Both a and B are orthogonal matrices. The determinant of a + B = 0. It is proved that the determinant of (a + b) is equal to 0

Both a and B are orthogonal matrices. The determinant of a + B = 0. It is proved that the determinant of (a + b) is equal to 0

Solution: known that a, B are n-order orthogonal matrix, so AA ^ t = a ^ TA = e, BB ^ t = B ^ TB = e, and the determinant of orthogonal matrix is equal to 1 OR-1, because | a | | + | B | = 0, so a 124124124124124124124124a = |a ^ t | a + B | B ^ t | = | a ^ t (a + b) B ^ t |

Proof of determinant |x -1 0 …… 0 0| |0 x -1 …… 0 0| |…… …… …… |=x^n+a_ 1x^n-1+…… A_ n-1x+a_n |0 0 0 …… x -1| |a_ n a_ n-1 a_ n-2…… A_ 2 x+a_ 1| A_ N means n is the subscript of a, and so on X ^ n is the nth power of X, and so on

The general form is too troublesome to write, write a 4-level, N-level, and so on. The dot is used to position, there is no other meaning. Dot is used to locate, there is no other meaning. The dot is used to position, there is no other meaning. The dot is used to position, there is no other meaning. The dot is used to position, there is no other meaning. The dot is used to position, there is no other meaning. (expand according to the first line).. | (expand according to the first line); (expand according to the first line); (expand according to the first line); (expand in line 1); (expand) 0.. X.; - 1. (expand); (expand according to the first line); (expand in line 1); (expand in line 1); (expand in line 1); (expand in line 1); (expand the first|0.. X. - 1. | A3. A2. A1 |

A proof of determinant |by+az bz+ax bx+ay| |x y z| |bx+ay by+az bz+ax| =(a^3+b^3)|z x y| |bz+ax bx+ay by+az| |y z x| A ^ 3 is the third power of A

The first column is to be demolished first, and the first column is to be demolished first, which is the first column

Determinant proof |b+c c+a a+b| | a b c| |a+b b+c c+a| = 2 |c a b| |c+a a+b b+c| | b c a|

c1+c2+c3
Column 1 proposes 2
c1-c2
c3-c1
c2-c3

Determinant proof questions a+b ab 0 ...0 0 1 a+b ab ...0 0 0 1 a+b ...0 0 .................. 0 0 0 ...a+b ab =(a^n+1-b^n+1)/a-b 0 0 0 ...1 a+b

0

The following determinant is proved, a+b ab 0 ...0 0 1 a+b ab...0 0 0 1 a+b...0 0 ..... ..... ..... 0 0 0...a+b ab 0 0 0...1 a+b This determinant is equal to B ^ (n + 1) + AB ^ n +... + A ^ (n + 1)

Solution: when D1 = a + B, D2 = a ^ 2 + AB + B ^ 2. N > 2, expand DN according to the first column to get DN = (a + b) DN-1 - abdn-2 (1), so dn-adn-1 = B (dn-1-adn-2) = B ^ 2 (dn-2-adn-3) - iteration =... = B ^ (n-2) (d2-ad1) = B ^ (n-2) B ^ 2 (2) = B ^ n

Prove the determinant below. Thank you! Let D (n) = | 0 a (12) a (13). A (1n) |, and prove that D (n) = 0 when n is the cardinal number |-a(12) 0 a(23).a(2n)| |-a(13) -a(23) 0.a(2n)| |. . . . .| |-a(1n) -a(2n) -a(3n)...0|

It's easy to pull. It's easy to use the property of determinant - "determinant transpose, its value remains unchanged". Proof: according to determinant transposition, the value of determinant remains unchanged. D (n) = | 0 a (12) a (13). A (1n) | = | 0 - A (12) - A (13). - A (1n) | - A (12) 0 a (23). A (2n) | a (12) 0 - A (23). - A (2n) | - A (1

Read the verses and say what farm tools are related to the bold words.      1. Work hard in the daytime and report hemp at night. （    ）   2. In the afternoon of weeding, sweat drips down the soil. （    ） 3. The children and grandchildren did not understand the farming and weaving. （    ）   4. One millet is planted in spring and ten thousand seeds are harvested in autumn. （    ）（    ）

It is proved that: (1) when n = 1, left = 12-22 = - 3, right = - 1 × (2 + 1) = - 3,
So left = right,
When n = 1, the equation holds;
(2) If n = k, the equation holds,
That is 12-22 + 32 - +(2k-1) 2 - (2k) 2 = - K (2k + 1) holds,
Then when n = K + 1, the left = 12-22 + 32 - +（2k+1）2-（2k+2）2
=-k(2k+1)+[2(k+1)-1]2-[2(k+1)]2
=-k(2k+1)+(2k+1)2-4(k+1)2
=(2k+1)[(2k+1)-k]-4(k+1)2
=（k+1）（-2k-3）
=-（k+1）[2（k+1）+1]
From (1) and (2), we know that equation 12-22 + 32-42 + + (2n-1) 2 - (2n) 2 = - n (2n + 1) holds for any positive integer

It is proved by mathematical induction that: - 1 + 3 - 5 + +（-1）n（2n-1）=（-1）nn．

It is proved that: (1) when n = 1, left = - 1, right = - 1,
/ / left = right
(2) Suppose n = k, the equation holds, that is: - 1 + 3 - 5 + +（-1）k（2k-1）=（-1）kk；
When n = K + 1, the left side of the equation = - 1 + 3 - 5 + +（-1）k（2k-1）+（-1）k+1（2k+1）
=（-1）kk+（-1）k+1（2k+1）
=（-1）k+1．（-k+2k+1）
=（-1）k+1（k+1）．
That is to say, when n = K + 1, the equation holds
To sum up (1) (2), we can know: - 1 + 3-5 + +(- 1) n (2n-1) = (- 1) NN holds for any positive integer

Common derivation formula in high school

c'=0 (x^n)'=nx^(n-1）
(sinx)'=cosx (cosx)'=-sinx
(a^x)'=a^xlna (e^x)'=e^x
(logax)'=1/(xlna) (lnx)'=1/x