Răspuns :
[tex]A=\left(\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right)[/tex]
a)
Facem diferenta dintre produsul diagonalelor
detA=1-0=1
b)
[tex]2A-A\cdot A=\left(\begin{array}{cc}2 & 0 \\ -4 &1\end{array}\right)-\left(\begin{array}{cc}1 & 0 \\ -4 & 1\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\0 & 1\end{array}\right)=I_2[/tex]
c)
[tex]\left(\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right)\cdot \left(\begin{array}{cc}x-2& y\\ z+1 & 1\end{array}\right)-\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{cc}0& 0 \\ 0 & 0\end{array}\right)\\\\\left(\begin{array}{cc}x-2& y\\ -2x+4+z+1 &-2y+ 1\end{array}\right)-\left(\begin{array}{cc}1& 0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{cc}0& 0 \\ 0 & 0\end{array}\right)\\\\[/tex]
[tex]\left(\begin{array}{cc}x-3& y\\ -2x+5-z&-2y\end{array}\right)=\left(\begin{array}{cc}0& 0 \\ 0 & 0\end{array}\right)[/tex]
x-3=0
x=3
y=0
-2x+5-z=0
-6+5-z=0
z=1
Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9835794
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