Question

Determinati numarul real x pentru care C(x) * A - A * C(x) = B.
A = (1 3 2 1)
B = (-4 0 0 4)
C = (x 1 2 3)
Doresc o rezolvare completa!

Answer 1

$\displaystyle \mathtt{A= \left(\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt2&\mathtt1\end{array}\right);~B= \left(\begin{array}{ccc}\mathtt{-4}&\mathtt0\\\mathtt0&\mathtt4\end{array}\right);~C= \left(\begin{array}{ccc}\mathtt x&\mathtt1\\\mathtt2&\mathtt3\end{array}\right)} \\ \\ \mathtt{C(x) \cdot A-A \cdot C(x)=B}$

$\displaystyle \mathtt{C(x)\cdot A=\left(\begin{array}{ccc}\mathtt x&\mathtt1\\\mathtt2&\mathtt3\end{array}\right)\cdot\left(\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt2&\mathtt1\end{array}\right)=\left(\begin{array}{ccc}\mathtt{x\cdot1+1\cdot 2}&\mathtt{x\cdot3+1\cdot1}\\\mathtt{2\cdot 1+3\cdot2}&\mathtt{2\cdot3+3\cdot1}\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{x+2}&\mathtt{3x+1}\\\mathtt{2+6}&\mathtt{6+3}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{x+2}&\mathtt{3x+1}\\\mathtt8&\mathtt9\end{array}\right)}$

$\displaystyle \mathtt{C(x) \cdot A=\left(\begin{array}{ccc}\mathtt{x+2}&\mathtt{3x+1}\\\mathtt8&\mathtt9\end{array}\right)}$

$\displaystyle\mathtt{A\cdot C(x)=\left(\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt2&\mathtt1\end{array}\right)\cdot\left(\begin{array}{ccc}\mathtt x&\mathtt1\\\mathtt2&\mathtt3\end{array}\right)=\left(\begin{array}{ccc}\mathtt {1\cdot x+3\cdot2}&\mathtt{1\cdot1+3\cdot3}\\\mathtt{2\cdot x+1\cdot2}&\mathtt{2\cdot1+1\cdot3}\end{array}\right) =}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{x+6}&\mathtt{1+9}\\\mathtt{2x+2}&\mathtt{2+3}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{x+6}&\mathtt{10}\\\mathtt{2x+2}&\mathtt5\end{array}\right)}$

$\displaystyle \mathtt{A\cdot C(x)=\left(\begin{array}{ccc}\mathtt{x+6}&\mathtt{10}\\\mathtt{2x+2}&\mathtt5\end{array}\right)}$

$\displaystyle \mathtt{C(x) \cdot A-A \cdot C(x)=\left(\begin{array}{ccc}\mathtt{x+2}&\mathtt{3x+1}\\\mathtt8&\mathtt9\end{array}\right) -\left(\begin{array}{ccc}\mathtt{x+6}&\mathtt{10}\\\mathtt{2x+2}&\mathtt5\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{x+2-(x+6)}&\mathtt{3x+1-10}\\\mathtt{8-(2x+2)}&\mathtt{9-5}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{x+2-x-6}&\mathtt{3x-9}\\\mathtt{8-2x-2}&\mathtt4\end{array}\right)=}$

$\mathtt{=\left(\begin{array}{ccc}\mathtt{-4}&\mathtt{3x-9}\\\mathtt{-2x+6}&4\end{array}\right)= \left(\begin{array}{ccc}\mathtt{-4}&\mathtt{3(x-3)}\\\mathtt{-2(x-3)}&\mathtt4\end{array}\right)} \\ \\ \mathtt{C(x) \cdot A-A \cdot C(x)=\left(\begin{array}{ccc}\mathtt{-4}&\mathtt{3(x-3)}\\\mathtt{-2(x-3)}&4\end{array}\right)}$

$\displaystyle \mathtt{C(x)\cdot A-A\cdot C(x)=B \Rightarrow \left(\begin{array}{ccc}\mathtt{-4}&\mathtt{3(x-3)}\\\mathtt{-2(x-3)}&4\end{array}\right)=\left(\begin{array}{ccc}\mathtt{-4}&\mathtt0\\\mathtt0&\mathtt4\end{array}\right)\Rightarrow}\\ \\ \mathtt{\Rightarrow x-3=0 \Rightarrow x=0+3\Rightarrow \mathbf{\underline{x=3}}}$