Wikipedista:Matrix Computations

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Motto

• Я знаю, как управлять Вселенной. И скажите - зачем же мне бежать за миллионом?!
• I know how to control the universe. So tell me, why should I run for a million?!
• Vím, jak ovládat vesmír. Tak mi řekněte, proč bych se měl hnát za milionem?!

Grigorij Perelman, Komsomolskaja Pravda [1]

Matice, značení

Značení matic, které je rozumné používat (v článcích, které jsou primárně o maticích)

• $A,\,B,\,C\,\ldots$ matice (ne tučné);
• $A^{T}$ transponovaná matice;
• $A^{H}={\bar {A}}^{T}$ herminovsky sdružená matice ($A^{*}$ se plete s pseudoinverzí);
• $A^{-1}$ inverzní matice;
• $A^{\dagger }$ Moore-Penroseova pseudoinverze (možno též $A^{+}$, první mi příjde hezčí).

Pahýly, které jsem vyplodil

Maticové

• Polární rozklad
• Jacobiho matice (třídiagonální)
• Stochastická matice

Ostatní matika

• Tupperův vzorec
• Antonín Sochor (matematik)
• Česká matematická společnost

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Aux

Veselá rovnost

$-1=(-1)^{1}=(-1)^{\frac {2}{2}}={\sqrt {(-1)^{2}}}={\sqrt {1}}=1$

Veselé úpravy

${\begin{array}{rcl}m\neq n:\qquad &&\\x&=&(m+n)/2\\2x&=&(m+n)\\2x(m-n)&=&(m+n)(m-n)\\2xm-2xn&=&m^{2}-n^{2}\\n^{2}-2xn&=&m^{2}-2xm\\n^{2}-2xn+x^{2}&=&m^{2}-2xm+x^{2}\\(n-x)^{2}&=&(m-x)^{2}\\n-x&=&m-x\\n&=&m\end{array}}$

Limita

$\lim _{x\longrightarrow s}{\frac {x^{x^{x}}-x^{x^{s}}}{s^{x^{s}}-s^{s^{x}}}}$

Mocniny

$a^{b^{c}}$

$a^{bc}$

Odmocniny

${\sqrt {2}}^{\sqrt {2}}=(2^{\frac {1}{2}})^{\sqrt {2}}=2^{{\frac {1}{2}}\cdot {\sqrt {2}}}=2^{\frac {\sqrt {2}}{2}}=2^{\frac {1}{\sqrt {2}}}={\sqrt[{\sqrt {2}}]{2}}$

${\sqrt {{\sqrt[{3}]{4\,}}-1\,}}+{\sqrt {{\sqrt[{3}]{16\,}}-{\sqrt[{3}]{4\,}}\,}}={\sqrt {3}}$

Logaritmy

${\begin{array}{rcl}\log _{a}(x)+\log _{a}(y)\!\!&\!\!=\!\!&\!\!\log _{a}(x\cdot y)\\[2mm]\log _{a}(x)-\log _{a}(y)\!\!&\!\!=\!\!&\!\!\log _{a}\!{\Big (}{\frac {\displaystyle x}{\displaystyle y}}{\Big )}\\[2mm]s\cdot \log _{a}(x)\!\!&\!\!=\!\!&\!\!\log _{a}(x^{s})\\[2mm]-\log _{a}(x)\!\!&\!\!=\!\!&\!\!\log _{a}\!{\Big (}{\frac {\displaystyle 1}{\displaystyle x}}{\Big )}\\[2mm]\end{array}}\qquad \qquad \qquad {\begin{array}{rcl}\log _{a}(s)\cdot \log _{s}(x)\!\!&\!\!=\!\!&\!\!\log _{a}(x)\\[2mm]\log _{a}(x)\cdot \log _{x}(a)\!\!&\!\!=\!\!&\!\!1\\[2mm]x\!\!&\!\!=\!\!&\!\!a^{\displaystyle \,\log _{a}(x)}\\[2mm]x^{\displaystyle \,\log _{a}(y)}\!\!&\!\!=\!\!&\!\!y^{\displaystyle \,\log _{a}(x)}\\[2mm]\end{array}}\qquad \qquad \qquad {\begin{array}{rlc}\log _{a}(x)+\log _{b}(x)=\log _{c}(x),\;\;{\text{kde}}\;\;{\frac {\displaystyle 1}{\displaystyle \log _{s}(c)}}={\frac {\displaystyle 1}{\displaystyle \log _{s}(a)}}+{\frac {\displaystyle 1}{\displaystyle \log _{s}(b)}}\end{array}}$

Matice

${\begin{array}{cc}\left[{\begin{array}{ccc}1&1&1\\1-{\sqrt {2\,}}&{\sqrt {2\,}}+1&1\\{\sqrt {2\,}}-1&1-{\sqrt {2\,}}&1\\\end{array}}\right]&\left[{\begin{array}{ccc}1&-3&-3\\1-{\sqrt {2\,}}&{\sqrt {2\,}}-3&-3\\{\sqrt {2\,}}-1&1-{\sqrt {2\,}}&1\\\end{array}}\right]\\\left[{\begin{array}{ccc}1&1&1\\{\sqrt {2\,}}+1&1-{\sqrt {2\,}}&1\\-{\sqrt {2\,}}-1&{\sqrt {2\,}}+1&1\\\end{array}}\right]&\left[{\begin{array}{ccc}1&-3&-3\\{\sqrt {2\,}}+1&-{\sqrt {2\,}}-3&-3\\-{\sqrt {2\,}}-1&{\sqrt {2\,}}+1&1\\\end{array}}\right]\\\left[{\begin{array}{ccc}-1&3&3\\-{\sqrt {2\,}}-1&{\sqrt {2\,}}+3&3\\{\sqrt {2\,}}+1&-{\sqrt {2\,}}-1&-1\\\end{array}}\right]&\left[{\begin{array}{ccc}-1&-1&-1\\-{\sqrt {2\,}}-1&{\sqrt {2\,}}-1&-1\\{\sqrt {2\,}}+1&-{\sqrt {2\,}}-1&-1\\\end{array}}\right]\\\left[{\begin{array}{ccc}-1&3&3\\{\sqrt {2\,}}-1&3-{\sqrt {2\,}}&3\\1-{\sqrt {2\,}}&{\sqrt {2\,}}-1&-1\\\end{array}}\right]&\left[{\begin{array}{ccc}-1&-1&-1\\{\sqrt {2\,}}-1&-{\sqrt {2\,}}-1&-1\\1-{\sqrt {2\,}}&{\sqrt {2\,}}-1&-1\\\end{array}}\right]\end{array}}$

Sinus

${\begin{array}{rll}&\sin \left({\frac {n(2k-1)\pi }{2n+1}}-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\sin \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(sinus souctu uhlu)}}\\=&\left[\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]\sin \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(roznasobeni)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\sin \left({\frac {(2k-1)\pi }{2n+1}}\right)-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\sin \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(rozsireni zlomkem 2/2)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\sin \left({\frac {2}{2}}\,{\frac {(2k-1)\pi }{2n+1}}\right)-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\sin \left({\frac {2}{2}}\,{\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(sinus dvojnasobneho uhlu)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\left[2\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\left[2\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]&{\text{(prerovnani vyrazu)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\,2\left[\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]^{2}-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\,2\left[\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]^{2}&{\text{(vztahy }}2[\cos(x)]^{2}=1+\cos(2x){\text{ a }}2[\sin(x)]^{2}=1-\cos(2x){\text{)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\left[1+\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)\right]-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\left[1-\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)\right]&{\text{(roznasobeni)}}\\=&\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)+\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)\\&-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)+\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(prerovnani a vytknuti)}}\\=&\left[\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)-\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]\\&+\left[\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)+\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(hratky se znamenky a prerovnani)}}\\=&-\left[\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)-\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left({\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]\\&+\left[\cos \left({\frac {n(2k-1)\pi }{2n+1}}\right)\cos \left(-{\frac {(2k-1)\pi }{2(2n+1)}}\right)-\sin \left({\frac {n(2k-1)\pi }{2n+1}}\right)\sin \left(-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\right]\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(kosinus souctu uhlu)}}\\=&-\cos \left({\frac {n(2k-1)\pi }{2n+1}}+{\frac {(2k-1)\pi }{2(2n+1)}}\right)+\cos \left({\frac {n(2k-1)\pi }{2n+1}}-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(spolecny jmenovatel v prvnim kosinu)}}\\=&-\cos \left({\frac {(2n+1)(2k-1)\pi }{2(2n+1)}}\right)+\cos \left({\frac {n(2k-1)\pi }{2n+1}}-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(kraceni v prvnim kosinu)}}\\=&-\cos \left((2k-1){\frac {\pi }{2}}\right)+\cos \left({\frac {n(2k-1)\pi }{2n+1}}-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(kosinus pi/2 je nula)}}\\=&\cos \left({\frac {n(2k-1)\pi }{2n+1}}-{\frac {(2k-1)\pi }{2(2n+1)}}\right)\cos \left({\frac {(2k-1)\pi }{2n+1}}\right)&{\text{(hurrraaa, hotovo!)}}\end{array}}$

Matice

$\underbrace {\left[{\begin{array}{cccc}1&&&\\-2&1&&\\-3&&1&\\-6&&&1\end{array}}\right]} _{\displaystyle E_{1}}\underbrace {\left[{\begin{array}{cccc}1&2&3&4\\2&3&4&5\\3&6&9&1\\6&7&1&1\end{array}}\right]} _{\displaystyle A}=\underbrace {\left[{\begin{array}{cccc}1&2&3&4\\0&-1&-2&-3\\0&0&0&-11\\0&-5&-17&-23\end{array}}\right]} _{\displaystyle E_{1}A}$

$\underbrace {\left[{\begin{array}{cccc}1&&&\\-1/3&1&&\\-1/2&&1&\\-1/6&&&1\end{array}}\right]} _{\displaystyle E_{1}}\underbrace {\left[{\begin{array}{cccc}0&&&1\\&1&&\\&&1&\\1&&&0\end{array}}\right]} _{\displaystyle \Pi _{1}}\underbrace {\left[{\begin{array}{cccc}1&2&3&4\\2&3&4&5\\3&6&9&1\\6&7&1&1\end{array}}\right]} _{\displaystyle A}=\underbrace {\left[{\begin{array}{cccc}1&&&\\-1/3&1&&\\-1/2&&1&\\-1/6&&&1\end{array}}\right]} _{\displaystyle E_{1}}\underbrace {\left[{\begin{array}{cccc}6&7&1&1\\2&3&4&5\\3&6&9&1\\1&2&3&4\end{array}}\right]} _{\displaystyle \Pi _{1}A}=\underbrace {\left[{\begin{array}{cccc}6&7&1&1\\0&2/3&11/3&14/3\\0&5/2&17/2&1/2\\0&5/6&17/6&23/6\end{array}}\right]} _{\displaystyle E_{1}\Pi _{1}A}$

Rovina

$A=\left[{\begin{array}{ccc}x_{A}&y_{A}&z_{A}\end{array}}\right],\quad B=\left[{\begin{array}{ccc}x_{B}&y_{B}&z_{B}\end{array}}\right],\quad C=\left[{\begin{array}{ccc}x_{C}&y_{C}&z_{C}\end{array}}\right]$

$\rho :\quad ax+by+cz+d=0$

$\left[{\begin{array}{cccc}x_{A}&y_{A}&z_{A}&1\\x_{B}&y_{B}&z_{B}&1\\x_{C}&y_{C}&z_{C}&1\end{array}}\right]\left[{\begin{array}{c}a\\b\\c\\d\end{array}}\right]=\left[{\begin{array}{c}0\\0\\0\end{array}}\right]$

$ax+by+cz+d=0\qquad \Longleftrightarrow \qquad (ap)x+(bp)y+(cp)z+(dp)=0,\quad p\neq 0$

${\begin{array}{ll}p=a^{-1}:\quad &x=(-{\frac {b}{a}})y+(-{\frac {c}{a}})z+(-{\frac {d}{a}})\\p=b^{-1}:\quad &y=(-{\frac {a}{b}})x+(-{\frac {c}{b}})z+(-{\frac {d}{b}})\\p=c^{-1}:\quad &z=(-{\frac {a}{c}})x+(-{\frac {b}{c}})y+(-{\frac {d}{c}})\\p=d^{-1}:\quad &1=(-{\frac {a}{d}})x+(-{\frac {b}{d}})y+(-{\frac {c}{d}})z\end{array}}$

Limity

$\lim _{x\longrightarrow 0^{+}}0^{x}={\text{“}}\,0^{0}\,{\text{”}}=0.$

---

$f(x)=\exp(-1/x),\qquad \lim _{x\longrightarrow 0^{+}}f(x)=0^{+},$

$g(x)=-\log(a)x,\;\;\;a>0,\qquad \lim _{x\longrightarrow 0^{+}}g(x)=\left\{{\begin{array}{ll}0^{+},&a\in (0,1),\\0,&a=1,\\0^{-},&a\in (1,\infty ),\end{array}}\right.$

$\lim _{x\longrightarrow 0^{+}}{\Big (}f(x){\Big )}^{g(x)}={\text{“}}\,0^{0}\,{\text{”}}=\lim _{x\longrightarrow 0^{+}}\exp(-1/x)^{-\log(a)x}=\lim _{x\longrightarrow 0^{+}}\exp(-(-\log(a)x)/x)=\exp(\log(a))=a.$

---

$f(x)=\exp(-1/x^{2}),\qquad \lim _{x\longrightarrow 0^{+}}f(x)=0^{+},$

$g(x)=-x,\qquad \lim _{x\longrightarrow 0^{+}}g(x)=0^{-},$

$\lim _{x\longrightarrow 0^{+}}{\Big (}f(x){\Big )}^{g(x)}={\text{“}}\,0^{0}\,{\text{”}}=\lim _{x\longrightarrow 0^{+}}\exp(-1/x^{2})^{-x}=\lim _{x\longrightarrow 0^{+}}\exp(-(-x)/x^{2})=\lim _{x\longrightarrow 0^{+}}\exp(1/x)=+\infty .$

Cauchy--Riemann

Consider following bijection and injection

${\begin{array}{ll}\phi :\mathbb {C} \longleftrightarrow \mathbb {R} ^{2},\qquad &\phi (a+b\mathbf {i} )=\left[{\begin{array}{c}a\\b\end{array}}\right]\\\psi :\mathbb {C} \longrightarrow \mathbb {R} ^{2\times 2},\qquad &\psi (a+b\mathbf {i} )=\left[{\begin{array}{cr}a&-b\\b&a\end{array}}\right]\end{array}}$

Notable properties

${\begin{array}{ll}\forall c=a+b\mathbf {i} ,\quad z=x+y\mathbf {i} \quad \in \mathbb {C} :\\[3mm]\phi (c+z)=\phi {\Big (}(a+x)+(b+y)\mathbf {i} {\Big )}=\left[{\begin{array}{c}a+x\\b+y\end{array}}\right]=\left[{\begin{array}{c}a\\b\end{array}}\right]+\left[{\begin{array}{c}x\\y\end{array}}\right]=\phi (c)+\phi (z)\\\phi (cz)=\phi {\Big (}(ax-by)+(ay+bx)\mathbf {i} {\Big )}=\left[{\begin{array}{c}ax-by\\ay+bx\end{array}}\right]=\left[{\begin{array}{cr}a&-b\\b&a\end{array}}\right]\left[{\begin{array}{c}x\\y\end{array}}\right]=\psi (c)\,\phi (z)\end{array}}$

Consider a general vector field ${\vec {f}}$ over $\mathbb {R} ^{2}$ with Jacobian derivative

${\begin{array}{ll}{\vec {z}}=\left[{\begin{array}{c}x\\y\end{array}}\right],\qquad {\vec {f}}\!\left({\vec {z}}\right)=\left[{\begin{array}{c}u(x,y)\\v(x,y)\end{array}}\right]\\{\vec {f}}\!\left({\vec {z}}+d{\vec {z}}\right)={\vec {f}}\!\left({\vec {z}}\right)+J_{f}({\vec {z}})\,d{\vec {z}}=\left[{\begin{array}{c}u(x,y)\\v(x,y)\end{array}}\right]+\left[{\begin{array}{cc}{\frac {\partial }{\partial x}}u(x,y)&{\frac {\partial }{\partial y}}u(x,y)\\{\frac {\partial }{\partial x}}v(x,y)&{\frac {\partial }{\partial y}}v(x,y)\end{array}}\right]\,\left[{\begin{array}{c}dx\\dy\end{array}}\right]\end{array}}$

Consider a function $f$ on $\mathbb {C}$ with complex derivative (whatever it is)

${\begin{array}{ll}z=x+y\mathbf {i} ,\qquad f(z)=u(z)+v(z)\mathbf {i} \\[3mm]f(z+dz)=f(z)+f'(z)\,dz\\[2mm]\phi {\Big (}f(z+dz){\Big )}=\phi {\Big (}f(z){\Big )}+\psi {\Big (}f'(z){\Big )}\,\phi (dz)=\left[{\begin{array}{c}u(z)\\v(z)\end{array}}\right]+\psi {\Big (}f'(z){\Big )}\left[{\begin{array}{c}dx\\dy\end{array}}\right]\\\end{array}}$

Linking both toghether gives

$\phi (z)={\vec {z}},\qquad \phi {\Big (}f(z){\Big )}={\vec {f}}({\vec {z}})$

${\begin{array}{rl}\psi {\Big (}f'(z){\Big )}=&J_{f}({\vec {z}})\\f'(z)=&\psi ^{-1}\left(\left[{\begin{array}{cc}{\frac {\partial }{\partial x}}u(x,y)&{\frac {\partial }{\partial y}}u(x,y)\\{\frac {\partial }{\partial x}}v(x,y)&{\frac {\partial }{\partial y}}v(x,y)\end{array}}\right]\right)\qquad \Longrightarrow \end{array}}$

$\left[{\begin{array}{cc}{\frac {\partial }{\partial x}}u(x,y)&{\frac {\partial }{\partial y}}u(x,y)\\{\frac {\partial }{\partial x}}v(x,y)&{\frac {\partial }{\partial y}}v(x,y)\end{array}}\right]\in \left\{\left[{\begin{array}{cr}a&-b\\b&a\end{array}}\right]\;:\;a,b\in \mathbb {R} \right\}\subsetneq \mathbb {R} ^{2\times 2}\qquad \Longleftrightarrow \qquad {\frac {\partial }{\partial x}}u(x,y)={\frac {\partial }{\partial y}}v(x,y)\quad \land \quad {\frac {\partial }{\partial x}}v(x,y)=-{\frac {\partial }{\partial y}}u(x,y)\qquad$

Matrix inverse

Let

$M_{n}=\left[{\begin{array}{ccccccc}m_{1}^{(n)}&m_{2}^{(n)}&m_{3}^{(n)}&\cdots &m_{n-2}^{(n)}&m_{n-1}^{(n)}&m_{n}^{(n)}\\\end{array}}\right]=\left[{\begin{array}{ccccccc}1&2&3&\cdots &n\!\!-\!\!2&n\!\!-\!\!1&n\\2&2&3&\cdots &n\!\!-\!\!2&n\!\!-\!\!1&n\\3&3&3&\cdots &n\!\!-\!\!2&n\!\!-\!\!1&n\\\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots \\n\!\!-\!\!2&n\!\!-\!\!2&n\!\!-\!\!2&\cdots &n\!\!-\!\!2&n\!\!-\!\!1&n\\n\!\!-\!\!1&n\!\!-\!\!1&n\!\!-\!\!1&\cdots &n\!\!-\!\!1&n\!\!-\!\!1&n\\n&n&n&\cdots &n&n&n\\\end{array}}\right]$

Denote

${\widetilde {M}}_{n}=\left[{\begin{array}{c|c|c}m_{0}^{(n)}&M_{n}&m_{n+1}^{(n)}\\\end{array}}\right]=\left[{\begin{array}{c|ccccccc|c}1&1&2&3&\cdots &n\!\!-\!\!2&n\!\!-\!\!1&n&n\!\!+\!\!1\\2&2&2&3&\cdots &n\!\!-\!\!2&n\!\!-\!\!1&n&n\!\!+\!\!1\\3&3&3&3&\cdots &n\!\!-\!\!2&n\!\!-\!\!1&n&n\!\!+\!\!1\\\vdots &\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots &\vdots \\n\!\!-\!\!2&n\!\!-\!\!2&n\!\!-\!\!2&n\!\!-\!\!2&\cdots &n\!\!-\!\!2&n\!\!-\!\!1&n&n\!\!+\!\!1\\n\!\!-\!\!1&n\!\!-\!\!1&n\!\!-\!\!1&n\!\!-\!\!1&\cdots &n\!\!-\!\!1&n\!\!-\!\!1&n&n\!\!+\!\!1\\n&n&n&n&\cdots &n&n&n&n\!\!+\!\!1\\\end{array}}\right]$

Note that

$m_{0}^{(n)}=m_{1}^{(n)}\,,\qquad (n+1)\cdot m_{n}^{(n)}=n\cdot m_{n+1}^{(n)}$

For $j=1,2,3,\ldots ,n$

$m_{j-1}^{(n)}+m_{j+1}^{(n)}=2\cdot m_{j}^{(n)}+e_{j}^{(n)}\qquad \Longleftrightarrow \qquad \left[{\begin{array}{ccc}m_{j-1}^{(n)}&m_{j}^{(n)}&m_{j+1}^{(n)}\\\end{array}}\right]\left[{\begin{array}{r}1\\-2\\1\end{array}}\right]=e_{j}^{(n)}$

where $e_{j}^{(n)}$ denotes the $j$-th column of identity matrix of order $n$. Thus

$\left[{\begin{array}{c|ccccccc|c}m_{0}^{(n)}&m_{1}^{(n)}&m_{2}^{(n)}&m_{3}^{(n)}&\cdots &m_{n-2}^{(n)}&m_{n-1}^{(n)}&m_{n}^{(n)}&m_{n+1}^{(n)}\\\end{array}}\right]\left[{\begin{array}{rrrrrrr}1\\\hline -2&1\\1&-2&1\\&1&-2&1\\&&\ddots &\ddots &\ddots \\&&&1&-2&1\\&&&&1&-2&1\\&&&&&1&-2\\\hline &&&&&&1\\\end{array}}\right]=\left[{\begin{array}{ccccccc}e_{1}^{(n)}&e_{2}^{(n)}&e_{3}^{(n)}&\cdots &e_{n-2}^{(n)}&e_{n-1}^{(n)}&e_{n}^{(n)}\\\end{array}}\right]=I_{n}$

Now for $j=1$ and $j=n$

${\begin{array}{rcl}j=1\,:\qquad m_{j-1}^{(n)}+m_{j+1}^{(n)}&=&2\cdot m_{j}^{(n)}+e_{j}^{(n)}\,,\qquad m_{0}^{(n)}=m_{1}^{(n)}\\[2mm]m_{0}^{(n)}+m_{2}^{(n)}&=&2\cdot m_{1}^{(n)}+e_{1}^{(n)}\\m_{1}^{(n)}+m_{2}^{(n)}&=&2\cdot m_{1}^{(n)}+e_{1}^{(n)}\\-m_{1}^{(n)}+m_{2}^{(n)}&=&e_{1}^{(n)}\\\left[{\begin{array}{cc}m_{1}^{(n)}&m_{2}^{(n)}\\\end{array}}\right]\left[{\begin{array}{r}-1\\1\end{array}}\right]&=&e_{1}^{(n)}\\\end{array}}\qquad \qquad \qquad \qquad {\begin{array}{rcl}j=n\,:\qquad m_{j-1}^{(n)}+m_{j+1}^{(n)}&=&2\cdot m_{j}^{(n)}+e_{j}^{(n)}\,,\qquad (n+1)\cdot m_{n}^{(n)}=n\cdot m_{n+1}^{(n)}\\[2mm]m_{n-1}^{(n)}+m_{n+1}^{(n)}&=&2\cdot m_{n}^{(n)}+e_{n}^{(n)}\\m_{n-1}^{(n)}+{\frac {n+1}{n}}\,m_{n}^{(n)}&=&2\cdot m_{n}^{(n)}+e_{n}^{(n)}\\m_{n-1}^{(n)}+{\frac {1-n}{n}}\,m_{2}^{(n)}&=&e_{n}^{(n)}\\\left[{\begin{array}{cc}m_{n-1}^{(n)}&m_{n}^{(n)}\\\end{array}}\right]\left[{\begin{array}{c}1\\{\frac {1-n}{n}}\end{array}}\right]&=&e_{n}^{(n)}\\\end{array}}$

Altogether

$M_{n}^{-1}=\left[{\begin{array}{rrrrrrc}-1&1\\1&-2&1\\&1&-2&1\\&&\ddots &\ddots &\ddots \\&&&1&-2&1\\&&&&1&-2&1\\&&&&&1&{\frac {1-n}{n}}\\\end{array}}\right]$