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Banded matrix

Linear algebraHard

Let $T$ be the $6 \times 6$ matrix whose diagonal entries are all $3$ , whose entries directly above or directly below the diagonal are all $1$ , and whose other entries are all $0$ . Compute $det (T)$ .

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Hint

Expand along the first row and look for a recurrence linking the determinants of the smaller matrices of the same pattern.

Solution

Let $D_{n}$ be the determinant of the $n \times n$ matrix of this pattern, so $D_{1} = 3$ and $D_{2} = 3 \times 3 - 1 \times 1 = 8$ .
Expanding along row $1$ gives $D_{n} = 3 D_{n - 1} - D_{n - 2}$ : the diagonal $3$ leaves the smaller matrix of the same pattern, and the $1$ next to it leaves a minor whose first column holds a single $1$ .
Then $D_{3} = 3 \times 8 - 3 = 21$ , $D_{4} = 3 \times 21 - 8 = 55$ and $D_{5} = 3 \times 55 - 21 = 144$ .
Finally $D_{6} = 3 \times 144 - 55 = 377$ .

Answer: $377$

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