# Determinant

BASIC CONCEPTS

Consider the equations a1x + b1y = 0 and a2x + b2y = 0.

The expression $\large \left| \begin{array}{cc} a_1 & b_1 \\ a_2 & b_2 \end{array} \right|$ is called a determinant of order two , and equal to $a_1 b_2 – a_2 b_1$

A determinant of order three consisting of 3 rows and 3 columns is written as

$\large \left| \begin{array}{ccc} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right|$ and is equal to

$\large = a_1 \left| \begin{array}{cc} b_2 & c_2 \\ b_3 & c_3 \end{array} \right| – b_1 \left| \begin{array}{cc} a_2 & c_2 \\ a_3 & c_3 \end{array} \right| + c_1 \left| \begin{array}{cc} a_2 & b_2 \\ a_3 & b_3 \end{array} \right|$

$\large = a_1 (b_2 c_3 – c_2 b_3) – b_1 (a_2 c_3 – c_2 a_3) + c_1(a_2 b_3 – b_2 a_3)$

* The numbers ai, bi, ci ( i = 1,2,3 ) are called the elements of the determinant.

* The determinant obtained by deleting the ith row and jth column is called the minor of element at the ith row and the jth column. The cofactor of this element is (−1)i+j (minor).

## PROPERTIES OF DETERMINANTS :

The determinant remains unaltered if its rows are changed into columns and the columns into rows.

* If all the elements of a row (or column) are zero, then the value of determinant is zero.

* If the elements of a row (column) are proportional (or identical) to the elements of any other row (column), then the determinant is zero.

* The interchange of any two rows (columns) of the determinant changes its sign.

* On rolling over n rows the determinant value Δ reduces to (−1)nΔ.

* If all the elements of a row (column) of a determinant are multiplied by a non-zero constant then the determinant gets multiplied by the same constant.

* A determinant remains unaltered under a column ( Ci) operation of the form Ci + α Cj + βCk ( j,k≠i) or a row (Ri) operation of the form Ri + α Rj + βRk ( j,k≠i).

* If each element in any row (column) is the sum of r terms, then the determinant can be expressed as the sum of r determinants.

* If determinant Δ = f(x) and f(a) = 0, then (x − a) is a factor of the determinant. In other word, if two rows (or two columns) becomes proportional (identical) for x = a then (x − a) is a factor of determinant. In general, if r rows become identical for x = a then (x − a)r−1 is factor of the determinant.

* If in a determinant (of order three or more) the elements in all the rows (columns) are in A.P. with same or different common difference, the value of the determinant is zero.

* The determinant value of an odd order skew symmetric determinant is always zero.

Remarks:

* It is important to know that all the properties applicable to rows are also equally applicable to columns but independently.

* Whenever rows are disturbed by applications of properties of determinants, at least one of the row shall remain in original shape. In other words all the rows shall not be disturbed at a time.

* It is always desirable to try to bring in as many zeros as possible in any row (or column) and then expand the determinant with respect to that row (column). Mere expansion from the outset should be avoided as far as possible

* We can express a determinant

$\large \left| \begin{array}{ccc} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right|$ as $\large \left| \begin{array}{ccc} c_1 & c_2 & c_3 \end{array} \right|$ or $\large \left| \begin{array}{ccc} R_1 \\ R_2 \\ R_3 \end{array} \right|$

Where Ci (i = 1,2, 3 ) are the columns and Rj ( j = 1 , 2 , 3) are the rows of the determinant.

Exercise: Prove that

$\large \left| \begin{array}{ccc} a-b-c & 2b & 2c\\ 2a & b-c-a & 2c \\ 2a & 2b & c-a-b \end{array} \right|$

= (a + b + c)3

Illustration : Without expanding to any stage, prove that $\large \Delta = \left| \begin{array}{ccc} 1 & a & a^2 -bc\\ 1 & b & b^2-ca \\ 1 & c & c^2 – ab \end{array} \right| = 0$

Solution: Δ = $\large \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right|$ – $\large \left| \begin{array}{ccc} 1 & a & bc\\ 1 & b & ca \\ 1 & c & ab \end{array} \right|$

= Δ1 − Δ2 (say)

$\large \Delta_2 = \left| \begin{array}{ccc} 1 & a & bc\\ 1 & b & ca \\ 1 & c & ab \end{array} \right|$

Applying $\large R_1 \rightarrow a R_1 , R_2 \rightarrow b R_2 , R_3 \rightarrow c R_3$

$\large \Delta_2 = \frac{1}{abc} \left| \begin{array}{ccc} a & a^2 & abc\\ b & b^2 & bca \\ c & c^2 & abc \end{array} \right|$

$\large \Delta_2 = \frac{abc}{abc} \left| \begin{array}{ccc} a & a^2 & 1\\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right|$

$\large \Delta_2 = \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| = \Delta_1$

Hence Δ = Δ1 − Δ1 = 0

Illustration : Using the factor property of determinants show that

$\large \Delta = \left| \begin{array}{ccc} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & c+b & -2c \end{array} \right|$

= k (a + b) (b + c) (c + a). Evaluate k.

Exercise :

(i) Evaluate Δ using the properties of Determinant

$\large \Delta = \left| \begin{array}{ccc} loga & p & 1 \\ logb & q & 1 \\ logc & r & 1 \end{array} \right|$

where a, b, c (> 0) are the pth , qth and rth terms of a G.P.

(ii) Show that :

$\large \left| \begin{array}{ccc} b+c & c+a & a+b \\ q+r & r+p & p+q \\ y+z & z+x & x+y \end{array} \right|$ $\large = 2\left| \begin{array}{ccc} a & b & c \\ p & q & r \\ x & y & z \end{array} \right|$

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