11. Write down the co-factors of a, ƒ and c in the expansion of the determinant A =

10.

1 1

1

1=1.

1 2

3

4

1

3 6

10

a

h g

h b f
gf c

Shew that, if A, B, &c. are the co-factors of a, b, &c. in the above determinant, and A', B', &c. the co-factors of A, B, &c. in the determinant

420. Theorem V. If the elements of one column of a determinant be multiplied in order by the co-factors of the corresponding elements of any other column; then the sum of the products will be zero.

Let the elements of the rth column be multiplied by the co-factors of the corresponding elements in the column; then the sum of the products will be

a. A ̧+b,.B ̧+...

Now consider the determinant which differs from the original determinant only in having the sth column identical with the rth; then A, B,, &c. will be the same in the new determinant as in the original one.

The value of the new determinant will therefore, by Art. 419, be equal to

=

since a = a, b = b ̧

± {a ̧ . A ̧+b ̧ . B ̧+...}

± {a,. A ̧+b. B ̧+.....},

&c.

But, from Art. 415, we know that the new determinant is zero.

Hence a,. A, + b, . B ̧ +..... = 0.

also

8

Thus in the determinant A= [a,b,c,d1], we have

421.

Theorem VI. If each element of any row (or column) of a determinant be the sum of two quantities, the determinant can be expressed as the sum of two determinants of the same order.

It will be sufficient to take as an example the determinant

By Art. 418, we have, if A,, A,, A, be the co-factors of

the elements of the first column,

3

▲ = (a1 + a ̧) A ̧ + (α, +α2) A ̧ + (ɑ3 +α3) Ag

2

422. Theorem VII. A determinant is not altered in value by adding to all the elements of any column (or row) the same multiples of the corresponding elements of any number of other columns (or rows).

Take as an example a determinant of the third order: we have to shew that

By Theorem VI, the last determinant is equal to

But each of the last two determinants is zero [Art. 416, Cor.]; this proves the theorem.

Take the second row from the third, and then the first from

It is in the first place clear that a term of the determinant of the sixth order will be obtained by taking any term of [a,b,c,] with any term of [ay]. Thus ▲=[a,b,c,]. [α12] together with terms involving l, m, n, &c.; and we have to shew that all terms involving any of the letters l, m, n, &c. will vanish.

Now, in every term of the minor of 1, three elements must be chosen from the last three rows, and two only of these can be chosen from the last two columns; hence one of the three elements must be zero, and therefore every term of A, is zero. Hence the minor of 1, and so also the minor of each of the elements m, n, &c. is zero; this proves that there are no terms involving any of the letters l, m, n, &c.

It can be proved in a similar manner that any determinant of the 2nth order is the product of two determinants of the nth order, provided every element of one of the nth minors of the original determinant is zero.

424. Multiplication of determinants. We shall consider the case of two determinants of the third order: the method is however perfectly general.

To express as a determinant of the third order, the product of the two determinants.

Multiply the first three rows by a1, B1, Y, and add the products to the fourth row; then multiply the first three rows by a, B, Y, respectively and add the products to the fifth row; and then multiply the first three rows by a,, B, Ya respectively and add the products to the sixth row. We shall then have the equivalent determinant