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Math 2270 Sec. 2 MAPLE Assignment #1: Solutions.
with(linalg):
Warning, the protected names norm and trace have been redefined and unprotected
You are to create a document in which you answer the following questions, using a mixture
of MAPLE computations and textual insertions (using # to comment or handwritten text.)
Each problem is worth 5 points.
1.a Define
( 2 3 4 ) ( 3 2 1 )
A = ( 5 6 7 ) , B = ( 4 3 2 ).
( 8 9 0 ) ( 5 4 3 )
Compute AB and BA. Are they the same?
Solution:
A:= matrix([[ 2 , 3, 4], [5, 6, 7], [8, 9, 0]]);
A :=
B:= matrix([[3, 2, 1 ], [4 , 3, 2], [5, 4, 3]]);
B :=
multiply(A,B);
multiply(B,A);
The products AB and BA are not the same.
1.b. Compute A + B and B + A. Are they the same?
Solution:
evalm(A+B);
evalm(B+A);
Therefore A+B and B+A are the same.
1.c. Define C to be A + B. Compute C^2 and compare it to
A^2 + 2AB + B^2.
Are they the same? Can you think of a small change you could make in the
expression A^2 + 2AB + B^2 in order to make it equal to C^2?
Solution.
C:=evalm(A+B);
C :=
evalm(C^2);
evalm(A^2+B^2+2A&B);
evalm(A^2+B^2+A&B + B&A);
Therefore
C^2= A^2+B^2+A.B + B.A;
C =
2 A + + +
2 B
2 ( A. B ) ( B. A )
1.d. Compute the transpose of AB and compare it to the product of the transpose of A
and the transpose of B multiplied in the correct order to get equality.
transpose(A&*B);
J :=
col(J, 4);# The last column of the matrix J is the solution
vector.
Solution using linsolve:
x:=linsolve(A,v);
x :=
Solution using the inverse matrix: x=inverse(A)&*v.
evalm( inverse(A)&* v);
2.a Solve Bx = w where B is as above and w=(1,2,3). Verify that your
solution solves Bx = w.
Solution.
w:= vector([1,2,3]);
w :=[ 1 2 3, , ]
augment(B, w);
K:=rref(%);
K :=
We this see that the system has infinitely many solutions. We can find them all using linsolve;
note that the vector
col(K, 4);
[ -1 2 0, , ]
is only one solution of the system.
X:=linsolve(B, w);
X :=[ _t , , ] 1
− 2 _t 1
_t + 1
Hence MAPLE’s solution is: x=t, y=-2t, z=t+1. Let’s check that this solution really works:
evalm(B&*X);
[ 1 2 3, , ]
which is indeed the vector w of the right hand side.
2.b Repeat your work in order to solve Bx = z where z = (1,2,4). Explain
your answer.
Solution.
z:=vector([1,2,4]);
z :=[ 1 2 4, , ]
Y:=linsolve(B, z);
Y :=
So, MAPLE could not find a solution. Let’s see what is wrong by using RREF:
rref(augment(B, z));
Thus the last equation says 0=1, i.e. the system has no solutions. What is happening here is that the
vector
w is in the image of B, but the vector z is not.
3.a Plot the graph of the line x/2 + 3y=7 on the interval x=2..
using both plot and implicitplot commands
Solution.
with(plots):
solve(x/2+ 3*y=7, y);
x
plot(%, x=2..4); #Ordinary plot
3.b. By plotting lines (anyway you like) determine if the system of
equations 3x+2y=5, 2x-y=3 has a solution and if the solution is unique
Solution.
L1:=implicitplot(3x+2y=5, x=-1..3, y=-2..2):
L2:= implicitplot(2*x-y=3, x=-1..3, y=-2..2):
display({L1,L2});
0
1
2
y
0.5 1 1.5 2 2.5 3
x
Therefore the lines intersect in a single point, which means that the system has unique solution.
