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Linear Algebra - Sebastian Pancratz

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j det A = n<br />

i=1 (−1)i+j aij det(A ij )<br />

i det A = n<br />

j=1 (−1)i+j aij det(A ij )<br />

<br />

<br />

det A = d(A (1) , . . . , A (n) )<br />

=<br />

=<br />

n<br />

aij(−1) i+j−2 <br />

1 ∗<br />

d<br />

0 A ij<br />

i=1<br />

n<br />

i=1<br />

(−1) i+j aij det A ij ,<br />

<br />

A ∈ Mn(F) adj A n × n (i, j) <br />

(−1) i+j det A ij <br />

(adj A)A = (det A)I<br />

A A−1 = 1<br />

det A adj A<br />

<br />

j < k<br />

det A =<br />

n<br />

(adj A)jiaij = (adj A · A)jj.<br />

i=1<br />

0 = det(A (1) , . . . , A (k) , . . . , A (k) , . . . , A (n) )<br />

=<br />

n<br />

(adj A)jiaik<br />

i=1<br />

= (adj A · A)jk.<br />

1<br />

A det A = 0 <br />

adj A<br />

A −1 = 1<br />

det A<br />

det A<br />

adj A · A = I <br />

Ax = b m n <br />

A m × n b F m rank A =<br />

rank(A| b) n = rank A = rank(A| b) <br />

x = A −1 b <br />

m = n <br />

A ∈ Mn(F) Ax = b x =<br />

(x1, . . . , xn) T xi = 1<br />

detA det A îb i = 1, . . . , n A îb <br />

A i b

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