Mathematics 10 — Exercise 3.3 · Q4 (iv)

Full question

Verify 3(A+B) = 3A+3B.

Step-by-step solution

1. Compute the left side first, following the bracket order.

2. The orders match. Combine only entries in the same row and column.

   $$(\left[\begin{array}{cc}2& 3\\ -3& 2\end{array}\right],\left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right])$$

3. Entry (1,1): add the corresponding entries.

   $$2+3=5$$

4. Entry (1,2): add the corresponding entries.

   $$3+4=7$$

5. Entry (2,1): add the corresponding entries.

   $$-3+5=2$$

6. Entry (2,2): add the corresponding entries.

   $$2+6=8$$

7. Place the calculated entries in their original row and column positions.

   $$\left[\begin{array}{cc}5& 7\\ 2& 8\end{array}\right]$$

8. Multiply every matrix entry by the scalar.

   $$(3,\left[\begin{array}{cc}5& 7\\ 2& 8\end{array}\right])$$

9. Entry (1,1): multiply its value by the scalar.

   $$3\times 5=15$$

10. Entry (1,2): multiply its value by the scalar.

    $$3\times 7=21$$

11. Entry (2,1): multiply its value by the scalar.

    $$3\times 2=6$$

12. Entry (2,2): multiply its value by the scalar.

    $$3\times 8=24$$

13. Place the calculated entries in their original row and column positions.

    $$\left[\begin{array}{cc}15& 21\\ 6& 24\end{array}\right]$$

14. Now compute the right side independently.

15. Multiply every matrix entry by the scalar.

    $$(3,\left[\begin{array}{cc}2& 3\\ -3& 2\end{array}\right])$$

16. Entry (1,1): multiply its value by the scalar.

    $$3\times 2=6$$

17. Entry (1,2): multiply its value by the scalar.

    $$3\times 3=9$$

18. Entry (2,1): multiply its value by the scalar.

    $$3\times \left(-3\right)=-9$$

19. Entry (2,2): multiply its value by the scalar.

    $$3\times 2=6$$

20. Place the calculated entries in their original row and column positions.

    $$\left[\begin{array}{cc}6& 9\\ -9& 6\end{array}\right]$$

21. Multiply every matrix entry by the scalar.

    $$(3,\left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right])$$

22. Entry (1,1): multiply its value by the scalar.

    $$3\times 3=9$$

23. Entry (1,2): multiply its value by the scalar.

    $$3\times 4=12$$

24. Entry (2,1): multiply its value by the scalar.

    $$3\times 5=15$$

25. Entry (2,2): multiply its value by the scalar.

    $$3\times 6=18$$

26. Place the calculated entries in their original row and column positions.

    $$\left[\begin{array}{cc}9& 12\\ 15& 18\end{array}\right]$$

27. The orders match. Combine only entries in the same row and column.

    $$(\left[\begin{array}{cc}6& 9\\ -9& 6\end{array}\right],\left[\begin{array}{cc}9& 12\\ 15& 18\end{array}\right])$$

28. Entry (1,1): add the corresponding entries.

    $$6+9=15$$

29. Entry (1,2): add the corresponding entries.

    $$9+12=21$$

30. Entry (2,1): add the corresponding entries.

    $$-9+15=6$$

31. Entry (2,2): add the corresponding entries.

    $$6+18=24$$

32. Place the calculated entries in their original row and column positions.

    $$\left[\begin{array}{cc}15& 21\\ 6& 24\end{array}\right]$$

33. Every corresponding entry is equal, so the stated identity holds.

    $$(\left[\begin{array}{cc}15& 21\\ 6& 24\end{array}\right],\left[\begin{array}{cc}15& 21\\ 6& 24\end{array}\right])$$

Answer

Every corresponding entry is equal, so the stated identity holds.