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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 $$\left[\begin{array}{@{}c@{}} 1 \\ 0 \\ 0 \\ 0 \\ \end{array}\right],\... 3 Note that$$\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} $$3 Hint: Expand along the rows that have the most zeros. Expanding along row 2, we have:$$\begin{vmatrix} 1 & 0 & 0 & a\\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 1 \end{vmatrix}$$Note, what I mean by expanding along row 2 using Laplace Expansion (see link below), is:$$0 \begin{vmatrix} 2 & 0 &...

2

The formula reads: $$|A|=\sum_{i=1}^{n}(-1)^{i+j}a_{ij}M_{ij}$$ So the sign is positive/negative when the sum of the row index and the column index ($i+j$) is even/odd.

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 ...

2

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

2

When you perform Laplacian expansion, remember to switch sign from entry to entry. I think it should be -3 rather than +3 $$0\times \det\begin{bmatrix}1 & 0 \\ 1 & -\lambda\end{bmatrix}+(-4-\lambda)\times \det\begin{bmatrix}-5-\lambda & 0 \\ 5 & -\lambda\end{bmatrix}-3\times \det\begin{bmatrix}-5-\lambda & 1 \\ 5 & 1\end{bmatrix}$$

1

Starting with dim. 2 and 3, it is not difficult to deduce, and then demonstrate that it holds for all 2 <= dim., that the LU decomposition of your matrix is $$\left[ {\begin{array}{*{20}c} 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \... 1 I see two errors. -0\times \det\begin{bmatrix}1 & 0 \\ 1 & -\lambda\end{bmatrix}+(-4-\lambda)\times \det\begin{bmatrix}-5-\lambda & 0 \\ 5 & -\lambda\end{bmatrix}-3\times \det\begin{bmatrix}-5-\lambda & 1 \\ 5 & 1\end{bmatrix} and (-4-\lambda)(-5-\lambda)(-\lambda)-3(-5-\lambda-5) 1 You need not integrate by parts to evaluate this integral. In fact, one would need to integrate by parts twice. See the section following the highlighted SPOILER ALERT So, I thought it would be instructive to present a "trick" that we can use to quickly evaluate the integral of interest and other similar integrals. HERE IS A HINT:$$\sin(t)=\text{Im}\...

1

To compute $$\begin{vmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 1 & 1 & 1 & \dots & 1 & 0 \\ 1 & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ \end{... 1 You should convince yourself that det(A)=det(A^{T}) and that swapping two rows of A multiplies the determinant by -1. Then you can always expand by the 'first' row. 1 Your recursive approach is fine; just follow it through. Let D_n(a,b) be the determinant of the matrix with diagonal elements a and all other elements b; clearly D_1(a,b)=a. For n>1, multiplying the first row by -b/a and adding it to every other row gives 0's in the first column (except for an a in the upper left), a - b^2/a along the ... 1$$\left|\begin{array}{cc}1 & -3\\1 & 0\end{array}\right|=(1\cdot 0)-(1\cdot-3)=0-(-3)=3$$1 This is fairly straightforward to check for say a 3\times 3 matrix, but the details become messy to do this in general. This should give you enough idea of how to do the general case to convince you it is true though: Suppose N=3. Suppose v_i=\left(\begin{matrix} a_{i1} \\ a_{i2} \\ a_{i3}\end{matrix}\right). Then B_{11}=det(e_1, v_2, v_3)=a_{22}a_{... 1$$\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 ...

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In the case $\Bbb K = \Bbb R$, you can useM_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

The pattern $+-+-$ works in the link you gave since they are expanding along the first row; it would be the same pattern when expanding along the first column. But then it changes as you change what column/row you're expanding along. For expanding along the second column it would be $-+-+$, along the third $+-+-$, and along the fourth $-+-+$.

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