Mathematics 10 — Exercise 3.6 · Q4

Full question

Three equilibrium forces are F₁=(8,x), F₂=(−2,−7), F₃=(y,−1). Find x and y.

Step-by-step solution

1. Equilibrium makes each component sum zero: 8−2+y=0 and x−7−1=0.

2. Write the coefficients in A, the unknowns in X, and constants in B, giving AX = B.

   $$(\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],\left[\begin{array}{c}x\\ y\end{array}\right],\left[\begin{array}{c}-6\\ 8\end{array}\right])$$

3. For a 2 × 2 determinant, multiply the main diagonal and subtract the other diagonal product.

   $$D=0\times 0-1\times 1$$

4. Calculate the first diagonal product.

   $$0\times 0=0$$

5. Calculate the second diagonal product.

   $$1\times 1=1$$

6. Subtract the second product, including its sign.

   $$D=0-1$$

7. Simplify the subtraction.

   $$D=-1$$

8. For a 2 × 2 determinant, multiply the main diagonal and subtract the other diagonal product.

   $$D=0\times 0-1\times 1$$

9. Calculate the first diagonal product.

   $$0\times 0=0$$

10. Calculate the second diagonal product.

    $$1\times 1=1$$

11. Subtract the second product, including its sign.

    $$D=0-1$$

12. Simplify the subtraction.

    $$D=-1$$

13. The determinant is nonzero, so the inverse exists.

    $$-1\ne 0$$

14. Swap the two diagonal entries to begin the adjoint.

    $$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$$

15. Change the signs of the two off-diagonal entries.

    $$\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]$$

16. Calculate this operation before combining the remaining terms.

    $$\frac{1}{-1}=-1$$

17. Calculate this operation before combining the remaining terms.

    $$0\times \left(-1\right)=0$$

18. Divide adjoint entry (1,1) by the determinant.

    $$\frac{0}{-1}=0$$

19. Calculate this operation before combining the remaining terms.

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

20. Divide adjoint entry (1,2) by the determinant.

    $$-\frac{1}{-1}=1$$

21. Divide adjoint entry (2,1) by the determinant.

    $$-\frac{1}{-1}=1$$

22. Divide adjoint entry (2,2) by the determinant.

    $$\frac{0}{-1}=0$$

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

    $$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$$

24. Premultiply AX = B by A⁻¹. Since A⁻¹A = I, X = A⁻¹B.

25. 2 columns in the first matrix match 2 rows in the second. The product has 2 rows and 1 columns.

    $$(\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],\left[\begin{array}{c}-6\\ 8\end{array}\right])$$

26. Entry (1,1): pair row 1 with column 1 and add their products.

    $$0\times \left(-6\right)+1\times 8$$

27. Calculate this individual product, including its signs.

    $$0\times \left(-6\right)=0$$

28. Calculate this individual product, including its signs.

    $$1\times 8=8$$

29. Add the calculated products to obtain this entry.

    $$0+8=8$$

30. Entry (2,1): pair row 2 with column 1 and add their products.

    $$1\times \left(-6\right)+0\times 8$$

31. Calculate this individual product, including its signs.

    $$1\times \left(-6\right)=-6$$

32. Calculate this individual product, including its signs.

    $$0\times 8=0$$

33. Add the calculated products to obtain this entry.

    $$-6+0=-6$$

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

    $$\left[\begin{array}{c}8\\ -6\end{array}\right]$$

35. Calculate this operation before combining the remaining terms.

    $$0-6=-6$$

36. Check in this original equation by multiplying the coefficients by the calculated values.

    $$0\times 8+1\times \left(-6\right)=-6$$

37. Calculate this operation before combining the remaining terms.

    $$8+0=8$$

38. Check in this original equation by multiplying the coefficients by the calculated values.

    $$1\times 8+0\times \left(-6\right)=8$$

Answer

The final answer is shown below.