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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

For the third part, note that we know that: $$\mathcal{L}(t)=\frac{1}{s^2},~~\mathcal{L}(\cos t)=\frac{s}{s^2+1},~~\mathcal{L}(\sin t)=\frac{1}{s^2+1}$$ so $$x(t)=\mathcal{L}^{-1}(X(s))=\mathcal{L}^{-1}\left(\frac{1}{s^2}+\frac{s}{s^2+1}-3\frac{1}{s^2+1}\right)=t+\cos t-3\sin t$$


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.


2

Plot integrand for few values of $x$: It is apparent that the maximum shifts closer to the origin as $x$ grows. Let's rewrite the integrand as follows: $$ \int_0^{\pi/2} \sqrt{\sin(t)} \exp\left(-x \sin^4(t)\right) \mathrm{d}t = \int_0^{\pi/2} \exp\left(\frac{1}{2} \log(\sin(t))-x \sin^4(t)\right) \mathrm{d}t $$ The maximum of the integrand is ...


2

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

Take the Laplace transform of both sides. We have $$L(x'' + x) = L(t)$$ Now, use linearity of $L$ and the fact that $L(x'') = s^2X - sx(0) - x'(0)$. We then have taking the transform that $$s^2X-s+2 + X = \frac{1}{s^2}$$ Now do some algebra and solve for X.


2

$L\{f'(t)\}=\int_0^\infty e^{-st}f'(t)dt$ Integrating by parts we have, $L\{f'(t)\}=e^{-st}f(t)|_0^\infty+s\int_0^\infty e^{-st}f(t)dt$ $L\{f'(t)\}=e^{-s(\infty)}f(\infty)-e^{-s(0)}f(0)+sL\{f(t)\}$ If $e^{-st}$ grows more rapidly than $f(t)$, we have $e^{-st}f(t)\to0$ when $t\to\infty$ $L\{f'(t)\}=sL\{f(t)\}-f(0)$ Since $f(0)=0$, this reduces to ...


2

I am just going to evaluate $L(\sqrt t)$. Just substitute $\sqrt{st} = x$. So, $L(\sqrt t) = \int_0^\infty e^{-st} \cdot t^{-1/2} dt$ $= \frac{2}{\sqrt s}\cdot \int_0^\infty e^{-x^2} dx = \frac{2}{\sqrt s} \cdot \frac{\sqrt \pi }{2} = \sqrt {\frac{\pi}{s}}$


2

In Laplace's method, we recognize that the main contribution to the integral is in the neighborhood of the minimum of the exponent, here at $t=0$. It appears that the slowly varying part of the integrand has a singularity there, but this singularity is integrable. Thus, we may approximate as follows, noting that $\sinh{t} \sim t$ and $\cosh{t} \sim ...


1

This is not as difficult as it might look at first. The calculation is long, but when you don't forget anything, it works fine. First, you need to find all the derivatives in Eq. (3). It's important to remember, that also $X$ and $Y$ are functions of $z$, so that the $z$ derivative gives $i XYe^{iZ}A_G\partial_z Z + XYe^{iZ}\partial_z A_G + ...


1

For an initial value problem, don't forget to incorporate the initial values into the Laplace transform! $$\mathcal{x''}=s^2X(s)-sx(0)-x'(0)$$ The differential equation is transformed into the Laplace domain as follows: $$\mathcal{L}(x'')+\mathcal{L}(x)=\mathcal{L}(t)\\ s^2X(s)-sx(0)-x'(0)+X(s)=\mathcal{L}(t)$$ The partial fraction decomposition makes it ...


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

$$\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

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


1

As you probably already know, the Laplace Transform of a function $\operatorname{x}$ is given by $$\mathcal{L}[\operatorname{x}](s) := \int_0^{\infty} \!\!\operatorname{x}(t)\operatorname{e}^{-st} \, \operatorname{d}\!t$$ One really nice property of the Laplace Transform is that, if $\dot{\operatorname{x}}$ is shorthand for ...


1

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: $$j=j_0 , k=k_0 ~~ (j_0>k_0): \\ j>k\colon\; (-1)^{i+j_0}a_{ij_0}(-1)^{1+k_0}a_{1k_0}\\$$ $$ j=k_0, k=j_0:\\ j<k\colon\; (-1)^{i+k_0}a_{ik_0}(-1)^{j_0}a_{1j_0} $$


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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