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t r inant

l t ach,

• 1 0 0

n

n n

( 3 ) (^0 0 ) 0 0 0 0 0 0

a b c d e f g h i j l m n x y z

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

1 2 3 1 2 3 ( ) 1 2 3 1 2 3

a a b b c c c c d d

1 2 3

1 1 1 1 1 1 1 1 1 1 ( 5 ) 1 1 1 1 1

(^1 1 1 1 1) n

a a a

a

⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯

Hint: Observe that the last three rows are linear dependent.

det = 0

) 0 0 0 0 0 0 0 0 0

c d e f g h i j l x z

Hint: Observe that the columns are linear dependent. Or we can use additive property to calculate.

det = 0

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

1 2 3 1 2 3 ( ) 1 2 3 1 2 3

a a b b c c c c d d

Hint: Use elementary operations to simplify the form and generate rows (columns) that contains only one non-zero element, or get the diagonal form.

det 1

n

n

i i

a a a

= a

1 2 3

1 1 1 1 n

a

a

a

a

1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 2 2 3 3

- - - -^1 -^ -^ -^ -

n

n n

a a a

a a a a a a a^ a^ a^ a

a a a

a a a a a a

r 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2

= 2

c c a a b a b c c c a a b a b c c c a a b a b c

Hint: additive property and 3 nd type column operation

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

c c a a b b c a a b

c c a a b b c a a b

c c a a b b c a a b

c a a c a b b c a a b c

c a a c a b b c a a b c

c a a c a b b c a a b c

fi , and all are distinct. Please find the

t f r t e function.

• 1 - 1 - 1

n

n

n

n n n

x x x a a a g x

a a a

a i

i t: If xi is the i-th root of g(x), then g(xi)= 0.

i ans the n×n matrix is rank deficient.

i s, e have , i eans the last n- 1 rows are linear independent. f r t ander onde Matrix, xi should be chosen from. conclude that the roots for g(x) are.

2 - 1

• 1 2 - 1 1 1 1

2 - 1

• 1 - 1 - 1

1 1

1

n i i i n

n n n n

x x x a a a

a a a

⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦

⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ a i ≠ a (^) j , ∀ i , j = 1 , 2 , ⋯, n - 1 , i ≠ j

{ a i } a 1 (^) , a 2 (^) , ⋯, a n - 1

f r ,we have

t.

subtract row 1 from each of the other rows

t.

can subtract, in order, x 1 times column n- 1 from column n, x ti es column n- 2 from column n- 1 , and so on, till we subtract x ti es column 1 from column 2.

f r ,we have

xtract all these as factors

t. (^) re rite the form

t.

clude that

⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

1111 2222

mmmm

0 0 000 0 0000 0 0 A

1 2

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

1111 1111 1111 1111

AAAA 0000 AAAA 0000 IIII 0000 AAAA ==== ==== 0000 CCCC 0000 IIII 0000 CCCC

consider

r , 1111 ,, I,III^2222 are identity matrices whose dimension are the same with AAAA 11 11 ,,,,CCCC 1111

is result was proved by Cauchy in 1815.

r f can be found:

tt :// .proofwiki.org/wiki/Determinant_of_Matrix_Product

det( B)=det(A)det(B)

(I eneral (^) det(A+B) ≠det(A)+det(B))

⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

1111 2222

mmmm

0 0 000 0 0000 0 0 A

1 2

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

1111 1111 1111 1111

AAAA 0000 AAAA 0000 IIII 0000 AAAA ==== ==== 0000 CCCC 0000 IIII 0000 CCCC

det(AB)=det(A)det(B)

r f re, e have 1

2

det =det det

=det det

⎛ (^) ⎡ ⎤ ⎞ ⎛ (^) ⎡ ⎤⎞ ⎜ (^) ⎢ ⎥ ⎟ ⎜ (^) ⎢ ⎥⎟ ⎝ ⎣^ ⎦^ ⎠ ⎝ ⎣^ ⎦⎠ ⋅

1111 1111

1111 1111

AAAA 0000 IIII 0000 AAAA 0000 IIII 0000 CCCC

AAAA CCCC

i ans

det =det 1 det

⎛ (^) ⎡ ⎤ ⎞ ⎛ (^) ⎡ ⎤⎞ ⎜ (^) ⎢ ⎥ ⎟ ⎜ (^) ⎢ ⎥⎟ ⎜ (^) ⎢ ⎥ ⎟ (^) ⋅ ⎜ (^) ⎢ ⎥⎟ ⎜ (^) ⎢ ⎥ ⎟ ⎜ (^) ⎢ ⎥⎟ ⎜⎜ (^) ⎢ ⎥ ⎟⎟ ⎜⎜ (^) ⎢ ⎥⎟⎟ ⎝ ⎣^ ⎦^ ⎠ ⎝ ⎣^ ⎦⎠

mmmm mmmm

AAAA 0000 0000 AAAA 0000 0000 0000 AAAA 0000 0000 AAAA 0000 AAAA 0000 0000 0000 0000 0000 AAAA 0000 0000 0000 AAAA

⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯

fi lly, e get (^) det AAA = det AAAA 1 (^) ⋅ det A (^) 2 ⋯ AAA det A m