A. If X & sup2; + 4x + Y & sup2; - 6y + 13 = 0, then x =? Y =? B. It is proved by factorization that the seventh power of 25 and the twelfth power of 5 can be divisible by 120 A. If X & sup2; + 4x + Y & sup2; - 6y + 13 = 0, then x =? Y =? B. It is proved that the seventh power of 25 and the twelfth power of 5 can be divisible by 120 by factorization

A. If X & sup2; + 4x + Y & sup2; - 6y + 13 = 0, then x =? Y =? B. It is proved by factorization that the seventh power of 25 and the twelfth power of 5 can be divisible by 120 A. If X & sup2; + 4x + Y & sup2; - 6y + 13 = 0, then x =? Y =? B. It is proved that the seventh power of 25 and the twelfth power of 5 can be divisible by 120 by factorization

1. X & sup2; + 4x + Y & sup2; - 6y + 13 = (X & sup2; + 4x + 4) + (Y & sup2; - 6y + 9) = (x + 2) & sup2; + (Y-3) & sup2; = 0, so x = - 2 and y = 3
2. The seventh power of 25 - the twelfth power of 5 = the tenth power of 5 - the twelfth power of 5 = the tenth power of 5 × (1-25) = the ninth power of 5 × [5 × (- 24)] = - the ninth power of 120 × 5, so it can be divisible by 120
A. (X+2)^2+(Y-3)^2=0 X=-2 Y=3
B. 25 ^ 7-5 ^ 12 = 5 ^ 14-5 ^ 12 = 5 ^ 12 (25-1) = 5 ^ 12 * 120, so it can be divided by 12
A. If X & sup2; + 4x + Y & sup2; - 6y + 13 = x & sup2; + 4x + 4 + Y & sup2; - 6y + 9 = (x + 2) & sup2; + (Y-3) & sup2; = 0
So x = - 2, y = 3
B. The seventh power of 25 to the twelfth power of 5 = 5 ^ 14-5 ^ 12 = 25 * 5 ^ 12-5 ^ 12 = (25-1) * 5 * 5 ^ 11 = 24 * 5 * 5 ^ 11 = 120 * 5 ^ 11
So the seventh power of 25 - the twelfth power of 5 can be divided by 120
Is the determinant of a triangular matrix equal to the multiplication of the principal elements on its diagonal?
To add:
Such a matrix: 300
-5 1 0 0 0
3 8 0 0 0
0 -7 2 1 0
-4 1 9 -2 3
His determinant is 0. It shows that it is irreversible. The eigenvalue of an irreversible matrix should be 0, but there is another theorem that says that the eigenvalue of a triangular matrix is an element on its diagonal - what is the eigenvalue of this matrix? But why
Yes. An irreversible matrix has at least one 0 in its eigenvalues. This matrix has five eigenvalues. One of them is 0. No problem
yes. An irreversible matrix has at least one 0 in its eigenvalues, which has five eigenvalues. One of them is 0, no problem.
Given that real numbers x and y satisfy x 2 + y 2 + 4x - 6y + 13 = 0, find the value of Y X
The known equation is transformed into: (x + 2) 2 + (Y-3) 2 = 0, then x + 2 = 0, Y-3 = 0, that is, x = - 2, y = 3, then YX = 3-2 = 19
If the multiplication of two matrices A and B is a zero matrix, then must the determinant values of a and B be 0? Why?
Not necessarily, because the multiplication of a matrix is the multiplication of the number of each row by the number of another determinant, and then forms a new determinant
If the solution of the binary linear equations {x + y = 5K {X - y = 9K} about X, y is also the solution of the binary linear equation 2x + 2Y = 6, then the value of K is
Note that 2x + 2Y = 6 is not all 3Y on 3Y
x+y=5k (1)
x-y=9k (2)
(1)+(2)=> 2x=14k => x=7k
(1)-(2)=> 2y=-4k => y=-2k
Substituting
2x+2y=6 => x+y=3
=> 7k-2k=3 => 5k=3 => k =3/5
How to multiply the determinant of 3 columns by the determinant of 2 rows?
If it's a determinant, calculate it separately and multiply it. It doesn't matter how many rows and columns
If it is a matrix, according to the algorithm, it cannot be multiplied
For example, if a matrix A with five rows and three columns wants to be multiplied by a matrix B, that is ab, then matrix B must be three rows, that is, it must be equal to the number of columns of a before it can be multiplied
If the solutions X and y of the equations 3x + y = - 4K + 1, x + 3Y = 3 satisfy 2x + Y > 2, the value range of K is obtained
Let the two formulas be ① and ② respectively. If ① + ②, x + y = - K + 1 is denoted as ③
Then ① + ③ get 2x + y = - 5K / 2 + 1, because 2x + Y > 2
So the solution of - 5K / 2 + 1 > 2 is K
3x+y=-4k+1 (1)
x+3y=3 (2)
(1)*5+(2)
15x+5y+x+3y=-20k+5+3
16x+8y=-20k+8
2x+y=-5k/2+1
2x+y>2
-5k/2+1>2
5k/2 -3k+1+k/2>2
=> -5k/2>1
=> k
The determinant is transformed into upper triangle determinant and its value is calculated
Change determinant into upper triangle determinant and calculate its value
First line - 2 2 - 4 0
The second line 4 - 1 3 5
The third line 3 1 - 2 - 3
The fourth line is 2051
For beginners, I hope it's more common and clear,
If you want to change it to the main diagonal, it's all zero,
R2 + 2r1, R4 + R1, R1 * (1 / 2) [line 1 proposes 2], R3 + 3R1
-1 1 -2 0
0 3 -5 5
0 4 -8 -3
0 2 1 1
r2-r4,r3-2r4
-1 1 -2 0
0 1 -6 4
0 0 -10 -5
0 2 1 1
r4-2r2
-1 1 -2 0
0 1 -6 4
0 0 -10 -5
0 0 13 -7
= 2*(-1)*1*(10*7+5*13)
= -2*135
= -270.
If the solution of the system ax-3y = 5 2x by = 1 is x = 1 / 2, y = - 1, then the value of the square of a + the square of B is
1 / 2a-3 * (- 1) = 5
1/2*2-(-b)=1
The solution is a = 4
B=0
Then the original formula = 4 ^ 2 + 0 ^ 2 = 16
How to divide a determinant into the product of two determinants? What is the law of calculation?
It's actually matrix multiplication
According to the characteristics of elements in determinant | a |
Write matrix A as the product of two matrices A = BC
The results were obtained from | a | = | BC | = | B | C |