Let A = [1 3 0 0 0 -3 2 6 0 0 0 -6 0 0 -1 -2 3 -2 0 0 2 1 3 1 0 0 -1 0 0 -3], Fi

ID: 3112969 • Letter: L

Question

Let A = [1 3 0 0 0 -3 2 6 0 0 0 -6 0 0 -1 -2 3 -2 0 0 2 1 3 1 0 0 -1 0 0 -3], Find a basis for the row space of A, a basis for the column space of A, a basis for the null space of A, the rank of A, and the nullity of A. (Note that the reduced row echelon form of A is [1 3 0 0 0 -3 0 0 1 0 0 3 0 0 0 1 0 -2 0 0 0 0 1 -1 0 0 0 0 0 0].) Row Space basis: (1, 3, 0, 0, 0, -3), (0, 0, 1, 0, 0, 3) Column Space basis: (1, 2, 0, 0, 0), (0, 0, -12, -1), (0, 0, -2, 1, 0) (Null Space basis: Rank: 3 Nullity: 4

Explanation / Answer

The reduced row echelon form of the augmented matrix is

which corresponds to the system

The leading entries in the matrix have been highlighted in yellow.

A leading entry on the (i,j) position indicates that the j-th unknown will be determined using the i-th equation.

Those columns in the coefficient part of the matrix that do not contain leading entries, correspond to unknowns that will be arbitrary.

The system has infinitely many solutions:

The solution can be written in the vector form:

c2 +

1 3 0 0 0 -3 0 0 1 0 0 3 0 0 0 1 0 -2 0 0 0 0 1 -1 0 0 0 0 0 0

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Let A = [1 3 0 0 0 -3 2 6 0 0 0 -6 0 0 -1 -2 3 -2 0 0 2 1 3 1 0 0 -1 0 0 -3], Fi

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