# Tag Info

4

Note that your matrix can be written as $$\text{diag}(a_1,a_2,\ldots,a_n) + \begin{bmatrix} 1\\1\\1\\ \vdots\\1\end{bmatrix} \begin{bmatrix} b_1 & b_2 & \cdots & b_n\end{bmatrix}$$ This is a rank $1$ update to a diagonal matrix, whose determinant can be computed using the Sylvester determinant theorem: $$\det(I+UV^T) = \det(I+V^TU)$$ I will leave ...

3

Taking the first column, and substracting to it $\alpha$ times the column 2, we get $\det C_{n,n}=(1-\alpha^2)\det C_{n-1,n-1}$, hence we can conclude by Sylvester's criterion.

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1

\begin{array}{ll} D_n&=\begin{vmatrix} a_n+b_n & b_{n-1} & b_{n-2} & \dots & b_1& \\ b_n & a_{n-1} + b_{n-1} & b_{n-2} & \dots & b_1 \\ b_n & b_{n-1} & a_{n-2} + b_{n-2} & \dots & b_1 \\ \vdots & \vdots & \vdots& & \vdots \\ b_n & b_{n-1} & b_{n-2} &\dots & a_1 + b_1 ...

1

This is fairly straightforward to check for say a $3\times 3$ matrix, but the details become messy to do this in general. This should give you enough idea of how to do the general case to convince you it is true though: Suppose $N=3$. Suppose $v_i=\left(\begin{matrix} a_{i1} \\ a_{i2} \\ a_{i3}\end{matrix}\right)$. Then \$B_{11}=det(e_1, v_2, ...

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