Matrix addition and subtraction

LAB A08 · 01

Understand the task.

Aim: Matrix addition and subtraction.

Review the related theory in Unit 3.

LAB A08 · 02

Know what goes in and what comes out.

Rows and columns, each 1–5. Then all A entries row by row, followed by all B entries row by row. Entries are integers from -20 to 20. Both matrices use the same supplied dimensions.

2 2
1 2
3 4
5 6
7 8

Sum:
6 8
10 12
Difference (A - B):
-4 -4
-4 -4

Source programs read the documented input directly, without prompt text, so the same sample works in the embedded compiler’s STDIN field.

LAB A08 · 03

Plan the steps before writing code.

1. Read and validate the shared dimensions.
2. Read A using nested row and column loops.
3. Read B in the same way.
4. For each position, display A[i][j] + B[i][j].
5. For each position, display A[i][j] - B[i][j].

LAB A08 · 04

One complete program. Ready to run.

#include <stdio.h>
#define MAX 5
int main(void)
{
    int a[MAX][MAX], b[MAX][MAX], rows, columns, i, j;
    if (scanf("%d%d", &rows, &columns) != 2 || rows < 1 || rows > MAX || columns < 1 || columns > MAX) {
        puts("Enter rows and columns from 1 to 5.");
        return 1;
    }
    for (i = 0; i < rows; i++) {
        for (j = 0; j < columns; j++) {
            if (scanf("%d", &a[i][j]) != 1 || a[i][j] < -20 || a[i][j] > 20) {
                puts("Enter matrix values from -20 to 20."); return 1;
            }
        }
    }
    for (i = 0; i < rows; i++) {
        for (j = 0; j < columns; j++) {
            if (scanf("%d", &b[i][j]) != 1 || b[i][j] < -20 || b[i][j] > 20) {
                puts("Enter matrix values from -20 to 20."); return 1;
            }
        }
    }
    puts("Sum:");
    for (i = 0; i < rows; i++) {
        for (j = 0; j < columns; j++) printf("%s%d", j ? " " : "", a[i][j] + b[i][j]);
        putchar('\n');
    }
    puts("Difference (A - B):");
    for (i = 0; i < rows; i++) {
        for (j = 0; j < columns; j++) printf("%s%d", j ? " " : "", a[i][j] - b[i][j]);
        putchar('\n');
    }
    return 0;
}

gcc A08-matrix-add-subtract.c -o program

LAB A08 · 05

Connect each statement to its purpose.

• Two indexes identify an element: i is the row and j is the column.
• Corresponding cells are combined; this is different from matrix multiplication.
• Subtraction order is A - B, so reversing the operands changes the sign.
• Only the entered rows and columns are processed; unused capacity is not accessed.

LAB A08 · 06

Follow the sample calculation.

LAB A08 · 07

Check more than one input.

Time and storage

O(rows × columns) time. The two input matrices occupy O(rows × columns) logical storage within fixed 5 × 5 capacities.

Start with the sample, then test a boundary and a case that changes the result. A correct sample alone does not establish that every input is handled correctly.

LAB A08 · 08

Use the real C compiler.

Choose an example and load it into the embedded C compiler. Select I/O inside the compiler, check the sample STDIN, and press Run. Change the input or edit the source to explore a different case.

2 2
1 2
3 4
5 6
7 8

Sum:
6 8
10 12
Difference (A - B):
-4 -4
-4 -4

#include <stdio.h>
#define MAX 5
int main(void)
{
    int a[MAX][MAX], b[MAX][MAX], rows, columns, i, j;
    if (scanf("%d%d", &rows, &columns) != 2 || rows < 1 || rows > MAX || columns < 1 || columns > MAX) {
        puts("Enter rows and columns from 1 to 5.");
        return 1;
    }
    for (i = 0; i < rows; i++) {
        for (j = 0; j < columns; j++) {
            if (scanf("%d", &a[i][j]) != 1 || a[i][j] < -20 || a[i][j] > 20) {
                puts("Enter matrix values from -20 to 20."); return 1;
            }
        }
    }
    for (i = 0; i < rows; i++) {
        for (j = 0; j < columns; j++) {
            if (scanf("%d", &b[i][j]) != 1 || b[i][j] < -20 || b[i][j] > 20) {
                puts("Enter matrix values from -20 to 20."); return 1;
            }
        }
    }
    puts("Sum:");
    for (i = 0; i < rows; i++) {
        for (j = 0; j < columns; j++) printf("%s%d", j ? " " : "", a[i][j] + b[i][j]);
        putchar('\n');
    }
    puts("Difference (A - B):");
    for (i = 0; i < rows; i++) {
        for (j = 0; j < columns; j++) printf("%s%d", j ? " " : "", a[i][j] - b[i][j]);
        putchar('\n');
    }
    return 0;
}

LAB A08 · 09

Explain why the program works.

LAB A08 · 10

Build the next skill.

Read the viva answers, try the extra cases, and return to the lab index to mark this exercise as practised.