Question

Considere as matrizes $A=\begin{pmatrix}1&0&-1\\0&-1&2\end{pmatrix}$, $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$, $x=\begin{pmatrix}x\\y\end{pmatrix}$ e $B=\begin{pmatrix}1\\2\end{pmatrix}$. Sabendo que $(AA^t-3I)x=B$, calcule x + y.

Answer 1

$\mathbf{A}=\left[\begin{array}{ccc} 1&0&-1\\ 0&-1&2 \end{array} \right ],\;\;\; \mathbf{I}=\left[\begin{array}{cc} 1&0\\ 0&1 \end{array} \right ],\;\;\; \mathbf{x}=\left[\begin{array}{c} x\\ y \end{array} \right ],\;\;\; \mathbf{B}=\left[\begin{array}{c} 1\\ 2 \end{array} \right ]$


$\mathbf{A}\mathbf{A}^{t}=\left[\begin{array}{ccc} 1&0&-1\\ 0&-1&2 \end{array}\right]\cdot\left[\begin{array}{cc} 1&0\\ 0&-1\\ -1&2 \end{array}\right]\\ \\ =\left[\begin{array}{cc} 1\cdot 1+0\cdot 0+\left(-1 \right )\cdot \left(-1 \right )\;\;&\;\;1 \cdot 0+0\cdot \left(-1 \right )+\left(-1 \right )\cdot 2\\ 0 \cdot 1+\left(-1 \right )\cdot 0+2\cdot \left(-1 \right )\;\;&\;\;0 \cdot 0+\left(-1 \right )\cdot \left(-1 \right )+2 \cdot 2 \end{array}\right]\\ \\ =\left[\begin{array}{cc} 1+0+1\;\;&\;\;0+0-2\\ 0+0-2\;\;&\;\;0+1+4 \end{array}\right]\\ \\ =\left[\begin{array}{cc} 2&-2\\ -2&5 \end{array}\right]$


$\mathbf{A}\mathbf{A}^{t}-3\mathbf{I}\\ \\ =\left[\begin{array}{cc} 2&-2\\ -2&5 \end{array}\right]-3\cdot\left[ \begin{array}{cc} 1&0\\ 0&1 \end{array} \right ]\\ \\ =\left[ \begin{array}{cc} 2-3\cdot 1\;\;&\;\;-2-3 \cdot 0\\ -2-3\cdot 0\;\;&\;\;5-3\cdot 1 \end{array} \right ]\\ \\ =\left[ \begin{array}{cc} 2-3\;\;&\;\;-2-0\\ -2-0\;\;&\;\;5-3 \end{array} \right ]\\ \\=\left[ \begin{array}{cc} 2-3\;\;&\;\;-2-0\\ -2-0\;\;&\;\;5-3 \end{array} \right ]\\ \\ =\left[ \begin{array}{cc} -1&-2\\ -2&2 \end{array} \right ]$


$\left(\mathbf{A}\mathbf{A}^{t}-3\mathbf{I} \right )\mathbf{x}=\mathbf{B}\\ \\ \left[ \begin{array}{cc} -1&-2\\ -2&2 \end{array} \right ]\left[ \begin{array}{c} x\\y \end{array} \right ]=\left[ \begin{array}{c} 1\\2 \end{array} \right ]$


Vou resolver este sistema pela Regra de Cramer. Calculado o determinante da matriz $\mathbf{A}\mathbf{A}^{t}-3\mathbf{I}$, temos

$\det\left(\mathbf{A}\mathbf{A}^{t}-3\mathbf{I} \right )\\ \\ =\det \left[ \begin{array}{cc} -1&-2\\ -2&2 \end{array} \right ]\\ \\ =\left(-1 \right ) \cdot 2-\left(-2 \right )\cdot \left(-2 \right )\\ \\ =-2-4\\ \\ =-6$


O determinante $\Delta x$ da incógnita $x$ é obtido substituindo a primeira coluna de $\mathbf{A}\mathbf{A}^{t}-3\mathbf{I}$ pelo vetor $\mathbf{B}$:

$\Delta x=\det\left[\begin{array}{cc} 1&-2\\ 2&2 \end{array} \right ]\\ \\ =1 \cdot 2- 2 \cdot \left(-2 \right )\\ \\ =2+4\\ \\ =6$


O determinante $\Delta y$ da incógnita $y$ é obtido substituindo a segunda coluna de $\mathbf{A}\mathbf{A}^{t}-3\mathbf{I}$ pelo vetor $\mathbf{B}$:

$\Delta y=\det\left[\begin{array}{cc} -1&1\\ -2&2 \end{array} \right ]\\ \\ =\left(-1 \right )\cdot 2-\left(-2 \right )\cdot 1\\ \\ =-2+2\\ \\ =0$


$x=\dfrac{\Delta x}{\det\left(\mathbf{A}\mathbf{A}^{t}-3\mathbf{I} \right )}\\ \\ x=\dfrac{6}{-6}\\ \\ \boxed{x=-1}\\ \\ \\ y=\dfrac{\Delta y}{\det\left(\mathbf{A}\mathbf{A}^{t}-3\mathbf{I} \right )}\\ \\ y=\dfrac{0}{6}\\ \\ \boxed{y=0}$


Então,

$x+y=-1+0\\ \\ \boxed{x+y=-1}$