Find a, b, c and d if [a + 2b, 3a − c; b − d, 2c + d] = [7, 1; −1, 8].

1. 1 · Write what is given and what is to be found. Given: two equal 2 × 2 matrices. To find: a, b, c and d.
2. 2 · Write the rule. Equal matrices have the same order and equal corresponding entries.
3. 3 · Write the four equations.
   • a + 2b = 7 ... (1)
   • 3a − c = 1 ... (2)
   • b − d = −1 ... (3)
   • 2c + d = 8 ... (4)
4. 4 · Write c and d in terms of a and b. From (2), c = 3a − 1. From (3), d = b + 1.
5. 5 · Substitute in (4). 2(3a − 1) + (b + 1) = 8, so 6a + b = 9 ... (5)
6. 6 · Solve (1) and (5).
   • From (5), b = 9 − 6a. Put this in (1): a + 18 − 12a = 7, so 11a = 11 and a = 1.
   • Then b = 9 − 6 = 3, c = 3 − 1 = 2 and d = 3 + 1 = 4.
7. 7 · Check and box the answer. 1 + 6 = 7, 3 − 2 = 1, 3 − 4 = −1 and 4 + 4 = 8. All four match: a = 1, b = 3, c = 2, d = 4.

Scoring tip: Number your equations. It shows the examiner your plan and makes each substitution easy to follow.