How do you find the rank of this 3x3 matrix?

Question

(1 3 1
2 5 4
1 0 7)

Answer 1

The rank is the maximal number of linearly independant rows in the matrix.

1. Subtract twice the first row from the second:

1 3 3
0 -1 2
1 0 7

2. Subtract The first row from the third:
1 3 3
0 -1 2
0 -3 6

3. Subtract 3 times the 2nd row from the 3rd row

1 3 3
0 -1 2
0 0 0

At the end of the process you have one row of zeros, and two linearly independant rows, thus the rank is 2.

Answer 2

TRUE Call f the mapping on R^3 such that f(X) = AX, call E_j = Im f^j and K = Kef f Then Rank(A^j) = Rank(A^(j+1)) <===> E_j and F are disjoint. But if k>j, then E_k and F are disjoint. Hence Rank(A^j) = Rank(A^(j+1)) ====> Rank(A^k) = Rank(A^(k+1)) Now, Rank f^0, f^1,f^2,f^3,f^4 can't all distinct since there are only 4 possible values for the ranks. Hence for one j <= 3, Rank(A^j) = Rank(A^(j+1)). Hence Rank(A^k) is constant for k >= 3.

Answer 3

How do you mean rank? Well if I'm interpreting your qusetion correctly, square it. And if you get ties, you cube it. still getting ties? that plus that squared. if you're stlil getting ties, then that plus that squared plus that cubed. not sure if you can do it on a normal graphic calculator, usually i'd do it on excel.