5 (6 points) Follow Steps 0 - 3 for the following three matrices. 2 Points per matrix. A = 0 0 1 -2 1 0 -1 1 2-3 0 1 0 0 -1 1 -1 1 121 2-3 N -2 1 0 0 1 -2 -1 1 B = 0 0 -1 0 0 T TO T -1 -1 -1 -1 0 0 -1 0 0 2 0 132 0 0 0 0 0 0 0 1 2 1 1 0 1 -1 -1 1 -1 3 1 1 0 1 2 -1 0 0 -1 1 0 2 2 -1 -1 2 -2 2 2 0 3 -1 -3 2 -2 -1 0 0 -1 0 -1 0 0 0 2 0 3 -1 0 0 -2 0 0 2 2 0 0 1 0 0 1 -3 0 78 -1 -1 -3 -1 -2 0 -1 -1 0 200 0 HONN 2 -1 2-1 -1 -1 0 O C 006 -18 -12 18 6 3 6 -5 -1 -2 7 -3 -5 -15 -9 3 -5 15 7 2 4 -4 0 -20 -4 Step 0: Either by hand or using Octave's or Matlab's rref() function, find the reduced row echelon for this matrix. Write down the solution to this problem up to the first 4 numbers after the decimal point. (You do not need to turn in code or show your work. Just input the matrix into Octave/Matlab and then run the code.) Step 1: What is the rank of that matrix? Step 2: Find a subset of the columns which form a basis for the column space. List those columns. Step 3: Now using rref ( ) function in Octave or Matlab (or by hand), find a basis for the Null Space. Does the Rank-Nullity Theorem hold true