# proj2 - MATLAB Project 2 for MAS 3114 Due Friday October 14...

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Unformatted text preview: MATLAB Project 2 for MAS 3114 Due Friday, October 14 Input the function ﬁle lead.m given below into your MATLAB directory: function L=lead(A) [m,n]=size(A); R=rref(A); L=logical(zeros(1,n)); row=1; col=1; while col<=n & row<=m, if R(row,col)==1, L(col)=1; row=row+1; end col=col+1; end 1. Apply lead to the following matrices: (a) A=[1 2 0 4 0 3; 9 18 3 54 0 12; 9 18 0 36 1 20; 3 6 1 18 0 4]. (b) B=rand(4,6). (c) C=vander(1:5). (d) D=magic(8). Describe what lead(M) is for an arbitrary matrix M. (Hint: Compare lead(M) with rref(M).) 2. Use lead(A) to write a function v=nullrank(A) whose output is a 1 × 2 matrix v such that v(1,1) is the number of pivot variables in the solution to Ax = 0, and v(1,2) is the number of free variables in the solution to Ax = 0. Your ﬁle nullrank.m should not use the MATLAB functions rank or null. Apply nullrank to the matrices in problem 1 above. 3. Use the MATLAB function rref and the function lead above to write a MATLAB function N=nullbase(A) which computes a matrix N whose columns form a basis for the nullspace of A. Your ﬁle nullbase.m should not use the MATLAB functions rank or null. Compute nullbase(A) and nullbase(A’*A) for the matrices given in problem 1 above. Make a reasonable conjecture based on these computations. MATLAB has some methods for constructing matrices which you may ﬁnd useful. For instance, B=A(:,[1,5,4]) creates a matrix B whose columns are the 1st, 5th, and 4th columns of A. If the second input is a vector with the “logical” data type then MATLAB interprets it diﬀerently: If A has 6 columns then A(:,[0,1,0,0,1,1]) is a matrix whose columns are the 2nd, 5th, and 6th columns of A. ...
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## This note was uploaded on 12/27/2011 for the course MAS 3114 taught by Professor Olson during the Fall '08 term at University of Florida.

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