Name: I. Suppose that A is an n × n matrix and that is a constant real number. D

Question

Name: I. Suppose that A is an n × n matrix and that is a constant real number. Denne W to be the set of all vectors that satisfy the equation A:r = . (a) Show that the set W is a subspace of R". (Hint: §4.2 Theorem 1) (b) Let = 1 and let A be given by 16 3 4 4 7 2 2 1 11 1 3 4 Find a basis for the subspace W, the set of solutions to AZ = Hint 1: Theorem 2 of §4.2 says that the set of solutions to systems of the form Bi = 0 form subspaces. Use the rules of matrix algebra to transform Ar = x into the form Br = 0, then solve for the solution space (via the algorithm in §4.4 p 258) and find a basis for the solution space of the transformed equation Bx 258) and find a basis for the solution space of the transformed equation B#- 0 Hint 2: The fact that li = a, where I is the 4 × 4 identity matrix, is useful. Hint 3: Use either Maple or MATLAB to compute RREF of matrix B. Maple: LinearA1gebra [ReducedRowEchelonForm] (B) MATLAB: rref (B)

Explanation / Answer

Reduced row echelon form of matrix (Gauss-Jordan elimination)

rref(sym(magic(4)))

ans =

[ 1, 0, 0, 1]

[ 0, 1, 0, 3]

[ 0, 0, 1, -3]

[ 0, 0, 0, 0]

Compute the reduced row echelon form of the following symbolic matrix:

syms a b c

A = [a b c; b c a; a + b, b + c, c + a];

rref(A)

ans =

[ 1, 0, -(- c^2 + a*b)/(- b^2 + a*c)]

[ 0, 1, -(- a^2 + b*c)/(- b^2 + a*c)]

[ 0, 0,                            0]

R = rref(A) produces the reduced row echelon form of A using Gauss Jordan elimination with partial pivoting. A default tolerance of (max(size(A))*eps *norm(A,inf)) tests for negligible column elements.

[R,jb] = rref(A) also returns a vector jb such that:

[R,jb] = rref(A,tol) uses the given tolerance in the rank tests.

Roundoff errors may cause this algorithm to compute a different value for the rank than rank, orth and null. Additionally, use mldivide to solve linear systems when high precision is required.

Examples

Use rref on a rank-deficient magic square:

A = magic(4), R = rref(A)

A =

   16    2   3   13

    5   11   10    8

    9    7    6   12

    4   14   15    1

R =

    1    0    0    1

    0    1    0    3

    0    0    1   -3

    0    0    0    0

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