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+//page 63
+//Example 2.22
+clear;
+clc;
+close;
+A = [1 2 0 3 0;1 2 -1 -1 0;0 0 1 4 0;2 4 1 10 1;0 0 0 0 1];
+disp(A,'A = ');
+//part a
+T = A;                  //Temporary storing A in T
+disp('Taking an identity matrix P:');
+P = eye(5,5);
+disp(P,'P = ');
+disp('Applying row transformations on P and A to get a row reduced echelon matrix R:');
+disp('R2 = R2 - R1 and R4 = R4 - 2* R1');
+A(2,:) = A(2,:) - A(1,:);
+P(2,:) = P(2,:) - P(1,:);
+A(4,:) = A(4,:) - 2 * A(1,:);
+P(4,:) = P(4,:) - 2 * P(1,:);
+disp(A,'A = ');
+disp(P,'P = ');
+disp('R2 = -R2 , R3 = R3 - R1 + R2 and R4  = R4 - R1 + R2');
+A(2,:) = -A(2,:);
+P(2,:) = -P(2,:);
+A(3,:) = A(3,:) - A(2,:);
+P(3,:) = P(3,:) - P(2,:);
+A(4,:) = A(4,:) - A(2,:);
+P(4,:) = P(4,:) - P(2,:);
+disp(A,'A = ');
+disp(P,'P = ');
+disp('Mutually interchanging R3, R4 and R5');
+x = A(3,:);
+A(3,:) = A(5,:);
+y = A(4,:);
+A(4,:) = x;
+A(5,:) = y - A(3,:);
+x = P(3,:);
+P(3,:) = P(5,:);
+y = P(4,:);
+P(4,:) = x;
+P(5,:) = y - P(3,:);
+R = A;
+A = T;
+disp(R,'Row reduced echelon matrix R = ');
+disp(P,'Invertible Matrix P = ');
+disp('Invertible matrix P is not unique. There can be many that depends on operations used to reduce A');
+disp('-----------------------------------------');
+//part b
+disp('For the basis of row space W of A, we can take the non-zero rows of R');
+disp('It can be given by p1, p2, p3');
+p1 = R(1,:);
+p2 = R(2,:);
+p3 = R(3,:);
+disp(p1,'p1 = ');
+disp(p2,'p2 = ');
+disp(p3,'p3 = ');
+disp('-----------------------------------------');
+//part c
+disp('The row space W consists of vectors of the form:');
+disp('b = c1p1 + c2p2 + c3p3');
+disp('i.e. b = (c1,2*c1,c2,3*c1+4*c2,c3) where, c1 c2 c3 are scalars.');
+disp('So, if b2 = 2*b1 and b4 = 3*b1 + 4*b3  =>  (b1,b2,b3,b4,b5) = b1p1 + b3p2 + b5p3');
+disp('then,(b1,b2,b3,b4,b5) is in W');
+disp('-----------------------------------------');
+//part d
+disp('The coordinate matrix of the vector (b1,2*b1,b2,3*b1+4*b2,b3) in the basis (p1,p2,p3) is column matrix of b1,b2,b3 such that:');
+disp('  b1');
+disp('  b2');
+disp('  b3');
+disp('-----------------------------------------');
+//part e
+disp('Now, to write each vector in W as a linear combination of rows of A:');
+disp('Let b = (b1,b2,b3,b4,b5) and if b is in W, then');
+disp('we know,b = (b1,2*b1,b3,3*b1 + 4*b3,b5)  =>  [b1,b3,b5,0,0]*R');
+disp('=> b = [b1,b3,b5,0,0] * P*A  =>  b = [b1+b3,-b3,0,0,b5] * A');
+disp('if b = (-5,-10,1,-11,20)');
+b1 = -5;
+b2 = -10;
+b3 = 1;
+b4 = -11;
+b5 = 20;
+x = [b1 + b3,-b3,0,0,b5];
+disp(']',A,'[','*',')',x,'(','b = ');
+disp('-----------------------------------------');
+//part f
+disp('The equations in system RX = 0 are given by R * [x1 x2 x3 x4 x5]');
+disp('i.e., x1 + 2*x2 + 3*x4');
+disp('x3 + 4*x4');
+disp('x5');
+disp('so, V consists of all columns of the form');
+disp('[','X=');
+disp('  -2*x2 - 3*x4');
+disp('  x2');
+disp('  -4*x4');
+disp('  x4');
+disp('  0');
+disp('where x2 and x4  are arbitrary',']');
+disp('-----------------------------------------');
+//part g
+disp('Let x2 = 1,x4 = 0 then the given column forms a basis of V');
+x2 = 1;
+x4 = 0;
+disp([-2*x2-3*x4; x2; -4*x4; x4; 0]);
+disp('Similarly,if x2 = 0,x4 = 1 then the given column forms a basis of V');
+x2 = 0;
+x4 = 1;
+disp([-2*x2-3*x4; x2; -4*x4; x4; 0]);
+disp('-----------------------------------------');
+//part h
+disp('The equation AX = Y has solutions X if and only if');
+disp('-y1 + y2 + y3 = 0');
+disp('-3*y1 + y2 + y4 -y5 = 0');
+disp('where, Y = (y1 y2 y3 y4 y5)');
+//end