Advanced Algebra Linear Algebra Problems, Advanced Math, Linear Algebra Problems?

Upstairs is my problem, forgot to log in, give me extra points!

This question is very simple and easy to understand:

When A removes 1 row to get the B matrix, there is always a relationship.

rank(a) >= rank(b) >= rank(a)-1

Now, any submatrix A1 composed of s rows, which is actually a (m-s) row removed, is known by the above relation.

rank(a) >= rank(a1) >= rank(a)-(m-s)

Because rank(a)=r

So rank(a1) > = r-(m-s) = r-m+s

I don't know if this explanation can be understood, but the key point is to think of a1 as the matrix obtained by subtracting the (m-s) rows from a, and rank(a) is the number of non-zero rows after the elementary transformation of a.

So. If the (m-s) rows deleted by a are all "0 rows", the value of rank(a1) remains unchanged and remains rank(a) 0

If the (m-s) rows deleted by a are all "non-0 rows", the value of rank(a1) is reduced to rank(a) (m-s).

Otherwise, the rank(a1) value is always somewhere between the two extremes mentioned above, i.e.

rank(a)－(m-s) <= rank(a1) <= rank(a)－0

The left half inequality is proved. Actually, as long as you understand it, it is self-evident :)

This question is very simple and easy to understand:

When A removes 1 row to get the B matrix, there is always a relationship.

Rank(a) >= Rank(B) >= Rank(A)-1 Now, any submatrix A1 composed of S rows, which actually removes the (m-s) row of A, is known from the above relation.

rank(a) >= rank(a1) >= rank(a)-(m-s)

Because rank(a)=r

So rank(a1) >= r-(m-s) = r-m+s because r(a)=r

The coefficient matrix is:

2 a 3 ]

The second line is subtracted twice from the first line, and the third chapter is subtracted from the first line

0 a-4 -3 ]

Divide the third line by -2, and scatter the royal wanton:

0 a-4 -3 ]

The first row subtracts 2 times from the third row, and the second row subtracts (a-4) times from the third row

The first line is added to the second line, and then the second and third lines are swapped to get:

It is easy to know that the dismantling rank of the coefficient matrix is 3, which is equal to the number of unknowns, full rank.

Therefore, no matter what the value of a is, there is no nonzero solution to this system of homogeneous linear equations.

by question x(a-i)=a;

x=a(a-i)

In each of these operations, the matrix should be enclosed in parentheses. As for the last multiplication, the corresponding number of each row of the first matrix is multiplied by the 1, 2, and 3 columns of the other matrix, respectively. For example, the first number in the result is 1 = 1x0 + 0x0 + 1x1

If you are staring at the ruler of the figure, the world will show the limb of the tomb.

I'm really drunk on this question. It's so hard ...

The 6,9 upstairs did it wrong!

Question 6: You make an extended matrix and list the system of linear equations.

x+y=1x+2y=0

x+y+z=-2

x=2, y=-1, z=1

The answer is =2 1- 2+ 3

Question 8: False.

The correct one should be |2a|=2^n|a|

Question 9: False.

The column vector that should be a is linearly independent.

Question 10: False.

a, b is not true if it is an irreversible matrix.

Question 11: False.

Counter-examples can be given.

Question 12: Correct.

This question requires a little skill, specifically, let's experience it, see the answer below.