Mathematics 243, section 1 – Algebraic Structures Solutions for Midterm Exam 1 – September 29, 2006

I. For this problem,

U = Z
A = {x ∈ Z : x is odd and − 10 ≤ x ≤ −4}
B = {−2, −1, 0, 1}
C = {x ∈ Z : x < 0}

and f : Z → Z is the mapping defined by

f(x) = \{ \begin{matrix}x + 3 & if x is even \\ x − 1 & if x is odd \end{matrix} .

A) (10) What is the set A ∪ (B ∩ C′)?

Solution: C′ = {x ∈ Z : x ≥ 0}, so B ∩ C′ = {0, 1}, and

A ∪ (B ∩ C′) = {−9, −7, −5, 0, 1}.

B) (10) What is the set f(B)?

Solution: f(B) = {f(−2), f(−1), f(0), f(1)} = {1, −2, 3, 0} (using the first part of the definition to find f(−2), f(0) and the second to get f(−1), f(1)).

C) (10) Given: f is a permutation of Z (you don’t need to show this). Find the inverse mapping f^{−1} : Z → Z. Solution: Since x + 3 is odd when x is even, and x − 1 is even when x is odd:

f^{−1}(x) = \{ \begin{matrix}x − 3 & when x is odd \\ x + 1 & when x is even \end{matrix}

II. Let f : A → B be any mapping. A) (15) Prove that for any subsets T_{1} and T_{2} of B, f^{−1}(T_{1} ∪ T_{2}) = f^{−1}(T_{1}) ∪ f^{−1}(T_{2}).

Solution: by the definitions of the inverse image and union of sets we have

x ∈ f^{−1}(T_{1} ∪ T_{2}) ⇔ f(x) ∈ T_{1} ∪ T_{2}
⇔ f(x) ∈ T_{1} or f(x) ∈ T_{2}
⇔ x ∈ f^{−1}(T_{1}) or x ∈ f^{−1}(T_{2})
⇔ x ∈ f^{−1}(T_{1}) ∪ f^{−1}(T_{2})

Hence f^{−1}(T_{1} ∪ T_{2}) = f^{−1}(T_{1}) ∪ f^{−1}(T_{2}) because each of these sets is contained in the other.

B) (10) Is f^{−1}(f(S)) always equal to S for all subsets S ⊆ A? If you say no, give a counterexample; if you say yes, say why.

Solution: This is not always true. It can fail when the mapping f is not injective (one-to-one). For instance, let f : R → R be defined by f(x) = x^{2}. Then if S = {1}, we have f(S) = {1}. But f^{−1}(f(S)) = {−1, 1}, since both f(−1) = 1 and f(1) = 1.

III. Let ∗ be the binary operation on R defined by x ∗ y = xy − 2x. A) (10) Show with specific counterexamples that ∗ is not associative, and not commuta-

tive.

Solution: ∗ is not associative because for instance (1 ∗ 2) ∗ 1 = (2 − 2) ∗ 1 = 0 ∗ 1 = 0, but 1 ∗ (2 ∗ 1) = 1 ∗ (2 − 4) = 1 ∗ (−2) = −2 − 2 = −4. Similarly, ∗ is not commutative: 1 ∗ 2 = 0 but 2 ∗ 1 = −2.

B) (15) An element e is said to be a left identity for a binary operation if e ∗ x = x for all x. Similarly, e is a right identity if x ∗ e = x for all x. Does the ∗ operation have a left identity? Does it have a right identity? If so, say what e is in each case. If not, explain how you reach your conclusions.

Solution: ∗ does have a right identity e = 3 since x ∗ 3 = 3x − 2x = x for all x. On the other hand, it does not have a left identity because if e ∗ y = y for all y, then we have ey − 2e = y for all y ∈ R This equation says e = \frac{y}{y−2}, so while there is some e that works for each y individually, there is no single e that works for all y.

IV. In both parts of this problem,

A = ( \begin{matrix}1 & 1 \\ 3 & 4 \end{matrix} ) B = ( \begin{matrix}4 & −1 \\ −3 & 1 \end{matrix} ) .

A) (10) Compute the matrix product AB or say why it doesn’t exist.

Solution: A and B are 2 × 2 matrices, so the product is defined, and

AB = ( \begin{matrix}1 & 1 \\ 3 & 4 \end{matrix} ) ( \begin{matrix}4 & −1 \\ −3 & 4 \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )

(the 2 × 2 identity matrix). Note that

BA = ( \begin{matrix}4 & −1 \\ −3 & 1 \end{matrix} ) ( \begin{matrix}1 & 1 \\ 3 & 4 \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )

also.

B) (10) Define a mapping

F : M_{2×2}(R) → M_{2×2}(R)

for X a 2 × 2 matrix by by F (X) = BX (matrix product). Prove that F is one-to-one (injective) and give its left inverse mapping.

Solution: We show F is injective first. Assume F (X) = F (Y) for two matrices X, Y. Then BX = BY. Multiply A on the left of both sides in this equation: A(BX) = A(BY). By associativity of matrix multiplication, this says (AB)X = (AB)Y. Since AB = I_{2} (the 2 × 2 identity matrix), this implies I_{2}X = I_{2}Y, so X = Y.

The left inverse mapping of F is G : M_{2×2}(R) → M_{2×2}(R) defined by G(X) = AX. This follows because for all X

(G ∘ F)(X) = G(F (X)) = A(BX) = (AB)X = I_{2}X = X

(The fact that F is injective also follows from the fact that there is a left inverse.)