Question

Va rog sa ma ajutati la b de la exercitiul cu matrice! Va rog sa explicati cum ati facut.

E de la subiectul 2 de bac MI

Multumesc anticipat!

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$A(m,n) = \left(\begin{array}{ccc}1&1&1\\1&m&m^2\\1&n&n^2\end{array}\right)$

$B(m,n) = \left(\begin{array}{c}1&m^3&n^3\end{array}\right)$

$\Delta = (m-1)(n-1)(n-m)$

$\\\Delta_x = \left|\begin{array}{ccc}1&1&1\\m^3&m&m^2\\n^3&n&n^2\end{array}\right|=m\cdot n\cdot \left|\begin{array}{ccc}1&1&1\\m^2&1&m\\n^2&1&n\end{array}\right| =$

$=m\cdot n \cdot (-1)\cdot (-1)\cdot \left|\begin{array}{ccc}1&1&1\\1&m&m^2\\1&n&n^2\end{array}\right| = m\cdot n\cdot \Delta$

$\Rightarrow x = \dfrac{\Delta x}{\Delta} = \dfrac{m\cdot n\cdot \Delta}{\Delta} \Rightarrow \boxed{x = mn}$

$\\\Delta_y = \left|\begin{array}{ccc}1&1&1\\1&m^3&m^2\\1&n^3&n^2\end{array}\right|\overset{\begin{array}{l}C_2\to C_2 - C_3\\C_3\to C_3 - C_1\end{array}}{=}$

$=\left|\begin{array}{ccc}1&1-1&1-1\\1&m^3-m^2&m^2-1\\1&n^3-n^2&n^2-1\end{array}\right| =$

$=\left|\begin{array}{ccc}1&0&0\\1&m^2(m-1)&m^2-1\\1&n^2(n-1)&n^2-1\end{array}\right| =$

$=m^2(m-1)(n^2-1)-n^2(n-1)(m^2-1) =$

$=m^2(m-1)(n-1)(n+1)-n^2(n-1)(m-1)(m+1)=$

$=(m-1)(n-1)\Big[m^2(n+1)-n^2(m+1)\Big] =$

$=(m-1)(n-1)\Big(m^2n+m^2-n^2m-n^2\Big) =$

$=(m-1)(n-1)\Big(m^2n-n^2m+m^2-n^2\Big)$

$=(m-1)(n-1)\Big[mn(m-n)+(m-n)(m+n)\Big]=$

$=(m-1)(n-1)(m-n)(mn+m+n)$

$\Rightarrow y = \dfrac{\Delta_y}{\Delta} = \dfrac{-\Delta\cdot(mn+m+n)}{\Delta} \Rightarrow \boxed{y = -mn-m-n}$

$\\\Delta_z = \left|\begin{array}{ccc}1&1&1\\1&m&m^3\\1&n&n^3\end{array}\right|\overset{\begin{array}{l}L_2\to L_2 - L_1\\L_3\to L_3 - L_1\end{array}}{=}$

$=\left|\begin{array}{ccc}1&1&1\\1-1&m-1&m^3-1\\1-1&n-1&n^3-1\end{array}\right|=$

$=\left|\begin{array}{ccc}1&1&1\\0&m-1&(m-1)(m^2+m+1)\\0&n-1&(n-1)(n^2+n+1)\end{array}\right|=$

$=(m-1)(n-1)\left|\begin{array}{ccc}1&1&1\\0&1&m^2+m+1\\0&1&n^2+n+1\end{array}\right|=$

$=(m-1)(n-1)\Big[(n^2+n+1) - (m^2+m+1)\Big]=$

$=(m-1)(n-1)(n^2-m^2+n-m) =$

$=(m-1)(n-1)\Big[(n-m)(n+m)+n-m\Big]=$

$= (m-1)(n-1)(n-m)(m+n+1)$

$\Rightarrow z= \dfrac{\Delta_z}{\Delta}= \dfrac{\Delta\cdot (m+n+1)}{\Delta}\Rightarrow \boxed{z = m+n+1}$