Step 1: Finding Eigenvalues of Matrix \( A \)
- We first find the eigenvalues of matrix \( A \). To do this, we solve the characteristic equation: \[ \det(A - \lambda I) = 0 \] \[ A - \lambda I = \begin{bmatrix} 1-\lambda & 1 & 1
1 & 1-\lambda & 1
1 & 1 & -1-\lambda \end{bmatrix} \] - The determinant of the above matrix is calculated, and we solve for \( \lambda \). The eigenvalues of \( A \) are \( \lambda_1 = 3, \lambda_2 = -1, \lambda_3 = -1 \).

Step 2: Finding Eigenvalues of \( A^{13} \)
- If \( \lambda \) is an eigenvalue of matrix \( A \), then \( \lambda^{13} \) will be the corresponding eigenvalue of matrix \( A^{13} \). - Therefore, the eigenvalues of \( A^{13} \) are: \[ \lambda_1^{13} = 3^{13}, \quad \lambda_2^{13} = (-1)^{13} = -1, \quad \lambda_3^{13} = (-1)^{13} = -1 \]

Step 3: Conclusion
- The eigenvalues of \( A^{13} \) are \( 3^{13}, -1, -1 \).

Quick Tip: \textbf{Eigenvalues of Matrix Powers:} The eigenvalues of \( A^n \) are the eigenvalues of \( A \) raised to the power \( n \).

Step 1: Understanding the layers and their function- Networking Layer (Layer 3): This layer is responsible for data packet transfer across different networks, typically by routing the packets between source and destination.
- Transport Layer (Layer 4): It provides host-to-host communication by managing end-to-end connections, flow control, and error recovery.
- Data Link Layer (Layer 2): This layer ensures that data is properly formatted and transmitted over a physical link. It handles error detection and correction to ensure reliable communication between devices on the same network.

Step 2: Explanation of the matches Networking Layer: It is responsible for the routing and forwarding of data packets between devices, making "Data packet transfer" the correct match. Transport Layer: It ensures the reliable communication between two hosts by managing error recovery, flow control, and segmenting the data into smaller packets, which is why "Host-to-host communication" is the correct function. Data Link Layer: This layer detects and corrects errors that may occur at the physical layer, making "Error detection and correction" the most appropriate function.

Quick Tip: OSI Model Layers: Layer 2 (Data Link Layer) deals with the physical link. Layer 3 (Networking Layer) is concerned with routing data packets across networks. Layer 4 (Transport Layer) focuses on communication between hosts.

Step 1: Understanding the Code - In the `main` function, two integer variables \(a\) and \(b\) are initialized with values 20 and 25, respectively. - A pointer variable \(z\) is declared, and \(z\) is assigned the address of \(a\) using the statement \(z = \&a\).

Step 2: Understanding the Function Call `foo(z, b)` - The `foo` function takes two arguments: a pointer to an integer (`p`) and an integer (`x`). - Inside the `foo` function, the statement `*p = x;` modifies the value stored at the memory address pointed to by \(p\). In this case, \(p = z\), which points to \(a\), and \(x = b = 25\). - Therefore, the value of \(a\) is updated to 25, as \( *z = 25 \).

Step 3: Conclusion - After the call to `foo`, the value of \(a\) becomes 25. - The `printf` function then prints the value of \(a\), which is 25. Final Output: \[ \text{The output of the program is } 25. \]

Quick Tip: In C, when a pointer is passed to a function, any changes made to the value pointed by the pointer will affect the original variable in the calling function.

Step 1: Understanding Signed Binary Representation - In signed binary representation, numbers are represented in two's complement form. To find the signed binary expansion of a number, we need to follow a few steps. - The two's complement representation uses the most significant bit (MSB) to represent the sign of the number, where `0` represents a positive number and `1` represents a negative number.

Step 2: Find Binary Expansion for 6 - First, we find the binary representation of the positive number 6 in standard binary format. - 6 in binary is \( 0110_2 \) for a 4-bit binary.

Step 3: Apply Two's Complement to Get -6 - To get the two's complement representation for -6, we follow these steps: 1. First, write the binary representation of +6: \( 0110_2 \). 2. Invert all the bits: \( 1001_2 \). 3. Add 1 to the result: \( 1001_2 + 1 = 1010_2 \). - So, the 2's complement representation of -6 for 4-bit binary is \( 1010_2 \). Step 4: Extending to 5 Bits for Signed Expansion - For the signed expansion in 5 bits, we need to add an additional bit to the left of the representation. The extra bit is 1 to maintain the negative sign. - Thus, the 5-bit signed binary representation of -6 is \( 11010_2 \).

Final Answer: \[ \text{The signed binary expansion of -6 in 5-bit representation is } 11010_2. \] Quick Tip: To convert a negative number into its signed binary representation using two's complement: Write the binary of the positive number. Invert the bits and add 1.

Step 1: Convert the decimals to fractions - The repeating decimal \( 0.\overline{6} \) (i.e., \(0.6666\ldots\)) is equivalent to \( \frac{2}{3} \). - The decimal \( 0.666 \) is a terminating decimal that equals \( \frac{2}{3} \). - The fraction \( \frac{5}{6} \) is already in its simplest form. - The fraction \( \frac{2}{3} \) is also in its simplest form.

Step 2: Comparison - \( 0.\overline{6} = \frac{2}{3} \), which is less than \( 0.666 \), since \( 0.666 \) is slightly greater than \( \frac{2}{3} \) in the decimal representation. - \( 0.666 \) is actually the greatest value among all the options, as \( 0.666 \) represents a decimal value that is slightly greater than the repeating decimal and is closer to \( \frac{5}{6} \).

Step 3: Conclusion - Therefore, \( 0.666 \) is the greatest of the four choices. Final Answer: \[ \boxed{0.666} \] Quick Tip: To compare repeating and terminating decimals, convert them to fractions and compare their values.

Step 1: Understanding the Address Breakdown - The logical address space (L.A.) is 32 bits. This means the total number of bytes that can be addressed is \( 2^{32} \) bytes. - The physical address space (P.A.) is 20 bits. This means the total number of bytes that can be addressed in physical memory is \( 2^{20} \) bytes.

Step 2: Page Size Calculation - The page size is 2048 bytes, or \( 2^{11} \) bytes.

Step 3: Number of Pages in the Logical Address Space - The number of pages in the logical address space is: \[ \frac{2^{32}}{2^{11}} = 2^{21} \]

Step 4: Number of Frames in the Physical Address Space - The number of frames in the physical address space is: \[ \frac{2^{20}}{2^{11}} = 2^{9} \]

Step 5: Number of Entries in the Page Table - The number of entries in the page table is the number of pages in the logical address space, which is \( 2^{21} \). Thus, the maximum number of entries in the page table is \( 2^{21} \). Final Answer: \[ \boxed{2^{21}} \] Quick Tip: To calculate the number of entries in a page table, divide the total logical address space by the page size to get the number of pages. The number of entries is equal to the number of pages.

Step 1: Understanding the 3-Way Handshake The 3-way handshake is used to establish a TCP connection between a client and a server. The steps are as follows:
- Step 1: SYN (Synchronize)
The client sends a SYN (synchronize) packet to the server to begin the connection. This packet contains the initial sequence number.
- Step 2: SYN-ACK (Synchronize-Acknowledgment)
The server responds with a SYN-ACK packet. This packet acknowledges the SYN request from the client and sends back its own SYN to the client.
- Step 3: ACK (Acknowledgment)
The client sends an ACK (acknowledgment) packet back to the server, confirming that the connection has been established.


Step 2: Conclusion Thus, the correct order of the 3-way handshake is \(\text{SYN} \rightarrow \text{SYN-ACK} \rightarrow \text{ACK}\).

Final Answer: \[ \boxed{\text{SYN} \rightarrow \text{SYN-ACK} \rightarrow \text{ACK}} \]
Quick Tip: The 3-way handshake establishes a reliable TCP connection between a client and server, ensuring that both sides are ready to send and receive data.

Step 1: Understanding the Algorithm
- The outer loop runs \( n \) times.
- The inner loop also runs \( n \) times for each iteration of the outer loop.
- The total number of executions of the \texttt{printf statement is \( n \times n = n^2 \). Step 2: Conclusion
- The time complexity is \( O(n^2) \). Final Answer: \[ \boxed{O(n^2)} \] Quick Tip: To determine the time complexity of an algorithm, consider the number of iterations of each loop: - The time complexity is the product of the iterations for nested loops. - For a loop running \( n \) times inside another loop running \( n \) times, the time complexity is \( O(n^2) \).

Step 1: Stack Operations Breakdown
- Push 10: Stack = [10]
- Push 20: Stack = [10, 20]
- Pop: Removes 20, Stack = [10]
- Push 30: Stack = [10, 30]
- Pop: Removes 30, Stack = [10]
- Push 40: Stack = [10, 40] Step 2: Conclusion
- The final content of the stack is [10, 40]. Final Answer: \[ \boxed{10, 40} \] Quick Tip: A stack operates on a Last-In, First-Out (LIFO) principle. - When you "push," you add an element to the top of the stack. When you "pop," you remove the top element. - Carefully follow the sequence of push and pop operations to track the stack’s content.

Step 1: Understanding Binary Search
- Binary search works by dividing the search space in half at each step.
- It requires the array to be sorted in ascending order for proper functioning. Step 2: Conclusion
- The correct statement is: "It divides the search space into two equal halves on each iteration." Final Answer: \[ \boxed{3. \text{It divides the search space into two equal halves on each iteration.}} \] Quick Tip: Binary search only works on sorted arrays. - It divides the search space in half each time, making it an efficient \( O(\log n) \) algorithm. - Always ensure that the array is sorted before applying binary search. If the array is unsorted, binary search will not function correctly.