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]]);
 A := [ 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);
 B := [ 20 7 ] [ 14 6 ]

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.