Test 1.

1. Prove that if A and B are invertible matrices of the same size then a) AB is invertible and b) the inverse of AB is equal to the inverse of B times the inverse of A.

Solution

Let A' be the inverse of A, B' be the inverse of B. Multiply (AB) by (B'A'): (AB)(B'A')=(associativity of the matrix product)= A(BB')A'=(since B' is the inverse of B) = AIA'= (since AI=A)= AA'=(since A' is the inverse of A)=I. Thus (AB)(B'A')=I. Similarly (B'A')(AB)=I. By the definition of the inverse, these two equalities mean that AB is invertible and B'A' is the inverse of it.

2. Is the product of two symmetric matrices of the same size

a) always symmetric?

b) never symmetric?

Solution

a) No. Example:
Let A be the following matrix:

                           1 2
                           2 1

and let B be the following matrix:

                           2 0
                           0 1

Then both A and B are symmetric but their product AB:

                           2 2             
                           4 1

is not symmetric.

b) No. Indeed, let A=B=0 (the zero matrix) then A=B=AB=0 are all symmetric.

3. For which values of the parameter a is the following matrix invertible?

                          a 1 1
                          a a 1
                          a a a

In the case when it is invertible, find the inverse.

Solution

Augment this matrix with the identity matrix:

                        a 1 1 1 0 0          
                        a a 1 0 1 0
                        a a a 0 0 1

We shall reduce this matrix to the reduced row echelon form trying to avoid fractions. First subtract the first row from the other two rows:

                        a 1   1    1 0 0
                        0 a-1 0   -1 1 0
                        0 a-1 a-1 -1 0 1

Now subtract the second row from the third row:

                       a 1   1    1  0 0
                       0 a-1 0   -1  1 0
                       0 0   a-1  0 -1 1

Now we need to divide by a and by a-1, so we have to consider three cases.
Case 1. a=0. In this case the matrix has the following form

                      0  1  1  1  1 0
                      0 -1  0 -1  1 0
                      0  0 -1  0 -1 1

Divide the second and the third rows by -1 and subtract them from the first row:

                      0 0 0 0  1 -1
                      0 1 0 1 -1  0
                      0 0 1 0  1 -1

We see that the reduced row echelon form of the original matrix is not the identity matrix. So by a theorem about invertible matrices, this matrix is not invertible.

Case 2. Let a=1. Then the matrix has the following form:

                      1 1 1  1  0 0             
                      0 0 0 -1  1 0
                      0 0 0  0 -1 1

Again the reduced for of our matrix is not the identity matrix so the matrix is not invertible.

Case 3. Let a be neither 0 nor 1. Then divide the first row by a and the other two rows by a-1:

                      1 1/a 1/a   1/a       0         0
                      0  1   0  -1/(a-1)  1/(a-1)     0
                      0  0   1     0     -1/(a-1)  1/(a-1)

Now subtract the second and the third rows multiplied by 1/a from the first row:

                     1 0 0  1/(a-1)   0     -1/a(a-1)
                     0 1 0 -1/(a-1) 1/(a-1)   0
                     0 0 1    0     -1/(a-1) 1/(a-1)

The left half of this matrix is the identity matrix, so the original matrix was invertible. The right half is the inverse. So in this case the matrix is invertible and its inverse is the following:

                        1/(a-1)    0     -1/(a(a-1))
                       -1/(a-1)  1/(a-1)    0
                          0     -1/(a-1)  1/(a-1)

4. Represent the following matrix as a product of elementary matrices:

                          3 4
                          5 6

Solution

Reduce this matrix to the identity matrix:

     [3 4]   [1 4/3]  [1  4/3]  [1 4/3]  [1 0] 
     [5 6] , [5  6 ], [0 -2/3], [0  1 ], [0 1]

Here are the elementary matrices corresponding to these row operations:

         [1/3 0]   [ 1 0]    [1 0   ]  [1 -4/3]
         [0   1],  [-5 1],   [0 -3/2], [0   1 ]

Here are the inverses of these matrices (they correspond to the inverse row operations):

         [3 0]  [1 0]  [1   0 ]  [1 4/3]
         [0 1], [5 1], [0 -2/3], [0  1 ]

The product of these 4elementary matrices is equal to the original matrix.

5. Find the general solution of the following system of equations:

              2x+2y+3z+2t = 3
              x+y+2z+t+u  = 2
              x+y+t-u     = 0

Solution

Here is the augmented matrix of our system:

             2 2 3 2  0 3
             1 1 2 1  1 2
             1 1 0 1 -1 0

Swap the first two rows:

             1 1 2 1  1 2
             2 2 3 2  0 3   
             1 1 0 1 -1 0

Subtract twice the first row from the second row and the first row from the third row:

             1 1  2 1  1  2
             0 0 -1 0 -2 -1
             0 0 -2 0 -2 -2

Divide the second row by -1:

             1 1  2 1  1  2
             0 0  1 0  2  1
             0 0 -2 0 -2 -2

Add the second row multiplied by 2 to the third row:

             1 1 2 1  1 2
             0 0 1 0  2 1
             0 0 0 0  2 0

Divide the third row by 2:

            1 1 2 1  1 2
            0 0 1 0  2 1
            0 0 0 0  1 0

Subtract twice the second row from the first row:

            1 1 0 1 -3 0
            0 0 1 0  2 1
            0 0 0 0  1 0

Subtract twice the third row from the second row and add three times the third row from the first row:

            1 1 0 1 0 0
            0 0 1 0 0 1
            0 0 0 0 1 0

This matrix is in the row echelon form. Thus x, z and u are leadinf unknowns, y, t are free unknowns. The general solution is:

          y=p
          t=q
          x =-p-q
          z = 1
          u = 0