Page 335 - 35Linear Algebra

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Thus
     
3 1 −1 1 2
X =  0 −2 1    =  −1  .
2
−2 1 0 3 0
Finally,
 
1 − λ 1 1
P A (λ) = − det  2 2 − λ 3 
4 5 6 − λ
h i
= − (1 − λ)[(2 − λ)(6 − λ) − 15] − [2.(6 − λ) − 12] + [10 − 4.(2 − λ)]

3
2
= λ − 9λ − λ + 1 .

a b
3. Call M = . Then detM = ad − bc, yet
c d
2
1 2 1 2 1 a + bc ∗ 1 2
− tr M + (tr M) = − tr 2 − (a + d)
2 2 2 ∗ bc + d 2
1 2 2 1 2 2
= − (a + 2bc + d ) + (a + 2ad + d ) = ad − bc ,
2 2
which is what we were asked to show.

4.
 
1 2 3
perm  4 5 6  = 1 · (5 · 9 + 6 · 8) + 2 · (4 · 9 + 6 · 7) + 3 · (4 · 8 + 5 · 7) = 450 .
7 8 9

(a) Multiplying M by λ replaces every matrix element M i in the formula
σ(j)
i
for the permanent by λM σ(j) , and therefore produces an overall factor
n
λ .
(b) Multiplying the i th row by λ replaces M i in the formula for the
σ(j)
permanent by λM i . Therefore the permanent is multiplied by an
σ(j)
overall factor λ.
(c) The permanent of a matrix transposed equals the permanent of the
original matrix, because in the formula for the permanent this amounts
to summing over permutations of rows rather than columns. But we
σ(1) σ(2) σ(n)
could then sort the product M M . . . M n back into its original
1 2
order using the inverse permutation σ −1 . But summing over permuta-
tions is equivalent to summing over inverse permutations, and therefore
the permanent is unchanged.

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