Mathematics 10 — Exercise Review · Q4 (iii)

Full question

For the following vectors, find c+b−4a.

Step-by-step solution

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

   $$(\left[\begin{array}{c}4\\ 1\end{array}\right],\left[\begin{array}{c}3\\ 1\end{array}\right])$$

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

   $$4+3=7$$

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

   $$1+1=2$$

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

   $$\left[\begin{array}{c}7\\ 2\end{array}\right]$$

5. Multiply every matrix entry by the scalar.

   $$(4,\left[\begin{array}{c}2\\ -1\end{array}\right])$$

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

   $$4\times 2=8$$

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

   $$4\times \left(-1\right)=-4$$

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

   $$\left[\begin{array}{c}8\\ -4\end{array}\right]$$

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

   $$(\left[\begin{array}{c}7\\ 2\end{array}\right],\left[\begin{array}{c}8\\ -4\end{array}\right])$$

10. Entry (1,1): subtract the corresponding entries.

    $$7-8=-1$$

11. Entry (2,1): subtract the corresponding entries.

    $$2+4=6$$

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

    $$\left[\begin{array}{c}-1\\ 6\end{array}\right]$$

13. The first component multiplies î, the second multiplies ĵ.

    $$-î+6\times ĵ$$

Answer

The final answer is shown below.