9b. First, it is necessary to open up the linalg package. This is done
as follows.
> with(linalg):
We must define the matrix to Maple. This is done as follows.
> A:=matrix([[3,1],[2,1]]);
We must now substitute matrix a into the equation 2x2 - x + 1,
using the evalm command. This will give us 2A2 - A + 1. The 1 that we
added, means that we add (1 times) the identity matrix. This is done as
follows.
> B:=evalm(2*A^2-A+1);
13. Consider the matrix
| A = |
[ a11 |
0 |
0 |
... |
0 ] |
| [ 0 |
a22 |
0 |
... |
0 ] |
| ... | ... | ... | ... |
| [ 0 |
0 |
0 |
... |
ann ] |
where a11, a22, ...., ann are not equal to 0. To show that A is invertible,
we need to find a matrix, B, such that A*B = I. Multiplying these matrices
we get that [A]ii * [B]ii must equal
1 and, [A]ii * [B]ij must be equal
to 0 for every j which is not equal to i. So
[B]ii=1/aii, and [B]ij=0 for every
i and j such that i is not equal to j. Thus
| B = |
[ 1/a11 |
0 |
0 |
... |
0 ] |
| [ 0 |
1/a22 |
0 |
... |
0 ] |
| ... | ... | ... | ... |
| [ 0 |
0 |
0 |
... |
1/ann ] |
Now it is easy to check that indeed if B is this matrix then AB=I=BA.
14: Show that if matrix A satisfies A2 - 3*A + I = 0, then A-1 = 3*I - A.
Manipulate with the given equality:
-A2+3*A=I
Use the fact that A*I=A
-A2+3*A*I=I
Use the distributive law:
A(-A+3I)=I
Similarly we can prove that (3I-A)*A=I,
so by the definition of the inverse matrix, 3I - A is the inverse of A. This shows, in particular, that A is invertible.
16. Is the sum of two invertible matrices necessarily invertible?
No. Indeed, take A=I, B=-I. Then A is invertible (every identity matrix is
invertible: see examples after the definition of invertible matrices in the
notes). B is also invertible because if we multiply an
invertible matrix by a no-zero number, we get an invertible matrix (see the
Theorem about inverses). But A+B=I+(-I)=0 (easy to check) is not invertible
(see the properties of matrix operations in the notes).
23. To see if the given matrix is invertible we have to check if AB=I=BA
for some matrix B.
> A:=matrix([[1,0,1],[1,1,0],[0,1,1]]);
| A := |
[ 1 |
0 |
1 ] |
| [ 1 |
1 |
0 ] |
| [ 0 |
1 |
1 ] |
We want to find a 3 by 3 matrix B such that AB=I
> B:=matrix(3,3,[[x11,x12,x13],[x21,x22,x23],[x31,x32,x33]]);
| B := |
[ x11 |
x12 |
x13 ] |
| [ x21 |
x22 |
x23 ] |
| [ x31 |
x32 |
x33 ] |
> C:=evalm(A*B);
| C := |
[ x11 + x31 |
x12 + x32 |
x13 + x33 ] |
| [ x11 + x21 |
x12 + x22 |
x13 + x23 ] |
| [ x21 + x31 |
x22 + x32 |
x23 + x33 ] |
This matrix must be equal to I, that is the diagonal entries are equal to 1,
all other entries are equal to 0. This gives us a set of equations. To solve
this system of equations I will form the augmented matrix.
> D:=matrix(9,10,[[1,0,0,0,0,0,1,0,0,1],[0,1,0,0,0,0,0,1,0,0],[0,0,1,0,0,0,0,0,1,0],[1,0,0,1,0,0,0,0,0,0],[0,1,0,0,1,0,0,0,0,1],[0,0,1,0,0,1,0,0,0,0],[0,0,0,1,0,0,1,0,0,0],[0,0,0,0,1,0,0,1,0,0],[0,0,0,0,0,1,0,0,1,1]]);
| D := |
[ 1 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 ] |
| [ 0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 ] |
| [ 0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
0 ] |
| [ 1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 ] |
| [ 0 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 ] |
| [ 0 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 ] |
| [ 0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
0 ] |
| [ 0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
0 ] |
| [ 0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 ] |
Use the Gauss-Jordan method:
> F:=gaussjord(D);
| F := |
[ 1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1/2 ] |
| [ 0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1/2 ] |
| [ 0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
-1/2 ] |
| [ 0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
-1/2 ] |
| [ 0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1/2 ] |
| [ 0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1/2 ] |
| [ 0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1/2 ] |
| [ 0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
-1/2 ] |
| [ 0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1/2 ] |
This solution gives us the values for the entries of the matrix B. It is the
following matrix.
> B:=matrix(3,3,[[1/2,1/2,-1/2],[-1/2,1/2,1/2],[1/2,-1/2,1/2]]);
| B := |
[ 1/2 |
1/2 |
-1/2 ] |
| [ -1/2 |
1/2 |
1/2 ] |
| [ 1/2 |
-1/2 |
1/2 ] |
Now we can multiply A*B and B*A to check that the products are indeed equal to
I (notice the sign &* that we use for the matrix product, this sign should
be used instead of * if you want to evaluate an expression involving matrices):
> evalm(A&*B);
| [ 1 |
0 |
0 ] |
| [ 0 |
1 |
0 ] |
| [ 0 |
0 |
1 ] |
> evalm(B&*A);
| [ 1 |
0 |
0 ] |
| [ 0 |
1 |
0 ] |
| [ 0 |
0 |
1 ] |
29a ---- We are asked to show that B=C when we know that AB=AC
and A is invertible. Knowing that AB=AC and A is invertible we can multiply
both sides by A-1. This yields us the equation A-1(AB)=A-1(AC). This is
the same as saying (A-1*A)B = (A-1A)C. Because we know that (A-1 * A) = I
then we know that IB = IC. This is the same as saying that B=C.
29b --- We are asked why our answer in part a does not contradict the
answer from Example 3. In Example 3 A is not invertible. In part a A is
invertible.