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Basic Concepts Of Determinant Part 1

Posted on - 04-01-2017

Math

IIT JEE

Basic concepts

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


&⇒

&⇒
a1b2 – a2b1 = 0

We express this eliminant as= 0.

q

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

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

q 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). .

Note that : D = = a1A1 + b1B1+c1C1

where A1, B1 and C1 are the cofactors of a1, b1 and c1 respectively.

q We can expand the determinant through any row or column. It means that we can write.

These results are true for determinants of any order.

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 D reduces to (–1)nD. .
  • 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 + a Cj + bCk ( j,k>i) or a row (Ri) operation of the form Ri + a Rj + bRk ( 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 D = 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 as
    • Where Ci ( i = 1,2, 3 ) are the columns and Rj ( j=1,2,3) are the rows of the determinant.

      Example.1

      Evaluate D only by using the properties of determinant where
      D = .

      Solution

      Operating C1 → C1 – C2 and C2 → C2 – C3 , we get

      D =

      Operating C1 → C1 – C2 and C3 → C3 – 10C2 , we get

      D =

      Operating R1 → R1 – R3 and R2 → R2 + 3R3 , we get

      D = = 0 (as first two rows are proportional).

      Example.2

      Without expanding to any stage, prove that

      D =

      Solution

      D =

      = D1 -D2 (say)

      D2 =

      = = D1

      Hence D = D1 -D1 = 0

      Example.3

      Show that D = 0 if D =

      Solution

      Operating C2 → C2 – (C1 + C3), we get

      D = = 0

      Note: Using the A.P. property one can immediately write D = 0 directly .

      Example.4

      Using the factor property of determinants show that

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

      Solution

      On checking, (with b = -a), we find that

      On operating C1 → C1 + C2 & C3 → C3 + C2 we get

      Taking 2 common from C1 and (a+c) from C3

      D = 2(a + c)

      D = 2(a + c)

      The operation R3 → R3 + R1 and R2 → R2 + R3 yields i.e. D= 0 .

      Similarly it can be proved that (b + c) and (c + a) are factor of the determinant. On putting a = 1 , b = 1, c =1. .

      R.H.S. = 8 k and L.H.S. = 32. Hence k = 4.

       
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