Răspuns:
Explicație pas cu pas:
a) Pentru a calcula A(1)*A(2) mai intai trebuie sa vedem cine este A(1) si cine este A(2). Pentru acest lucru pur si simplu il vom inlocui pe a cu valoarea 1 iar apoi cu valoarea 2.
A(1) = [tex]= \left[\begin{array}{ccc}2*1&2*1-3\\2*1-3&2*1\end{array}\right] =\left[\begin{array}{ccc}2&-1\\-1&2\end{array}\right][/tex]
A(2) = [tex]= \left[\begin{array}{ccc}2*2&2*2-3\\2*2-3&2*2\end{array}\right] = \left[\begin{array}{ccc}4&1\\1&4\end{array}\right][/tex]
A(1)*A(2) = [tex]\left[\begin{array}{ccc}2&-1\\-1&2\end{array}\right] * \left[\begin{array}{ccc}4&1\\1&4\end{array}\right] = \left[\begin{array}{ccc}7&-2\\-2&7\end{array}\right][/tex]
b) A(x) nu e inversabila <=> det(A(x) ) = 0 => 2x*2x - (2x-3)² = 0 => 4x²-4x²+12x-9=0 =>
12x=9 => x=9/12 = 3/4
c) X = [tex]\left[\begin{array}{ccc}x&y\\z&t\end{array}\right][/tex]
X*A(1)=A(2) => [tex]\left[\begin{array}{ccc}x&y\\z&t&\end{array}\right] * \left[\begin{array}{ccc}2&-1\\-1&2\end{array}\right] = \left[\begin{array}{ccc}4&1\\1&4\end{array}\right] => 2x-y = 4 , 2y-x = 1 , 2z-t=1 , 2t-z=4[/tex]
y=2x-4 => 2(2x-4)-x=1 => 3x=9 => x=3 => y=2
t = 2z-1 => 2(2z-1)-z=4 => 4z-2-z=4 => 3z=6 => z=2 => t=3
=> X = [tex]\left[\begin{array}{ccc}3&2\\2&3\end{array}\right][/tex]
