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season 3 to 6 in linear algebra
Typology: Exercises
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2 1 −k
k − 1 2
1 k 0
k 1 k
0 1 k
tan(x) 0 cot(x)
tan(x) 0 cot(x)
b 0 a
k − 1 4
0 k + 1 1
0 0 k − 3
k − 1 4
− 1 S
⊤ P + P A = −Q
− 1 B = B(A + B)
− 1 A = (A
− 1
− 1 )
− 1
− 1 A
n P = B
n
− 1
k 1 1 1
k k 1 1
k k k 1
= (1 − k)
3 , B =
1 + k 1 k 1 k 1 k 1
k 2 1 + k 2 k 2 k 2
k 3 k 3 1 + k 3 k 3
k 4 k 4 k 4 1 + k 4
= 1 + k2 + k3 + k 4