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.
