Calculating Determinants

1 Cofactors

By definition, if A is an n × n square matrix, its determinant,

| A | ...= ...

...
∑		ζ(π) n ∏ ai,π(i) i = 1
π ∊ Pn

Consider the elements of the r^th row of A (1 ≤ r ≤ n), then

| A | ...= ...

...
∑		ζ(π) ar,π(r) ... ∏ ai,π(i) i ≠ r
π ∊ Pn

Now π maps r to some s, where 1 ≤ s ≤ n. Considering each s in turn we can reorder the sum as

| A | ...= ...

n
∑		... ...ar s
s = 1

...
∑		ζ(π) ... ∏ ai,π(i) i ≠ r
π(r)=s

The inner sum in the expression is called the cofactor of a_{r s} and is written as A_{r s}:

A_{r s} ...= ...

...
∑		ζ(π) ... ∏ ai,π(i) i ≠ r
π(r)=s

1.1 Row Expansion

For any (fixed) r (1≤r≤n) we have

| A | ...= ...

n
∑		ar s Ar s
s = 1

This is called the expansion of the determinant of A by the r^th row.

1.2 Column Expansion

Similarly for any (fixed) s (1≤s≤n) we have

| A | ...= ...

n
∑		ar s Ar s
r = 1

called the expansion of the determinant of A by the s^th column.

2 Minors

Let B be any p × q matrix, B = (b'_{i j} )_{i =1 ⋯ p, j =1 ⋯ q} .

For ...m ≤ min{p, q} ...chose m rows (i(1) ⋯ i(m)) and m columns (j(1) ⋯ j(m)). ...We can form a new (square) matrix B' = (b'_{r s} )_{r =1 ⋯ m, s =1 ⋯ m} ...where b'_{r s} = b_{i(r), i(s)}, ...i.e. a matrix formed with only those elements of B which are both in one of the rows (i(1) ⋯ i(m)) and in one of the columns (j(1) ⋯ j(m)) of B.

The determinant of B' is called an m^th order minor of B.

2.1 Cofactor Expansion

For n × n square matrix A, ...let ...A^x_{r s} ...(r, s ≤ n) ...be the matrix formed by excluding the elements of the r^th row and the s^th column of A.
| A^x_{r s} | is an (n – 1)^th order minor of A. ...If A_{r s} is the cofactor of a_{r s} in A we have

A_{r s} ...= ...( – 1)^r+s | A^x_{r s} |

Proof:

Let B be the matrix obtained from A by interchanging successively the r^th and r+1^th rows, ⋯ the n-1^th and n^th rows, and then the s^th ⋯ n^th columns similarly. (n – r row operations and n – s column operations). ...So ...| B | ...= ...( – 1)^{n – r + n – s} | A | = ...( – 1)^{r + s} | A |

Now

B ...= ...

a1 1 ⋯ ⋯ a1 n a1 s : : : : : : an 1 ⋯ ⋯ an n an s ar 1 ⋯ ⋯ ar n ar s

...= ...

a1 s Axr s : : ar 1 ⋯ ⋯ ar s

Write B = (b_{i j} ), ...then the cofactor of a_{r s} in A is the cofactor of b_{n n} in B

| B | ...= ...

...
∑		ζ(σ) n – 1 ∏ bi,σ(i) i = 1
σ ∊ Pn

3 Inverse Matrix

3.1 Adjoint (or Adjugate)

If A_{i j} is the cofactor of a_{i j} in A, the adjoint or adjugate of A is the matrix defined by

adj A ...= ...(A_{i j} )^T

i.e. the transpose of the matrix of corresponding cofactors.

[ The term adjoint is also used to designate the transpose of the conjugate matrix of A, ...A ^T = A* , ...there is usually little scope for confusion, as the current type of adjoint is only used in the calculation of determinants. ]

Lemma: ......A(adj A) ...= ...(adj A)A ...= ...| A | I_n

Proof:
The (i, j)^th element of adj A: ...(adj A)_{i j} ...= ...A_{j i} , ...– the cofactor of a_{j i} in A. ...So the (r, p)^th element of A(adj A) is ...∑_s a_{r s} A_{p s} .
Now if r = p then this becomes ∑_s a_{r s} A_{r s} = | A | , ...– expansion of | A | by the r^th row.
If r ≠ p then ∑_s a_{r s} A_{p s} = 0 , ...as this is the determinant of the matrix obtained from A by replacing the p^th row of A by the r^th row, so this matrix will have two equal rows and so its determinant is zero.

Similar proof for ......(adjA)A ...= ...| A | I_n .

From the above if | A | ≠ 0 then

1) ...A^–1 ...= ...

	adj A
	| A |

2) ...| adj A | ...= ...| A | ^{n – 1}

Proof of 2): ...From 1) ...adj A = | A | A^–1 , ...so ...| adj A | ...= ...| A | ⁿ | A^–1 | ...= ...| A | ^{n – 1} .
[ Remember | λB | = λⁿ | B | ...and ...| A^–1 | = | A | ^–1 ]

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