same determinant properties as above

sure.
Note:
1. When the k-fold of a row is added to another row, it should be multiplied by K on the left, and by K on the right during column transformation
2. Block matrix does not satisfy diagonal rule
determinant
0 Am
Bn 0
= (-1)^mn |A||B|

How to prove that the square of determinant of orthogonal matrix is equal to one?

0

0

Orthogonal matrix has the property AA '= a'a = E;
So | AA '| = | e |;
Namely
|A||A'|=1,
And | a | = | a '|
therefore
|A|^2=1
|A | = 1 or - 1

It is proved that if a is an orthogonal matrix, then the determinant of a is equal to plus or minus 1

A is an orthogonal matrix|
A times a transpose matrix = identity matrix e
|A||A|=1
|A|2=1
|A | = plus or minus 1

How to prove that the determinant of an orthogonal matrix is plus or minus one

Let a be an orthogonal matrix
Then AA ^ t = E
The determinant on both sides
|AA^T| = |E| = 1
And | AA ^ t | = | a | a ^ t | = | a | a | 2
So | a ^ 2 = 1
So | a | = 1 or - 1

Let a and B be two orthogonal matrices of order n, and the determinant of AB is - 1. It is proved that the determinant of a + B is 0

A'a = AA '= e, b'b = b = BB' = e, a'a = a = AA '= e, b'b = BB' = E has a

Determinant proof questions Proof determinant |a11-0.5 a12 .a1n| |a21 a22-0.5...a2n| |a31 a32 a33-0.5...a3n| H. |an1.ann-0.5| Not equal to 0 Where all AIJ are integers

If | a-0.5e | = 0
It shows that 0.5 is the characteristic root of a, and | a-in-e | is an integral coefficient polynomial with the first coefficient of 1, and 0.5 can not be its root, so | a-0.5e | ≠ 0

Linear algebra (proof of matrix determinant) Let a be an n-order square matrix, and the transposition of a times a is equal to the unit matrix When the determinant of a is equal to - 1, the determinant of (a + I) is equal to 0 2. When the determinant of a is equal to 1 and N is odd, the determinant of (A-I) is equal to 0

(a + e) × a '= e + a', take the determinant on both sides, and the second question is the same