Description

1. Let .
is now upper triangular with two
pivots.
(b) Let b = (1,6,3)T (to match dimensions). The system A~x =~b can be solved by augmenting A with b:
.
This shows that the system has infinite solutions. If we let ~x = (x1,x2,x3)T, then from the reduce matrix above we know that the solutions have the form where x3 ∈ R is a free variable (although
everything here is real we could actually even say x3 ∈ C).
(c) Let b = (1,6,5)T (to match dimensions). The system A~x =~b can be solved by augmenting A with b:
.
This system has no solutions. The reason is that we have found through reduction that in order for ~x = (x1,x2,x3)T to be a solution, it must be the case that 0x3 = 0 = 2 according to the reduced matrix’s last row. But @x3 such that 0x3 = 2. So the system is inconsistent and has no solution.
(d) Since the system described in part (b) has infinite solutions, all ofthe form , we get two solutions by taking first x3 = 5 and then x3 = 0:
.
However, as we have seen, the system described in part (c) has no solutions to find.
2. Suppose AB = I,CA = I, where I is the nxn identity matrix.
(b) B = IB = (CA)B = C(AB) = CI = C. Therefore B = C.
(c) The matrix A is invertible iff there exists a matrix, denoted A−1 with the property that AA−1 = A−1A = I. From parts (b), (c) we see that the matrix B (= C) has this property. So A is invertible and A−1 = B.
3. Let A be a square matrix such that A2 = A (i.e. A is a projector). Then we have the following equalites:
(I − A)2 = I2 − 2IA + A2 = I − 2A + A = I − A
(I − A)7 = (I − A)(I − A)2(I − A)2(I − A)2
= (I − A)(I − A)(I − A)(I − A)
= (I − A)2(I − A)2
= (I − A)(I − A) (by thelastpart)
= (I − A)2
= (I − A)
4.
(a)
1 6  9 
−32 + 24 =  2 
3 2 −5
(b)
1 6 9  1 0 −3
A = 2 4 2  ∼ 0 1 2  3 2 −5 0 0 0
From this reduction (which I did on paper to solve part (a)) we see that there are two pivots.
5. Suppose A is a 6×20 matrix and B is a 20×7 matrix.
(a) The dimensions of C = AB are 6×7.
(b) Suppose A,B, and C are partitioned as shown below and that we know A11 is 2×10, B22 is 4×3 and C is ?x4.
A11 A12 A13
A21 A22 A23

A =
B11 B12
B21 B22
B31 B32

B = 
C11 C12
C21 C22

C =
Since A11 is 2×10 and C is ?x4, B11 must be 10×4. So C11 is 2×4. Since B22 is 4×3, B21 must be 4×4 and so A12 is 2×4. We can continue to fill in missing information making the assumption that if, for example, A11 has 2 rows then so must A12 and A13. Similarly, if B11 has 4 columns then so must both the blocks beneath it. This leads to the following dimensions:
2×10 2×4 2×6
4×10 4×4 4×6

A =
10×4 10×3
4×4 4×3
6×4 6×3

B = 
2×4 2×3
4×4 4×3

C =
where it should be clear that each block in the partition scheme from
before has the dimensions corresponding to its location in this second set.
(c) Writing out the blocks of C as the result of the block multiplications from the last part we have:
A11B11 + A12B21 + A13B31 A11B12 + A12B22 + A13B32
A21B11 + A22B21 + A23B31 A21B12 + A22B22 + A23B32

C =.
6. Let A be an m × n matrix.
(a) If QeRe = A, the smallest possible Qe is m × m and the smallest possible Re is m × n. Then QeRe still results in an m × n matrix. The reason that Qe is m × m is because its number or rows must equal the number of rows of A and since Qe is orthogonal (or perhaps unitary if we are dealing with complex matrices)it must be square. Because Qe is m × m this forces Re to have m rows and to also Re must have the same number of columns as A. (b)

Since the instructions say not to invert any matrices this is as far as I can go.