Mathematics 10 — Exercise 3.3 · Q2 (ii)

Full question

Find Y + 7Z.

Step-by-step solution

1. Multiply every matrix entry by the scalar.

   $$(7,\left[\begin{array}{cc}3& 0\\ 0& -2\end{array}\right])$$

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

   $$7\times 3=21$$

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

   $$7\times 0=0$$

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

   $$7\times 0=0$$

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

   $$7\times \left(-2\right)=-14$$

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

   $$\left[\begin{array}{cc}21& 0\\ 0& -14\end{array}\right]$$

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

   $$(\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],\left[\begin{array}{cc}21& 0\\ 0& -14\end{array}\right])$$

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

   $$1+21=22$$

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

   $$2+0=2$$

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

    $$3+0=3$$

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

    $$4-14=-10$$

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

    $$\left[\begin{array}{cc}22& 2\\ 3& -10\end{array}\right]$$

Answer

The final answer is shown below.