Question

Problema 1c , va rog frumos !

Answer 1

exercitiul e frumos, dar este mult mai avantajos sa il rezolvi pe tot, in ordinea care este dat
de fapt este un mersde calcul sugerat de catre autorul exerciţiului
cum ar fi determinara m,atricei A^ (-1) , nu clasic , ci din o ecuatie matriceala, cea data la punctul A

Answer 2

$\displaystyle \mathtt{1.~c)~A= \left(\begin{array}{ccc}\mathtt{5}&\mathtt{4}\\\mathtt{-4}&\mathtt{-3}\end{array}\right);~B=\left(\begin{array}{ccc}\mathtt{7}&\mathtt{6}\\\mathtt{-6}&\mathtt{-5}\end{array}\right);~AX=B}\\ \\ \mathtt{AX=B \Rightarrow X=A^{-1}B}$

$\displaystyle \mathtt{A^{-1}= \frac{1}{det(A)} \cdot A^*}\\ \\ \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt{5}&\mathtt{4}\\\mathtt{-4}&\mathtt{-3}\end{array}\right|=5 \cdot (-3)-4 \cdot (-4)=-15+16=1}\\ \\ \mathtt{det(A)=1}\\ \\ \mathtt{A=\left(\begin{array}{ccc}\mathtt{5}&\mathtt{4}\\\mathtt{-4}&\mathtt{-3}\end{array}\right)\Rightarrow A^T=\left(\begin{array}{ccc}\mathtt{5}&\mathtt{-4}\\\mathtt{4}&\mathtt{-3}\end{array}\right)}$

$\displaystyle \mathtt{D_{11}=(-1)^{1+1}\cdot (-3)=1\cdot (-3)=-3}\\ \\ \mathtt{D_{12}=(-1)^{1+2}\cdot 4=(-1) \cdot 4=-4}\\ \\ \mathtt{D_{21}=(-1)^{2+1}\cdot (-4)=(-1)\cdot (-4)=4}\\ \\ \mathtt{D_{22}=(-1)^{2+2}\cdot 5=1 \cdot 5=5}\\\\\mathtt{A^*=\left(\begin{array}{ccc}\mathtt{-3}&\mathtt{-4}\\\mathtt{4}&\mathtt{5}\end{array}\right)}$

$\displaystyle \mathtt{A^{-1}= \frac{1}{1}\cdot \left(\begin{array}{ccc}\mathtt{-3}&\mathtt{-4}\\\mathtt{4}&\mathtt{5}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{-3}&\mathtt{-4}\\\mathtt{4}&\mathtt{5}\end{array}\right)} \\ \\ \mathtt{A^{-1}=\left(\begin{array}{ccc}\mathtt{-3}&\mathtt{-4}\\\mathtt{4}&\mathtt{5}\end{array}\right)}$

$\displaystyle \mathtt{A^{-1}B=\left(\begin{array}{ccc}\mathtt{-3}&\mathtt{-4}\\\mathtt{4}&\mathtt{5}\end{array}\right) \cdot\left(\begin{array}{ccc}\mathtt{7}&\mathtt{6}\\\mathtt{-6}&\mathtt{-5}\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{-3\cdot7+(-4) \cdot (-6)}&\mathtt{-3\cdot 6+(-4) \cdot (-5)}\\\mathtt{4 \cdot 7+5 \cdot (-6)}&\mathtt{4 \cdot 6+5 \cdot (-5)}\end{array}\right)=}\\ \\ \mathtt{= \left(\begin{array}{ccc}\mathtt{-21+24}&\mathtt{-18+20}\\\mathtt{28-30}&\mathtt{24-25}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{3}&\mathtt2\\\mathtt{-2}&\mathtt{-1}\end{array}\right)}$ 

$\displaystyle \mathtt{X= \left(\begin{array}{ccc}\mathtt{3}&\mathtt2\\\mathtt{-2}&\mathtt{-1}\end{array}\right)}$