Question26Max. score: 2.00In Data Structures, John has executed the following pseudocode to perform an operation with the following input values. Then determine x with the respect to the following input values:Pseudocode:input A,B n*n matrixoutput C n*n matrixdefine i, j, kfor i from 0 to n for j from 0 to n C[i,j] = 0 end forend forfor i from 0 to n for j from 0 to n for k from 0 to n C[i, j] = C[i, j] + A[i, k] * B[k, j] end for end forend forInput values:N = 2A[N][N] = { {1, 2}, {3, 4} }B[N][N] = { {5, 4}, {1, 2} }Output:C[N][N] = xOptions:7 198 208 207 197 819 2019 207 81234

To determine the value of x, we need to execute the given pseudocode with the provided input values.

Step 1: Initialize the matrix C with zeros. Step 2: Iterate over the rows of matrix A and matrix B. Step 3: For each row, iterate over the columns of matrix A and matrix B. Step 4: For each element in the row, iterate over the columns of matrix A and matrix B. Step 5: Multiply the corresponding elements of matrix A and matrix B, and add the result to the corresponding element of matrix C. Step 6: Repeat steps 4 and 5 for all elements in the row. Step 7: Repeat steps 3 to 6 for all rows. Step 8: The resulting matrix C will contain the value of x.

Using the given input values: N = 2 A[N][N] = { {1, 2}, {3, 4} } B[N][N] = { {5, 4}, {1, 2} }

Let's execute the pseudocode step by step:

Step 1: Initialize matrix C with zeros: C[N][N] = { {0, 0}, {0, 0} }

Step 2: Iterate over the rows of matrix A and matrix B: For i = 0 to N: For j = 0 to N: // Perform steps 3 to 6

Step 3: Initialize C[i][j] with zero: C[i][j] = 0

Step 4: Iterate over the columns of matrix A and matrix B: For k = 0 to N: // Perform steps 5 and 6

Step 5: Multiply the corresponding elements of matrix A and matrix B, and add the result to C[i][j]: C[i][j] = C[i][j] + A[i][k] * B[k][j]

Step 6: Repeat steps 4 and 5 for all elements in the row.

Let's execute the pseudocode with the given input values:

For i = 0: For j = 0: For k = 0: C[0][0] = C[0][0] + A[0][0] * B[0][0] = 0 + 1 * 5 = 5 C[0][0] = C[0][0] + A[0][1] * B[1][0] = 5 + 2 * 1 = 7 For k = 1: C[0][0] = C[0][0] + A[0][0] * B[0][1] = 7 + 1 * 4 = 11 C[0][0] = C[0][0] + A[0][1] * B[1][1] = 11 + 2 * 2 = 15 For j = 1: For k = 0: C[0][1] = C[0][1] + A[0][0] * B[0][0] = 0 + 1 * 5 = 5 C[0][1] = C[0][1] + A[0][1] * B[1][0] = 5 + 2 * 1 = 7 For k = 1: C[0][1] = C[0][1] + A[0][0] * B[0][1] = 7 + 1 * 4 = 11 C[0][1] = C[0][1] + A[0][1] * B[1][1] = 11 + 2 * 2 = 15

For i = 1: For j = 0: For k = 0: C[1][0] = C[1][0] + A[1][0] * B[0][0] = 0 + 3 * 5 = 15 C[1][0] = C[1][0] + A[1][1] * B[1][0] = 15 + 4 * 1 = 19 For k = 1: C[1][0] = C[1][0] + A[1][0] * B[0][1] = 19 + 3 * 4 = 31 C[1][0] = C[1][0] + A[1][1] * B[1][1] = 31 + 4 * 2 = 39 For j = 1: For k = 0: C[1][1] = C[1][1] + A[1][0] * B[0][0] = 0 + 3 * 5 = 15 C[1][1] = C[1][1] + A[1][1] * B[1][0] = 15 + 4 * 1 = 19 For k = 1: C[1][1] = C[1][1] + A[1][0] * B[0][1] = 19 + 3 * 4 = 31 C[1][1] = C[1][1] + A[1][1] * B[1][1] = 31 + 4 * 2 = 39

The resulting matrix C is: C[N][N] = { {15, 19}, {31, 39} }

Therefore, the value of x is 39.