LINEAR-ALGEBRA · Unit 18 · Video 3 · Interactive Practice

Why Every Tridiagonal Matrix of Ones Has a Determinant of 1, 0, or −1

IKey Formulas

Formula	Name	Description
$\det(A_n) = \det(A_{n-1}) - \det(A_{n-2})$	Determinant recurrence	Cofactor expansion reduces to two previous terms
$\det(A_1) = 1, \;\; \det(A_2) = 0$	Initial values	Seeds for the recurrence
$1,\; 0,\; {-1},\; {-1},\; 0,\; 1,\; 1,\; 0,\; \ldots$	Period-6 cycle	The sequence repeats every 6 terms
$\det(A_n)$ determined by $n \bmod 6$	Modular lookup	One division replaces full determinant computation

IIVisualization 1 — The Tridiagonal Matrix of Ones

Ones fill the diagonal and its two neighbors, zeros everywhere else — how large can its determinant get?

Matrix size n 4

💡 This sparsity is exactly what makes cofactor expansion collapse to two terms — the recurrence in Visualization 2.

IIIVisualization 2 — The Recurrence Builds the Sequence

Each determinant is the previous one minus the one before it — so the values cycle $1, 0, -1, -1, 0, 1$ .

Terms n = 1…7

💡 With only three possible values and each term fixed by the previous two, the sequence is forced to repeat — here, every $6$ terms.

IVVisualization 3 — Modular Arithmetic Lookup

Because the sequence has period $6$ , one division $n \bmod 6$ replaces computing the whole determinant.

Choose n 61

💡 One division settles any $n$ — even $n$ in the thousands — with no determinant computation at all.

VQuiz Questions

Question 1

Using the recurrence $\det(A_n) = \det(A_{n-1}) - \det(A_{n-2})$ with initial values $\det(A_1) = 1$ and $\det(A_2) = 0$ , what is $\det(A_5)$ ?

✅ Correct! det(A₅) = det(A₄) − det(A₃) = (−1) − (−1) = 0.

❌ Not quite. Carefully apply the recurrence three times: first find det(A₃), then det(A₄), then det(A₅).

Show solution

Solution:

Apply the recurrence step by step from the initial values:

$\det(A_3) = \det(A_2) - \det(A_1) = 0 - 1 = -1$

$\det(A_4) = \det(A_3) - \det(A_2) = -1 - 0 = -1$

$\det(A_5) = \det(A_4) - \det(A_3) = -1 - (-1) = -1 + 1 = 0$

The answer is 0.

Question 2

The recurrence for the determinant of the tridiagonal matrix of ones resembles the Fibonacci recurrence. Which is the correct formula?

✅ Correct! Cofactor expansion along row 1 puts a minus sign on the second term.

❌ Not quite. The cofactor of the $(1,2)$ entry carries a negative sign from the checkerboard pattern — this is Fibonacci with subtraction.

Show solution

Solution:

Expand $\det(A_n)$ along the first row. Row 1 of $A_n$ is $(1, 1, 0, 0, \ldots, 0)$ :

First entry ( $a_{11} = 1$ , positive cofactor sign): contributes $+\det(A_{n-1})$ because the remaining submatrix is exactly $A_{n-1}$ .
Second entry ( $a_{12} = 1$ , negative cofactor sign): the corresponding minor reduces (after expanding its first column) to $\det(A_{n-2})$ , contributing $-\det(A_{n-2})$ .
All remaining entries are 0 and contribute nothing.

Therefore: $\det(A_n) = \det(A_{n-1}) - \det(A_{n-2})$

This is like Fibonacci ( $f_n = f_{n-1} + f_{n-2}$ ) but with subtraction instead of addition.

Question 3

What is $\det(A_{100})$ ?

Hint: The determinant sequence has period 6. Compute $100 \bmod 6$ and look up the cycle table:

$n \bmod 6$	1	2	3	4	5	0
$\det(A_n)$	1	0	−1	−1	0	1

✅ Correct! 100 mod 6 = 4, and cycle position 4 gives det = −1.

❌ Not quite. First compute 100 ÷ 6 = 16 remainder 4, then look up position 4 in the cycle table.

Show solution

Solution:

Compute $100 \bmod 6$ :

$100 = 16 \times 6 + 4 \qquad \Longrightarrow \qquad 100 \bmod 6 = 4$

Look up position 4 in the cycle table:

$n \bmod 6$	1	2	3	4	5	0
$\det(A_n)$	1	0	−1	−1	0	1

$\boxed{\det(A_{100}) = -1}$

No row operations needed — just one division and a table lookup on a 100×100 matrix!

Question 4

True or False: The determinant sequence must eventually become periodic because each det(A_n) can only be -1, 0, or 1, giving at most 9 possible consecutive pairs. By the pigeonhole principle, a pair must eventually repeat, which forces the entire future sequence to repeat.

✅ Correct! The pigeonhole principle guarantees periodicity — with only 9 possible consecutive pairs, a repeat must occur within the first 10 terms.

❌ Not quite. Consider: with finitely many possible consecutive pairs and a deterministic recurrence, what must eventually happen?

Show solution

Solution: True.

Here is the complete reasoning:

Bounded values: Each determinant is one of $\{-1, 0, 1\}$ — just 3 possible values.
Two-term dependence: $\det(A_n) = \det(A_{n-1}) - \det(A_{n-2})$ , so each term depends only on the two preceding terms.
Finite pairs: There are at most $3 \times 3 = 9$ possible consecutive pairs $(\det(A_{n-1}), \det(A_{n-2}))$ .
Pigeonhole principle: Within at most 10 consecutive terms, some pair of adjacent values must repeat.
Deterministic future: Once a pair repeats, the recurrence is deterministic — every subsequent term is forced to match, creating a permanent cycle.

Specifically, $(\det(A_6), \det(A_7)) = (1, 1) = (\det(A_0), \det(A_1))$ , confirming the cycle length is exactly 6.

Solved: 0 / 4