1. **Problem Statement:** We are given a universal set $U = \{1, 2, \ldots, 12\}$ and subsets $A, B, C, D$ defined as: - $A = \{1, 2, 3, 10, 12\}$ with bit-string $111000000101$. - $B = \{x \in U \mid x \text{ is odd}\}$ - $C = \{x \in U \mid x \text{ is even}\}$ - $D = \{x \in U \mid x \leq 5\}$ We need to find: **(a)** (i) $B \cap D$ (ii) $B \cup C \cup D$ 2. **Formula and Rules:** - The bit-string representation of a subset of $U$ is a binary string of length 12 where the $i^{th}$ bit is 1 if $a_i$ is in the subset, else 0. - Intersection $\cap$ means elements common to both sets. - Union $\cup$ means elements in either or both sets. 3. **Step-by-step Solution:** **(a)(i) Find $B \cap D$:** - $B = \{1, 3, 5, 7, 9, 11\}$ (all odd numbers in $U$) - $D = \{1, 2, 3, 4, 5\}$ (all numbers $\leq 5$) - Intersection $B \cap D = \{1, 3, 5\}$ - Bit-string: positions 1, 3, 5 are 1, others 0. - Bit-string: $101010000000$ **(a)(ii) Find $B \cup C \cup D$:** - $B \cup C = U$ since $B$ is odd and $C$ is even, covering all elements. - Union with $D$ does not add new elements. - So $B \cup C \cup D = U = \{1, 2, \ldots, 12\}$ - Bit-string: all ones $111111111111$ --- **(b) Venn Diagrams for sets $X, Y, Z$:** - (i) $X \cap Y \cap Z$: region common to all three circles. - (ii) $X \cap (Y - Z)$: elements in $X$ and $Y$ but not in $Z$. - (iii) $(X \cap Y) \cup Z$: union of intersection of $X$ and $Y$ with $Z$. (No numeric answer, conceptual only.) --- **(c) Sets $A_i = \{i+1, i+2, \ldots, i+10\}$ for $i=1$ to $5$:** - $A_1 = \{2,3,4,5,6,7,8,9,10,11\}$ - $A_2 = \{3,4,5,6,7,8,9,10,11,12\}$ - $A_3 = \{4,5,6,7,8,9,10,11,12,13\}$ - $A_4 = \{5,6,7,8,9,10,11,12,13,14\}$ - $A_5 = \{6,7,8,9,10,11,12,13,14,15\}$ (i) $X = \bigcup_{i=1}^5 A_i = \{2,3,4,5,6,7,8,9,10,11,12,13,14,15\}$ (ii) $Y = \bigcap_{i=1}^5 A_i$ is the intersection of all sets. - The smallest common elements are those in all $A_i$. - The largest lower bound is 6 (from $A_5$), smallest upper bound is 11 (from $A_1$). - So $Y = \{6,7,8,9,10,11\}$ --- **(d) Venn diagram proofs:** - (i) $(A \cup B)^c = A^c \cap B^c$ (De Morgan's law) - (ii) $(A \cap B)^c = A^c \cup B^c$ (De Morgan's law) --- **7(a) Matrix $M = \begin{bmatrix} x & -2 \\ 5 & 4 \end{bmatrix}$** (i) Find $x$ such that $\det(M) = -50$: - $\det(M) = x \cdot 4 - (-2) \cdot 5 = 4x + 10$ - Set equal to $-50$: $4x + 10 = -50$ - $4x = -60$ - $x = -15$ (ii) Find $x$ for which $M$ is singular (determinant zero): - $4x + 10 = 0$ - $4x = -10$ - $x = -\frac{10}{4} = -2.5$ (iii) Find $M^{-1}$ if $x=2$: - $\det(M) = 4 \cdot 2 + 10 = 8 + 10 = 18$ - $M^{-1} = \frac{1}{18} \begin{bmatrix} 4 & 2 \\ -5 & 2 \end{bmatrix}$ (swap diagonal, negate off-diagonal) --- **7(b) Given matrices:** $P = \begin{bmatrix} -3 & 5 & 7 \\ 6 & -2 & 0 \\ -5 & 10 & 21 \end{bmatrix}$ $Q = \begin{bmatrix} -1 & -3 & 5 \\ 6 & 2 & 0 \\ 5 & -5 & 6 \end{bmatrix}$ (i) Find $P + 2Q$: - Multiply $Q$ by 2: $2Q = \begin{bmatrix} -2 & -6 & 10 \\ 12 & 4 & 0 \\ 10 & -10 & 12 \end{bmatrix}$ - Add to $P$: $P + 2Q = \begin{bmatrix} -3-2 & 5-6 & 7+10 \\ 6+12 & -2+4 & 0+0 \\ -5+10 & 10-10 & 21+12 \end{bmatrix} = \begin{bmatrix} -5 & -1 & 17 \\ 18 & 2 & 0 \\ 5 & 0 & 33 \end{bmatrix}$ (ii) Find $P^T + Q^T$: - $P^T = \begin{bmatrix} -3 & 6 & -5 \\ 5 & -2 & 10 \\ 7 & 0 & 21 \end{bmatrix}$ - $Q^T = \begin{bmatrix} -1 & 6 & 5 \\ -3 & 2 & -5 \\ 5 & 0 & 6 \end{bmatrix}$ - Sum: $P^T + Q^T = \begin{bmatrix} -4 & 12 & 0 \\ 2 & 0 & 5 \\ 12 & 0 & 27 \end{bmatrix}$ (iii) Find $(P + Q)^T$: - $P + Q = \begin{bmatrix} -4 & 2 & 12 \\ 12 & 0 & 0 \\ 0 & 5 & 27 \end{bmatrix}$ - Transpose: $(P + Q)^T = \begin{bmatrix} -4 & 12 & 0 \\ 2 & 0 & 5 \\ 12 & 0 & 27 \end{bmatrix}$ (iv) Relationship: - $P^T + Q^T = (P + Q)^T$ - This shows that transpose distributes over addition. --- **7(c) Solve simultaneous equations using matrix method:** Equations: $2x + y - 10 = 0 \Rightarrow 2x + y = 10$ $4x - 2y + 12 = 0 \Rightarrow 4x - 2y = -12$ Matrix form: $AX = B$ where $A = \begin{bmatrix} 2 & 1 \\ 4 & -2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix}, B = \begin{bmatrix} 10 \\ -12 \end{bmatrix}$ Find $A^{-1}$: $\det(A) = 2 \times (-2) - 4 \times 1 = -4 - 4 = -8$ $A^{-1} = \frac{1}{-8} \begin{bmatrix} -2 & -1 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ \frac{1}{2} & -\frac{1}{4} \end{bmatrix}$ Compute $X = A^{-1} B$: $x = \frac{1}{4} \times 10 + \frac{1}{8} \times (-12) = 2.5 - 1.5 = 1$ $y = \frac{1}{2} \times 10 - \frac{1}{4} \times (-12) = 5 + 3 = 8$ **Final solution:** $x=1$, $y=8$.