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A problem determinant whose entries are from a equation involving kronecker delta Let $A_n$ be the $n × n$ matrix whose $(i, j )$-th entry is given by $2δ_{ij} – δ_{i+1,j} – δ_{i,j +1}$ where $δ_{ij}$ equals $1$ if $i = j$ and zero otherwise. Compute the determinant of $A_n$ how can I solve this proble.try to do by taking few examples but get confused |
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Notice that the matrix is tridiagonal ($2$ on principal diagonal and $-1$ on sub and super-diagonals). You can now show that the determinant is given by a recurrence relation $$ D_n = 2D_{n-1} - D_{n-2} $$ with $D_1=2$ and $D_2=3$. |
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Hints: 1) Present your matrix in explicit form: $\left( \begin{array}{ccccc} 2 & -1 & 0 & ... & 0 \\ -1 & 2 & ... & ... & ... \\ 0 & ... & ... & ... & 0 \\ ... & ... & ... & ... & -1 \\ 0 & 0 & ... & -1 & 2 \\ \end{array} \right)$ 2) Bring it to the triangular form and note that the diagonal elements are terms of the sequence $\{\frac{k+1}{k-1}\}_{k=1}^n$ 3) Using property of triangular matrix (The determinant of a triangular matrix equals the product of the diagonal entries.) you can obtain the answer $\det(A_n)=n+1$. |
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