(M) Let \(A = \left( {\begin{aligned}{*{20}{r}}{ - 6}&3&{ - 27}&{ - 33}&{ - 13}\\6&{ - 5}&{25}&{28}&{14}\\8&{ - 6}&{34}&{38}&{18}\\{12}&{ - 10}&{50}&{41}&{23}\\{14}&{ - 21}&{49}&{29}&{33}\end{aligned}} \right)\). Construct

a matrix \(N\) whose columns form a basis for \({\rm{Nul}}A\), and

construct a matrix \(R\) whose rows form a basis for \({\rm{Row}}A\) (see

Section 4.6 for details). Perform a matrix computation with

\(N\)and \(R\) that illustrates a fact from Theorem 3.

A linearly independent set that spans the whole space is called the basis of a space. There is more than one basis for a vector space.

Given that

\(A = \left( {\begin{aligned}{*{20}{r}}{ - 6}&3&{ - 27}&{ - 33}&{ - 13}\\6&{ - 5}&{25}&{28}&{14}\\8&{ - 6}&{34}&{38}&{18}\\{12}&{ - 10}&{50}&{41}&{23}\\{14}&{ - 21}&{49}&{29}&{33}\end{aligned}} \right)\)

Hence, by using MATLAB, construct the required matrix.

Enter the matrix

>> A=(-6 3 -27 -33 -13;6 -5 25 28 14; 8 -6 34 38 18 ; 12 -10 50 41 23 ; 14 -21 49 29 33);

Function to compute the Null basis:

(function N = nulbasis (A)

(R, pivcol) = rref(A, sqrt(eps));

(m, n) = size (A) ;

r = length(pivcol); freecol = l:n;

freecol(pivcol) = ();

N = zeros(n, n-r);

N (freecol, : ) = eye(n-r); N(pivcol, : ) = -R(l:r, freecol);

Compute the Null basis:

>> N = nulbasis(A)

We find that:

\(N = \left( {\begin{aligned}{*{20}{r}}{ - 5}&{0.33}\\{ - 1}&{1.33}\\1&0\\0&{ - 0.33}\\1&1\end{aligned}} \right)\)

Determine the row-reduce echelon form of the given matrix to find \(R\).

>> rref(A)

\(\left( {\begin{aligned}{*{20}{r}}1&0&5&0&{ - 0.33}\\0&1&1&0&{ - 1.33}\\0&0&0&1&{0.33}\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\)

Thus,

\(R = \left( {\begin{aligned}{*{20}{r}}1&0&5&0&{\frac{{ - 1}}{3}}\\0&1&0&1&{\frac{{ - 4}}{3}}\\0&0&0&1&{\frac{1}{3}}\end{aligned}} \right)\)

Compute the Product R and N:

>> R*N

Thus,

\(R \cdot N = \left( {\begin{aligned}{*{20}{r}}0&0\\0&0\\0&0\\0&0\\0&0\end{aligned}} \right)\)

Hence, \(R\) is orthogonal to \(N\).