Question

Buna seara!! Am nevoie de ajutor
Multumesc anticipat !

Answer 1

Explicație pas cu pas:

matricea:

$A(m) = \left[\begin{array}{ccc}0&0&-1\\1&0&1\\m&-m&2\end{array}\right]$

a)

$S = A(1) + A(2) + ... + A(10) = \\ = \left[\begin{array}{ccc}0 + 0 + ... + 0&0 + 0 + ... + 0&-1 - 1 - ... - 1\\1 + 1 + ... + 1&0 + 0 + ... + 0&1 + 1 + ... + 1\\1 + 2 + ... + 10&-1 - 2 - ... - 10&2 + 2 + ... + 2\end{array}\right] \\ = \left[\begin{array}{ccc}10 \times 0&10 \times 0&10(-1)\\10 \times 1&10 \times 0&10 \times 1\\\frac{10*11}{2} &-\frac{10*11}{2}&10 \times 2\end{array}\right] \\ = \left[\begin{array}{ccc}0&0&-10\\10&0&10\\55& - 55&20\end{array}\right]$

$detS = det\left[\begin{array}{ccc}0&0&-10\\10&0&10\\55& -55&20\end{array}\right] \\ = 10 \times ( - 55) \times ( - 10) = 5500$

b) matricea:

$A(1) = \left[\begin{array}{ccc}0&0&-1\\1&0&1\\1&-1&2\end{array}\right]$

$detA = det\left[\begin{array}{ccc}0&0&-1\\1&0&1\\1&-1&2\end{array}\right] \\ = 1( - 1)( - 1) = 1\neq 0 = > A \: este \: inversabila$

$A^{-1} = \frac{1}{detA} A^{*}$

$\frac{1}{detA} = \frac{1}{1} = 1 = > A^{-1} = A^{*}$

transpusa matricei A:

$^{t}A = \left[\begin{array}{ccc}0&1&1\\0&0& - 1\\ - 1&1&2\end{array}\right]$

adjuncta matricei A:

$A^{* } = \left[\begin{array}{ccc}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{array}\right]$

complemenții algebrici ai matricei transpuse:

$A_{11}=(-1)^{1+1}\left[\begin{array}{ccc}0&-1\\1&2\end{array}\right] = 1 \times (0 - ( - 1)) = 1 \\ A_{21}=(-1)^{2+1}\left[\begin{array}{ccc}0&-1\\-1&2\end{array}\right] =( - 1)(0 - 1) = 1 \\ A_{31}=(-1)^{3+1}\left[\begin{array}{ccc}0&0\\-1&1\end{array}\right] = 1 \times (0 - 0) = 0 \\ A_{12}=(-1)^{1+2}\left[\begin{array}{ccc}1&1\\1&2\end{array}\right] = ( - 1)(2 - 1) = - 1 \\ A_{22}=(-1)^{2+2}\left[\begin{array}{ccc}0&1\\-1&2\end{array}\right] = 1 \times (0 - ( - 1)) = 1 \\ A_{32}=(-1)^{3+2}\left[\begin{array}{ccc}0&1\\-1&2\end{array}\right] = ( - 1)(0 - ( - 1)) = - 1 \\ A_{13}=(-1)^{1+3}\left[\begin{array}{ccc}1&1\\0&-1\end{array}\right] = 1 \times (- 1 - 0) = - 1 \\ A_{23}=(-1)^{2+3}\left[\begin{array}{ccc}0&1\\0&-1\end{array}\right] = ( - 1)(0 - 0) = 0 \\ A_{33}=(-1)^{3+3}\left[\begin{array}{ccc}0&1\\0&0\end{array}\right] = 1 \times (0 - 0) = 0$

=>

$A^{*} = \left[\begin{array}{ccc}1&1&0\\-1&1&-1\\- 1&0&0\end{array}\right]$

$= > A^{-1} = \left[\begin{array}{ccc}1&1&0\\-1&1&-1\\- 1&0&0\end{array}\right]$