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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

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Since volume of a parallelipiped spanned by a set of vectors is invariant under the operation of adding a scalar multiple of one vector to another, it suffices to compute the volume of the parallelipiped spanned by $$ \left[\begin{array}{@{}c@{}} 1 \\ 0 \\ 0 \\ 0 \\ ...


3

Note that $$\bar M_{j_0k_0}^{i\,\,1}=\bar M_{k_0j_0}^{i\,\,1}$$ Thus, you always have two different coefficients for the same minor that add up: $$j=j_0 , k=k_0 ~~ (j_0>k_0): \\ j>k\colon\; (-1)^{i+j_0}a_{ij_0}(-1)^{1+k_0}a_{1k_0}\\$$ $$ j=k_0, k=j_0:\\ j<k\colon\; (-1)^{i+k_0}a_{ik_0}(-1)^{j_0}a_{1j_0} $$


2

If $A$ is your matrix, then $B=A+3I$ is the matrix all of whose entries are all $2$s. It is clear that the vector $(1,\dots,1)$ is an eigenvector of $B$ of eigenvalue $2n$. On the other hand, the rank of $B$ is obviously $1$, since the dimension of the vector space spanned by its rows is $1$: this means that $0$ is an eigenvalue of $B$ of multiplicity $n-1$. ...


2

Hint: the matrix $M = e e^T$ (where $e$ is a column vector consisting of $n$ $1$'s) satisfies $M^2 = n M$, so its eigenvalues are ...


1

Your recursive approach is fine; just follow it through. Let $D_n(a,b)$ be the determinant of the matrix with diagonal elements $a$ and all other elements $b$; clearly $D_1(a,b)=a$. For $n>1$, multiplying the first row by $-b/a$ and adding it to every other row gives $0$'s in the first column (except for an $a$ in the upper left), $a - b^2/a$ along the ...


1

You should convince yourself that $det(A)=det(A^{T})$ and that swapping two rows of A multiplies the determinant by -1. Then you can always expand by the 'first' row.


1

Never mind...I find this Jacobi's theorem from Prasolov's (yes...every time...) book: General case follows immediately from permutating the rows and columns.


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

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

$$\left|\begin{array}{cc}1 & -3\\1 & 0\end{array}\right|=(1\cdot 0)-(1\cdot-3)=0-(-3)=3$$



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