The sign of a12a53a41a24a35 in the fifth order determinant

The sign of a12a53a41a24a35 in the fifth order determinant

First adjust the order, A12, A24, A35, A41, A53. At this time, the column marks are arranged as 24513, and the reverse order number is 5, so the sign is negative;
Or without adjusting the order, the reverse number of row labels is t (15423) = 5, the reverse number of column labels is t (23145) = 2, and the total reverse order is 7,
And then take the minus sign

The symbol ABCD is called the second-order determinant, and its operation method is: ABCD = ad BC, 1 / 2 / 2 / 3 / 1

The answer is - 2 * 1 - (- 1 / 2) * 3 / 2 = - 2 + 3 / 4 = - 5 / 4

Third order square matrix A = (a, B, c) B = (a + B + C, a + 2B + C, a + 3B + 9C), |a | = m, find |b|

B=AC C=[1 1 1;1 2 3;1 1 9]
|C|=8
|B|=|A||C|=8m

Given that a and B are square matrices of order 3, | a | = 1, | B | = - 2, then the determinant | (2Ab * ^ (- 1) a | =?

Because | B | = - 2, so | b * | = | B | ^ (3-1) = (- 2) ^ 2 = 4
|(2AB*)^(-1)A|= | (1/2) (B*)^-1 A^-1 A|
= (1/2)^3 |(B*)^-1|
= (1/8) (1/|B*|)
= (1/8)*(1/4)
= 1/32.

Let a be a square matrix of order 3, B be a square matrix of order 4, and the determinant | a | = 1, | B | = - 2, then the value of the determinant | B | a |, the answer is: - 8 urgent process,

||B|A|
= |B|^3 |A|
= (-2)^3 * 1
= -8.
Knowledge point: | Ka | = k ^ n ||

Four numbers are arranged in two rows and two columns, and a vertical line is added on each side to mark them as | a B |. Define | a B | = ad BC, which is called the second-order determinant, | C D | C D|
Then | - 5 3x & # 178; + 5 | = ()
|2 x²-3|
The second way: a polynomial a minus the polynomial 2x & # 178; + 5x + 3, students will copy the minus sign into a plus sign, the calculation result is - X & # 178; + 3x-7, the correct result is ()
The third: calculation: - 8m & # 178; - [4m-2m & # 178; - (3m-m & # 178; - 7) - 8]
You don't have to answer the first question

1.-11x²+5；2.-3x²-2x-10；3.-7m²-m+1

|ABCD | is called the second-order determinant, its algorithm is ad BC, such as | 1234 | = 1 * 4-2 * 3 = 4-6 = - 2, calculate | 2 - 54 - 3|

=2*(-3)-(-5)*4
=14

If a, B, C and D are the four sides of the quadrilateral ABCD and satisfy a ^ 4 + B ^ 4 + C ^ 4 + D ^ 4-4abcd = 0, what kind of quadrilateral

If 2 / 3A = 6 / 5B = 5 / 4C = 3 / 4D, and ABCD is not equal to zero, please arrange the four numbers of ABCD from small to large

a=5/4b
a=15/8c
a=9/8d
a>d>b>c

If a and B are real numbers, and the root sign (1 + a) - (1-B) * root sign (B-1) is equal to 0, find the value of a to the power of 2008 - B to the power of 2008

The root sign (1 + a) = (1-B) * root sign (B-1)
The root sign (1 + a) must be greater than or equal to 0, and the root sign (B-1) must be greater than or equal to 0
We get: (1-B) greater than or equal to 0
Then B can only be equal to 1
In this case, a = - 1
So 2008 times of a - 2008 times of B = 1 - 1 = 0