Choosing an encoding key for the school notice board
The Digital Notice Board club at a senior secondary school encodes roll-number letters before posting absentee data on a screen visible in the corridor. Each pair of letters is written as a column vector of alphabet positions (A = 1, …, Z = 26) and multiplied by the 2 × 2 key matrix A = [[2, 3], [1, 2]]. Last week the pair GU (G = 7, U = 21) went up on the screen as the pair (77, 49), and a parent complained that the display was unreadable. A new member, wanting a key that is easier to remember, proposes B = [[2, 4], [3, 6]]. She runs it on the test vectors (4, 0) and (0, 2) and both produce exactly the same displayed pair. She insists B is acceptable because “no entry has gone missing”, and refuses to compute any determinant, calling it a formality. The club adviser wants a short written verdict, with calculations, before tomorrow's assembly, and a rule the club can apply to any future key.
1. Decode the displayed pair (77, 49) using the key A, and state the determinant condition a 2 × 2 key must satisfy if decoding is to return whole-number letter positions reliably.
Read the worked answer
Step 1 - determinant of A.
det A = (2)(2) − (3)(1) = 4 − 3 = 1. Since det A ≠ 0, A is invertible and the encoding loses no information.
Step 2 - inverse.
A⁻¹ = (1/det A)·[[2, −3], [−1, 2]] = [[2, −3], [−1, 2]].
Step 3 - decode.
First entry: 2(77) − 3(49) = 154 − 147 = 7 → G. Second entry: −1(77) + 2(49) = −77 + 98 = 21 → U.
The screen pair (77, 49) decodes to GU, which matches the roll-number record.
Step 4 - the rule.
det ≠ 0 is the condition for a unique decoding: a zero determinant destroys information. The stronger condition det = ±1 is what the club actually needs, because then A⁻¹ = ± adj A has integer entries, so every decoded entry is an exact integer with no division by a number other than 1. If |det A| ≥ 2, the inverse carries fractional entries and any hand or spreadsheet rounding can push the answer to a neighbouring letter. Rule: accept a key only when det = ±1.
2. Test the proposed key B = [[2, 4], [3, 6]] on the vectors (4, 0) and (0, 2). What does the result prove, and what is the verdict on B?
Read the worked answer
Step 1 - determinant.
det B = (2)(6) − (4)(3) = 12 − 12 = 0.
Step 2 - apply B.
B·(4, 0)ᵗ = (2·4 + 4·0, 3·4 + 6·0)ᵗ = (8, 12)ᵗ. B·(0, 2)ᵗ = (2·0 + 4·2, 3·0 + 6·2)ᵗ = (8, 12)ᵗ.
Step 3 - interpretation.
Two different inputs give the same output. So the map is many-to-one and no inverse matrix exists: given (8, 12) on the screen, the club cannot tell whether (4, 0) or (0, 2) was sent. That is not a cosmetic defect - the notice board would publish a wrong absentee record.
Step 4 - why it happens.
Column 2 of B is exactly 2 × column 1, so the columns are linearly dependent; the matrix collapses the plane onto a line. A non-zero determinant requires linearly independent columns.
Verdict: reject B. The member's argument (“no entry has gone missing”) confuses a blank cell with lost information; the loss is in the impossibility of recovery, and det B = 0 exposes it immediately.
3. The club later moves to a screen that stores letters modulo 26 and inverts keys by finding an integer matrix M with KM ≡ I (mod 26). One member proposes a 3 × 3 key with det K = −2 and another proposes a key with det K = 13. Which, if either, can work, and what is the general condition?
Read the worked answer
Step 1 - the trap.
Both matrices are perfectly invertible over the real numbers, because −2 ≠ 0 and 13 ≠ 0. If the club used ordinary arithmetic, both keys would decode uniquely. The modular screen changes the requirement: an integer matrix has an inverse mod 26 only when its determinant has an inverse mod 26, i.e. when a solution of (det K)·y ≡ 1 (mod 26) exists.
Step 2 - test the two proposals.
- det K = −2: we need y with −2y ≡ 1 (mod 26), i.e. 2y ≡ 25 (mod 26). The left side is always even and 25 is odd, so there is no solution.
- det K = 13: we need 13y ≡ 1 (mod 26). But 13y is always 0 or 13 modulo 26, never 1 - no solution.
Step 3 - general condition. A determinant is invertible modulo 26 exactly when it is coprime to 26 = 2 × 13. So the club needs det K odd and not a multiple of 13, e.g. det = ±1, ±3, ±5, ±7, ±9, ±11, ±15.
Verdict: neither proposal works. Divisibility of the determinant by 2 or by 13 is fatal mod 26, even though it is harmless over the reals - the arithmetic setting, not the determinant's size, decides the rule.