# 66. Determinants¶

Let $$a, b, c, d$$ be any four numbers, real or complex, the symbol

$\begin{split}\begin{vmatrix} a & b\\ c & d\\ \end{vmatrix}\end{split}$

denotes $$ad - bc$$ and is called a determinant of second order. $$a, b, c, d$$ are called elements of the determinant and $$ad - bc$$ is called value of the determinant.

As you can see, the elements of a determinant are positioned in the form of a square in its designation. The diagonal on which elements $$a$$ and $$d$$ lie is called the principal or primary diagonal of the determinant and the diagonal which is formed on the line of $$b$$ and $$c$$ is called the secondary diagonal.

A row is constituted by elements lying in the same horizontal line and a column is constituted by elements lying in the same vertical line.

Clearly, determinant of second order has two rows and two columns and its value is equal to the products of elements along primary diagonal minus the product of elements along the secondary diagonal. Thus, by definition

$\begin{split}\begin{vmatrix} 2 & 4\\ 3 & 9\\ \end{vmatrix} = 18 - 12 = 6\end{split}$

Let $$a_1, a_2, a_3, b_1, b_2, b_3, c_1, c_2, c_3$$ be any nine numbers, then the symbol

$\begin{split}\begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}\end{split}$

is another way of saying

$\begin{split}a_1\begin{vmatrix} b_2 & b_3\\ c_2 & c_3\\ \end{vmatrix} - a_2\begin{vmatrix} b_1 & b_3\\ c_1 & c_3 \end{vmatrix} + a_3\begin{vmatrix} b_1 & b_2\\ c_1 & c_2 \end{vmatrix}\end{split}$

i.e. $$a_1(b_2c_3 - b_3c_2)-a_2(b_1c_3-b_3c_1) + a_3(b_1c_2-b_2c_1)$$

Rule to put + or - before any element: Find the sum of number of rows and columns in which the considered element occus. If the sum is even put a $$+$$ sign before the element and if the sum is odd, put a $$-$$ sign before the element. Since $$a_1$$ occurs in first row and first column whose sum is $$1 + 1 = 2$$ which is an even number, therefore $$+$$ sign occurs for it. Since $$a_2$$ occurs in first row and second column whose sum is $$1+ 2 = 3$$ which is an odd number, therefore $$-$$ sign occurs before it.

We have expanded the determinant along first row in previous case. The value of determinant does not change no matter which row or column we expand it along.

Expanding the determinant along second row, we get

$\begin{split}\begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 \end{vmatrix} = -b_1\begin{vmatrix} a_2 & a_3\\ c_2 & c_3 \end{vmatrix} + b_2\begin{vmatrix} a_1 & a_3\\ c_1 & c_3 \end{vmatrix} - b_3\begin{vmatrix} a_1 & a_2\\ c_1 & c_2 \end{vmatrix}\end{split}$

$$= -b_1(a_2c_3 - a_3c_2) + b_2(a_1c_3 - a_3c_1) - b_3(a_1c_2 - a_2c_1)$$

$$= a_1(b_2c_3 - b_3c_2)-a_2(b_1c_3-b_3c_1) + a_3(b_1c_2-b_2c_1)$$

Thus, we see that value of determinant remains unchanged irrespective of the change of row and column against which it is expanded.

Usually, an element of a determinant is denoted by a letter with two suffices, first one indicating the row and second one indicating the column in which the element occcur. Thus, $$a_{ij}$$ element indicates that it has occurred in ith row and jth column.

We also denote the rows by $$R_1, R_2, R_3$$ and so on. $$R_i$$ denotes the ith row of determinant while $$R_j$$ denotes jth row. Columns are denoted by $$C_, C_2, C_3$$ and so on. $$C_i$$ and $$C_j$$ denote ith and jth column of determinant.

$$\Delta$$ is the usual symbol for a determinant. Another way of denoting the determinant $$\begin{vmatrix}a_1&b_1&c_1\\a_2&b^2&c_2\\a_3&b_3&c_3 \end{vmatrix}$$ is ($$a_1b_2c_3$$).

The expanded form of determinant has $$n!$$ terms where $$n$$ is the number of rows or columns.

Ex 1. Find the value of the determinant

$\begin{split}\Delta = \begin{vmatrix} 1 & 2 & 4\\ 3 & 4 & 9\\ 2 & 1 & 6 \end{vmatrix}\end{split}$

Expanding the determinant along the first row

$\begin{split}\Delta = 1\begin{vmatrix} 4 & 9\\ 1 & 6 \end{vmatrix} -2\begin{vmatrix} 3 & 9\\ 2 & 6 \end{vmatrix} + 4\begin{vmatrix} 3 & 4\\ 2 & 1 \end{vmatrix}\end{split}$

Expanding the determinant along first row $$= 1(24 -9) - 2(18 - 18) + 4(3 - 8) = -5$$

Ex 2. Find the value of the determinant

$\begin{split}\Delta = \begin{vmatrix} 3 & 1 & 7\\ 5 & 0 & 2\\ 2 & 5 & 3 \end{vmatrix}\end{split}$

Expanding the determinant along second row,

$\begin{split}\Delta = -5\begin{vmatrix} 1 & 7\\ 5 & 3 \end{vmatrix} + 0\begin{vmatrix} 3 & 7\\ 2 & 3 \end{vmatrix} -2\begin{vmatrix} 3 & 1\\ 2 & 5 \end{vmatrix}\end{split}$

$$= -5(3 - 35) -2(15 -2) = 134$$

## 66.1. Minors¶

Consider the determinant

$\begin{split}\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{31} & a_{33} \end{vmatrix}\end{split}$

If we leave the elements belonging to row and column of a particular element $$a_{ij}$$ then we will obtain a second order determinant. The determinant thus obtained is called minor of $$a_{ij}$$ and it is denoted by $$M_{ij},$$ since there are $$9$$ elements in the above determinant we will have $$9$$ minors.

For example, the minor of element $$a_{21}=\begin{vmatrix}a_{12} & a_{13}\\ a_{32} & a_{33}\end{vmatrix} = M_{21}$$

The minor of element $$a_{32} = \begin{vmatrix}a_{11} & a_{13}\\a_{21} & a_{23}\end{vmatrix} = M_{32}$$

If we want to write the determinant in terms of minors then following is the expression obtained if we expand it along first row

$$\Delta = (-1)^{1+1}a_{11}M_{11} + (-1)^{1 + 2}a_{12}M_{12} + (-1)^{1 + 3} a_{13}M_{13}$$

$$=a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}$$

## 66.2. Cofactors¶

The minor $$M_{ij}$$ multiplied with $$(-1)^{i+j}$$ is known as cofactor of the element $$a_{ij}$$ and is denoted like $$A_{ij}$$.

Thus, we can say that, $$\Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$$

## 66.3. Theorems on Determinants¶

Theorem I. The value of a determinant is not changed when rows are changed into corresponsing columns.

Let $$\Delta = \begin{vmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\ a_3 & b_3 & c_3\end{vmatrix},$$ $$\Delta^{\prime} = \begin{vmatrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\c_1 & c_2 & c_3\end{vmatrix}$$

The leading term of each determinant is $$a_1b_2c_3.$$

The remaning terms of $$\Delta$$ are derived from $$a_1b_2c_3$$ by keeping the letters in their natural order and arranging the suffixes in all possible ways, an interchange of two suffixes producing a change of sign.

The remaining terms of $$\Delta^{\prime}$$ are derived from $$a_1b_2c_3$$ by keeping the suffixes in their natural order and arranging the letters in all possible ways, an interchange of two letters producing a change of sign.

The result in the two cases are identical. The same argument applies to determinant of any order.

Theorem II. The interchange of two rows, or of two columne, change the sign of a determinant without altering its numerical value.

For the interchange of two row is equivalent to interchange of the suffixes, and the interchange of two columns is equivalent to the interchange of two letters. Hence in either case the sign of every term of determinant is changed.

Theorem III. A determinant in which two rows or two columns are identical is equal to zero.

IF two rows or or tow columns of a determinant $$\Delta$$ are identical the change of rows or columns does not change numerical vlaue but changes the sign. Thus, $$\Delta = -\Delta \Rightarrow \Delta = 0$$

Impoetant Identities. For the determinant $$\Delta = (a_1b_2c_3)$$ we have a number of identities of the following types:

1. $$a_2A_1 + b_2B_1 + c_2C_1 = 0$$

2. $$b_1A_1 + b_2A_2 + b_2A_3 = 0$$

where $$A_1, B_1, \ldots$$ are cofactors of $$a_, b_1, \ldots$$