Question

Vă rog frumos să rezolvați ex 1​

Answer 1

inversa unei matrici

$A - B = \left(\begin{array}{ccc}2-(-1)&-1-(-1)&-1-(-1)\\-1-(-1)&2-(-2)&-1-(-1)\\-1-(-1)&-1-(-1)&2-(-3)\end{array}\right)=$ $= \left(\begin{array}{ccc}3&0&0\\0&4&0\\0&0&5\end{array}\right)$

$M = (A - B)^{2} = \left(\begin{array}{ccc}3&0&0\\0&4&0\\0&0&5\end{array}\right) \cdot \left(\begin{array}{ccc}3&0&0\\0&4&0\\0&0&5\end{array}\right) =$ $= \left(\begin{array}{ccc}9&0&0\\0&16&0\\0&0&25\end{array}\right)$

• se calculează determinantul:

$det(M) = 9 \cdot 16 \cdot 25 \neq 0 \to inversabil\breve{a}$

• se construiește matricea transpusă:

$M^{t} = \left(\begin{array}{ccc}9&0&0\\0&16&0\\0&0&25\end{array}\right)$

• se construiește matricea adjunctă:

$(-1)^{1+1} = \left|\begin{array}{cc}16&0\\0&25\end{array}\right| = 16 \cdot 25$

$(-1)^{1+2} = \left|\begin{array}{cc}0&0\\0&25\end{array}\right| = 0$

$(-1)^{1+3} = \left|\begin{array}{cc}0&16\\0&0\end{array}\right| = 0$

$(-1)^{2+1} = \left|\begin{array}{cc}0&0\\0&25\end{array}\right| = 0$

$(-1)^{2+2} = \left|\begin{array}{cc}9&0\\0&25\end{array}\right| = 9 \cdot 25$

$(-1)^{2+3} = \left|\begin{array}{cc}9&0\\0&0\end{array}\right| = 0$

$(-1)^{3+1} = \left|\begin{array}{cc}0&0\\16&0\end{array}\right| = 0$

$(-1)^{3+2} = \left|\begin{array}{cc}9&0\\0&0\end{array}\right| = 0$

$(-1)^{3+3} = \left|\begin{array}{cc}9&0\\0&16\end{array}\right| = 9 \cdot 16$

$M^{*} = \left(\begin{array}{ccc}16 \cdot 25&0&0\\0&9 \cdot 25&0\\0&0&9 \cdot 16\end{array}\right)$

• se construiește matricea inversă:

$M^{-1} = \dfrac{1}{det M} \cdot M^{*} =$

$= \dfrac{1}{9 \cdot 16 \cdot 25} \cdot \left(\begin{array}{ccc}16 \cdot 25&0&0\\0&9 \cdot 25&0\\0&0&9 \cdot 16\end{array}\right)$

$M^{-1} = \left(\begin{array}{ccc}\dfrac{1}{9}&0&0\\0&\dfrac{1}{16}&0\\0&0&\dfrac{1}{25}\end{array}\right)$