Find the rank and the nullity of the matrix A rank A nulli

Find the rank and the nullity of the matrix A = []/rank (A) = nullity (A) = rank(A) + nullity (A) =

Solution

To find null space we will have following steps

First, let\'s put our matrix in Reduced Row Eschelon Form...

Add (1 * row1) to row2

Add (1 * row1) to row3

Add (2 * row3) to row2

Add (-1 * row3) to row1

Add (-1 * row2) to row1

The matrix has 3 pivot columns (hilighted in yellow) and 2 free columns; because the matrix has 3 pivots, the rank of the matrix is 3.

Let\'s take the \'free\' part of the reduced row echelon form matrix (hilighted below in yellow)...

and turn it into its own matrix:

Let\'s multiply this matrix by -1:

Now, we add the Identity Matrix to the rows in our new matrix which correspond to the \'free\' columns in the original matrix, making sure our number of rows equals the number of columns in the original matrix (otherwise, we couldn\'t multiply the original matrix against our new matrix):

Finally, the Null Space of our matrix is defined by scalar multiples of these column vectors:

to find rank we will have following steps

1 R₁ + R₂ R₂ (multiply 1 row by 1 and add it to 2 row); 1 R₁ + R₃ R₃ (multiply 1 row by 1 and add it to 3 row)

1100-1010-300000-3

R₃ / -3 R₃ (divide the 3 row by -3)

1100-1010-3000001

Answer. Since there is 3 non-zero rows, then Rank(A) = 3. answer

1	1	0	1	-1
0	1	0	-2	0
-1	-1	0	0	-2