# 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.

3


1

In the case $\Bbb K = \Bbb R$, you can use$$M_n(t)=\begin{vmatrix} a_1+tb_1 & tb_2 & tb_3 & \dots & tb_n& \\ tb_1 & a_2 + tb_2 & tb_3 & \dots & tb_n \\ tb_1 & tb_2 & a_3 + tb_3 & \dots & tb_n \\ \vdots & \vdots & \vdots& & \vdots \\ tb_1 & tb_2 & tb_3 &\dots & a_n + tb_n ... 1 This is fairly straightforward to check for say a 3\times 3 matrix, but the details become messy to do this in general. This should give you enough idea of how to do the general case to convince you it is true though: Suppose N=3. Suppose v_i=\left(\begin{matrix} a_{i1} \\ a_{i2} \\ a_{i3}\end{matrix}\right). Then B_{11}=det(e_1, v_2, ... 1$$\left|\begin{array}{cc}1 & -3\\1 & 0\end{array}\right|=(1\cdot 0)-(1\cdot-3)=0-(-3)=3

Only top voted, non community-wiki answers of a minimum length are eligible