(1) Sec 2.6, 1 (justify your answers), 10

(2) Show that if L1 and L2 are lower triangular,
then so is L1*L2. Show that if U1 and U2 are
upper triangular, so is U1*U2.

(3) Show that if L is lower triangular and invertible,
so is L^{-1}. Show that if U is upper triangular
and invertible, so is U^{-1}.

[Problems (2) and (3) together show that the set of
n by n invertible lower (or upper) triangular matrices
are closed under multiplication and inversion. In the
language of Math 113 (which some of you are taking), this
means the lower (or upper) invertible triangular matrices 
form a group.  (This dicussion of groups is not a homework 
problem, just an example for the Ma113 students.)]

(4) Show that if L1 and L2 are "unit" lower triangular,
i.e. their diagonal entries are all ones, then so is L1*L2.
Show that if U1 and U2 are unit upper triangular, so is U1*U2.

(5) Show that if L is unit lower triangular, so is L^{-1}.
Show that if U is unit upper triangular, so is U^{-1}.

[Problems (4) and (5) show that the lower (or upper) unit
triangular matrices also form a group, another example
for Ma113 students.]

(6) If P is a permutation matrix, and Y = X*P, show that
Y has the same columns as X but in a permuted order.

(7) Show that if P1 and P2 are permutation matrices, then
so is P1*P2.

(8) Show that if P is a permutation matrix, so is P^t,
and P^t = P^{-1}. 

[Problems (7) and (8) together show that the set of 
n by n permutation matrices also form a group,
and are indeed essentially the same as the group of 
permutations. This is another example for Ma113 students.]

(9) Use LU decomposition to compute an LU decomposition of
    A = [ 0 2 4 ]
        [ 0 1 2 ]
        [ 3 2 1 ]
    and use it to describe the complete solution set of
    A*x = [ 8]  and A*x = [ 0]
          [ 4]            [ 1]
          [-4;            [ 0]

(10) Let A be an m by n matrix and B be an n by p matrix,
so the m by p matrix C = A*B is defined. 
    Let m = m_1 + m_2 + ... + m_r,
        n = n_1 + n_2 + ... + n_s,
        p = p_1 + p_2 + ... + p_t,
and write A and B as block matrices:

           n_1  n_2      n_s
    A = [ A_11 A_12 ... A_1s ] m_1
        [ A_21 A_22 ... A_2s ] m_2
        [           ...      ]
        [ A_r1 A_r2 ... A_rs ] m_r

where block A_ij is m_i by n_j, and

           p_1  p_2      p_t
    B = [ B_11 B_12 ... B_1t ] n_1
        [ B_21 B_22 ... B_2t ] n_2
        [           ...      ]
        [ B_s1 B_s2 ... B_st ] n_s

where block B_ij is n_i by p_j.

Show that the product C = A*B can be written as 
the block matrix

           p_1  p_2      p_t
    C = [ C_11 C_12 ... C_1t ] m_1
        [ C_21 C_22 ... C_2t ] m_2
        [           ...      ]
        [ C_r1 C_r2 ... C_rt ] m_r

where C_ij = sum_{k=1 to s}  A_ik*B_kj.
When all the n_i = m_j = p_k = 1, so
all the blocks A_ij, B_ij and C_ij are
1 by 1 (scalars), this is just the usual
definition of matrix multiplication.