P2 EA Diurno Gab Algebra linear

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L : P2 ! P3 L(at2 + bt+ c) = (a+ 2b)t3 + (b+ c)t� (a+ 2b+ 3c)
P2 L(at2 + bt+ c) = 0 (a+ 2b)t3 + (b+ c)t�
(a+ 2b+ 3c) = 0
8<: a+ 2b = 0b+ c = 0
a+ 2b+ 3c = 0
0
Ker(L) = {0}, � =
L(at2+bt+c) = (a+2b)t3+(b+c)t�(a+
2b+3c) L(at2+bt+c) = at3+2bt3+bt+ct�a�2b�3c L(at2+bt+c) = a(t3�1)+b(2t3+t�2)+c(t�3)
� = {t3� 1; 2t3 + t� 2; t� 3}
(1, 4) (9, 10) 2
C = (5, 7). v = (8, 6) = 2(4, 3).
vu =
1
5 (4, 3) 2a = 10 a = 5 b = 2
c =
p
21 vu
x¯2
25
+
y¯2
4
= 1
⇢
x¯ = 15 (4x+ 3y � 41)
y¯ = 15 (�3x+ 4y � 13)
) (4x+ 3y � 41)
2
625
+
(�3x+ 4y � 13)2
100
= 1.
B1 = C + 2v?u =
1
5 (19, 43) B2 =
1
5 (31, 27) F1 = C +
p
21vu =
1
5
�
5 + 4
p
21, 7 + 3
p
21
�
F2 =
1
5
�
5� 4p21, 7� 3p21�
0 2 4 6 8 10
4
6
8
10
A =
24 �3 0 �10 �2 0
�1 0 �3
35 .
det(A � �I) = (� + 2)2(� + 4) = 0 �1 = �4 �2,3 = �2
(A � �I)X = 0 �1 = �4 v1 =
24 10
1
35
�2,3 = �2 v2 =
24 �10
1
35 v3 =
24 01
0
35
A =
24 �3 0 �10 �2 0
�1 0 �3
35 =
264
p
2
2 �
p
2
2 0
0 0 1p
2
2
p
2
2 0
375
24 �4 0 00 �2 0
0 0 �2
35
264
p
2
2 0
p
2
2
�
p
2
2 0
p
2
2
0 1 0
375 .
4x2 � y2 + z2 � 16x+ 8y � 6z + 16 = 0
4
�
x2 � 4x� � �y2 � 8y� + �z2 � 6z� = �16
4 (x� 2)2 � (y � 4)2 + (z � 3)2 = �7
(y � 4)2
7
� (x� 2)
2
7/4
� (z � 3)
2
7
= 1.
C = (2, 4, 3) XZ
H : x2 � y2 + 4x + 2y = �12
(y � 1)2
9
� (x+ 2)
2
9
= 1() x¯
2
9
� y¯
2
9
= 1.
x¯ = y � 1 y¯ = x + 2 C = (�2, 1) a = b = 3 c = 3p2
e = ca =
p
2 V1 = (�2,�2) V2 = (�2, 4) y¯ = ±x¯
x � y + 3 = 0 x + y + 1 = 0 Y
u = (1, 0)
F1 =
��2, 1� 3p2� , Extremo = F1 + ⇣ b2a ⌘u = ��5, 1� 3p2�