Mathematics 10 — Exercise Review · Q4 (iii)

Full question

Find A⁻¹.

Step-by-step solution

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

   $$D=7\times 1-3\times 2$$

2. Calculate the first diagonal product.

   $$7\times 1=7$$

3. Calculate the second diagonal product.

   $$3\times 2=6$$

4. Subtract the second product, including its sign.

   $$D=7-6$$

5. Simplify the subtraction.

   $$D=1$$

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

   $$1\ne 0$$

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

   $$\left[\begin{array}{cc}1& 3\\ 2& 7\end{array}\right]$$

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

   $$\left[\begin{array}{cc}1& -3\\ -2& 7\end{array}\right]$$

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

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

10. Calculate this operation before combining the remaining terms.

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

11. Calculate this operation before combining the remaining terms.

    $$1\times 1=1$$

12. Calculate this operation before combining the remaining terms.

    $$\left(-3\right)\times 1=-3$$

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

    $$-3\times \frac{1}{1}=-3$$

14. Calculate this operation before combining the remaining terms.

    $$\left(-2\right)\times 1=-2$$

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

    $$-2\times \frac{1}{1}=-2$$

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

    $$7\times \frac{1}{1}=7$$

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

    $$\left[\begin{array}{cc}1& -3\\ -2& 7\end{array}\right]$$

Answer

The final answer is shown below.