Question

Se considera matricea A=(2 2 3 x -1 0 -1 2 1).
a) Pentru x=1 sa se calculeze inversa matricei A.
b) Pentru x=1 sa se rezolve ecuatia X*A=B,unde B=(2 1 3).
Putin ajutor,va rog?

Answer 1

$\displaystyle \mathtt{A= \left(\begin{array}{ccc}\mathtt2&\mathtt2&\mathtt3\\\mathtt x&\mathtt{-1}&\mathtt0\\\mathtt{-1}&\mathtt2&\mathtt1\end{array}\right)}\\ \\ \mathtt{a)x=1 \Rightarrow A= \left(\begin{array}{ccc}\mathtt2&\mathtt2&\mathtt3\\\mathtt 1&\mathtt{-1}&\mathtt0\\\mathtt{-1}&\mathtt2&\mathtt1\end{array}\right)~~~~~~~~~~~~~~~~~~~~~~~~~A^{-1}= \frac{1}{det(A)}\cdot A^* }$
$\displaystyle \mathtt{det(A)= \left|\begin{array}{ccc}\mathtt2&\mathtt2&\mathtt3\\\mathtt 1&\mathtt{-1}&\mathtt0\\\mathtt{-1}&\mathtt2&\mathtt1\end{array}\right|=2 \cdot (-1) \cdot 1+3 \cdot 1 \cdot 2+2 \cdot 0 \cdot (-1)-} \\ \\ \mathtt{-3 \cdot (-1) \cdot (-1)-2 \cdot 1 \cdot 1-2 \cdot 0 \cdot 2=-2+6+0-3-2-0=-1} \\ \\ \mathtt{det(A)=-1 \not = 0}$
$\displaystyle \mathtt{A^t= \left(\begin{array}{ccc}\mathtt2&\mathtt1&\mathtt{-1}\\\mathtt 2&\mathtt{-1}&\mathtt2\\\mathtt{3}&\mathtt0&\mathtt1\end{array}\right)}$
$\displaystyle \mathtt{D_{11}=(-1)^{1+1} \cdot \left|\begin{array}{ccc}\mathtt{-1}&\mathtt{2}\\\mathtt{0}&\mathtt{1}\end{array}\right|}=-1 \\ \\ \mathtt{D_{12}=(-1)^{1+2}\cdot \left|\begin{array}{ccc}\mathtt{2}&\mathtt{2}\\\mathtt{3}&\mathtt{1}\end{array}\right|=4}\\ \\ \mathtt{D_{13}=(-1)^{1+3} \cdot\left|\begin{array}{ccc}\mathtt{2}&\mathtt{-1}\\\mathtt{3}&\mathtt{0}\end{array}\right|=3 }$
$\displaystyle \mathtt{D_{21}=(-1)^{2+1}\cdot \left|\begin{array}{ccc}\mathtt{1}&\mathtt{-1}\\\mathtt{0}&\mathtt{1}\end{array}\right|=-1}\\ \\ \mathtt{D_{22}=(-1)^{2+2 }\cdot \left|\begin{array}{ccc}\mathtt{2}&\mathtt{-1}\\\mathtt{3}&\mathtt{1}\end{array}\right| =5}\\ \\ \mathtt{D_{23}=(-1)^{2+3}\cdot \left|\begin{array}{ccc}\mathtt{2}&\mathtt{1}\\\mathtt{3}&\mathtt{0}\end{array}\right|=3}$
$\displaystyle \mathtt{D_{31}=(-1)^{3+1}\cdot \left|\begin{array}{ccc}\mathtt{1}&\mathtt{-1}\\\mathtt{-1}&\mathtt{2}\end{array}\right|=1}\\ \\ \mathtt{D_{32}=(-1)^{3+2} \cdot \left|\begin{array}{ccc}\mathtt{2}&\mathtt{-1}\\\mathtt{2}&\mathtt{2}\end{array}\right|=-6}\\ \\ \mathtt{D_{33}=(-1)^{3+3} \cdot \left|\begin{array}{ccc}\mathtt{2}&\mathtt{1}\\\mathtt{2}&\mathtt{-1}\end{array}\right|=-4}$
$\displaystyle \mathtt{A^*= \left(\begin{array}{ccc}\mathtt{-1}&\mathtt4&\mathtt3\\\mathtt{-1}&\mathtt5&\mathtt3&\mathtt1&\mathtt{-6}&\mathtt{-4}\end{array}\right)}$
$\displaystyle \mathtt{A^{-1}= \frac{1}{det(A)}\cdot A^*= \frac{1}{(-1)}\cdot \left(\begin{array}{ccc}\mathtt{-1}&\mathtt4&\mathtt3\\\mathtt{-1}&\mathtt5&\mathtt3&\mathtt1&\mathtt{-6}&\mathtt{-4}\end{array}\right)=}\\ \\\mathtt{= (-1)\cdot \left(\begin{array}{ccc}\mathtt{-1}&\mathtt4&\mathtt3\\\mathtt{-1}&\mathtt5&\mathtt3&\mathtt{1}&\mathtt{-6}&\mathtt{-4}\end{array}\right)= }$
$\displaystyle \mathtt{= \left(\begin{array}{ccc}\mathtt{(-1) \cdot (-1)}&\mathtt{(-1) \cdot 4}&\matht{(-1) \cdot 3}\\\mathtt{(-1) \cdot(-1)}&\mathtt{(-1) \cdot 5}&\mathtt{(-1) \cdot 3}&\mathtt(-1 )\cdot 1&\mathtt{(-1) \cdot (-6)}&\mathtt{(-1) \cdot (-4)}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{1}&\mathtt{-4}&\mathtt{-3}\\\mathtt{1}&\mathtt{-5}&\mathtt{-3}&\mathtt{-1}&\mathtt{6}&\mathtt{4}\end{array}\right)}$
$\displaystyle \mathtt{A^{-1}= \left(\begin{array}{ccc}\mathtt{1}&\mathtt{-4}&\mathtt{-3}\\\mathtt{1}&\mathtt{-5}&\mathtt{-3}&\mathtt{-1}&\mathtt{6}&\mathtt{4}\end{array}\right) }$
$\displaystyle \mathtt{b)x=1;~X A=B;~B= \left(\begin{array}{ccc}\mathtt2&\mathtt1&\mathtt3\end{array}\right)} \\ \\ \mathtt{XA=B \Rightarrow X=BA^{-1}}$
$\displaystyle \mathtt{X= \left(\begin{array}{ccc}\mathtt2&\mathtt1&\mathtt3\end{array}\right) \cdot \left(\begin{array}{ccc}\mathtt{1}&\mathtt{-4}&\mathtt{-3}\\\mathtt{1}&\mathtt{-5}&\mathtt{-3}&\mathtt{-1}&\mathtt{6}&\mathtt{4}\end{array}\right) \Rightarrow }\\ \\ \mathtt{\Rightarrow X= \left(\begin{array}{ccc}2 \cdot 1+1 \cdot 1+3 \cdot (-1)&2 \cdot (-4)+1 \cdot (-5)+3 \cdot 6&2 \cdot (-3) +1 \cdot (-3)+3 \cdot 4 \end{array}\right)\Rightarrow}$
$\displaystyle \mathtt{\Rightarrow X=\left(\begin{array}{ccc}\mathtt{2+1-3}&\mathtt{-8-5+18}&\mathtt{-6-3+12}\end{array}\right) \Rightarrow X=\left(\begin{array}{ccc}\mathtt0&\mathtt5&\mathtt3\end{array}\right) }\\ \\ \mathtt{\Rightarrow X=\left(\begin{array}{ccc}\mathtt0&\mathtt5&\mathtt3\end{array}\right)}$