Răspuns:
Explicație pas cu pas:
2)
a) det A = 2 · 5 - 3 · (-1) = 10 +3 = 13
b) B = [tex]A^{2}[/tex] - 3A + 2[tex]I_{2}[/tex]
B = [tex]\left[\begin{array}{ccc}2&3\\-1&5\\\end{array}\right][/tex] ·[tex]\left[\begin{array}{ccc}2&3\\-1&5\\\end{array}\right][/tex] - 3 · [tex]\left[\begin{array}{ccc}2&3\\-1&5\\\end{array}\right][/tex] + 2 ·[tex]\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right][/tex]
B = [tex]\left[\begin{array}{ccc}-1&-9\\-7&-22\\\end{array}\right][/tex] - [tex]\left[\begin{array}{ccc}6&9\\-3&15\\\end{array}\right][/tex] + [tex]\left[\begin{array}{ccc}2&0\\0&2\\\end{array}\right][/tex]
B = [tex]\left[\begin{array}{ccc}5&-18\\-4&-5\\\end{array}\right][/tex]
c) X · A = [tex]\left[\begin{array}{ccc}1&1\\-1&1\\\end{array}\right][/tex]
X · [tex]\left[\begin{array}{ccc}2&3\\-1&5\\\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}1&1\\-1&1\\\end{array}\right][/tex]
X se scrie ca [tex]\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right][/tex] · [tex]\left[\begin{array}{ccc}2&3\\-1&5\\\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}1&1\\-1&1\\\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}2a-b&3a+5b\\2c-d&3c+5d\\\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}1&1\\-1&1\\\end{array}\right][/tex]
2a - b = 1
3a + 5b = 1
2c - d = - 1
3c + 5d = 1
rezolvând sistemul avem a = 6/13, b = -1/13, c = - 4/13 și d = - 5/13
