How to find the first row of determinant ax + by ay + BZ AZ + BX, the second row ay + BZ AZ + BX ax + by, the third row AZ + BX ax + by ay + BZ? I know, but the problem is that after three lines are added and the common factor is raised, it can't be transformed into a triangular matrix to solve it. Let's talk about it

How to find the first row of determinant ax + by ay + BZ AZ + BX, the second row ay + BZ AZ + BX ax + by, the third row AZ + BX ax + by ay + BZ? I know, but the problem is that after three lines are added and the common factor is raised, it can't be transformed into a triangular matrix to solve it. Let's talk about it

Add three lines

In linear algebra, when AX = B or ax = 0 has only unique solution, can coefficient matrix a form determinant?
When AX = B or ax = 0 has only one solution, does the coefficient matrix A have a certain number of rows = number of columns, and the determinant is not equal to zero? If the number of equations is greater than the number of unknowns, what is the situation?
I basically understand. Only one thing I want to confirm is the example you gave. In the first deformation, that is, when two equations become three
"Next, increase the number of equations,
X1+X2=3
2X1+X2=4
2X1+2X2=6,
Obviously, the third equation is the deformation of the first one. After simplification, the rank of the augmented matrix is 2, which is equal to the number of unknowns, and the system of equations still has a unique solution.
”
Can this example show that "when the number of equations is greater than the number of unknowns, it cannot be determined by determinant" because it is a 3 * 2 matrix and cannot form determinant?

It must be that the number of rows is greater than or equal to the number of columns, and the rank of the augmented matrix (from coefficient matrix A to column matrix B) is equal to the number of columns of the coefficient matrix, that is, the rank of the augmented matrix must be equal to the number of unknowns, and the equation has a unique solution. Determinant is not equal to 0, which only applies to the case that the number of equations is equal to the number of unknowns. When the number of equations is greater than the number of unknowns, determinant cannot be used to judge
for instance:
X1+X2=3
2X1+X2=4
By this system of equations, the unique solutions X1 = 1, X2 = 2 are obtained,
Next, add the number of equations,
X1+X2=3
2X1+X2=4
2X1+2X2=6,
Obviously, the third equation is the deformation of the first one. After simplification, the rank of the augmented matrix is 2, which is equal to the number of unknowns
Another change
X1+X2=3
2X1+2X2=6
3X1+3X2=9
After simplifying the augmented matrix, it is found that the rank of the augmented matrix is 1, and the equations have infinite solutions
To sum up,
When the rank of the augmented matrix is equal to the number of unknowns, the equation has a unique solution
When the rank of the augmented matrix is less than the number of unknowns, there are infinite solutions
Forget an important premise, that is, when the rank of the coefficient matrix and the rank of the augmented matrix are equal, the equations may have solutions, otherwise there is no solution. Give an example to illustrate
X1+X2=3
0X1+0X2=6
Obviously, the rank of the coefficient matrix is 1, and the rank of the augmented matrix is 2. Generally speaking, when the rank of the augmented matrix is greater than the coefficient matrix, after linear transformation, there will be a situation similar to "0x1 + 0x2 = 6". There are so many words, I don't know if I can explain them clearly

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 |, is () A. - 8 B. - 2 C.2 d.8
Thank you very much!

|B | is equivalent to a constant, | B | a | = - 2A | = (- 2) ^ 3 | a | = - 8