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Properties of Determinant
Determinant: To each square matrix A we can associate a expression or number (real or complex)
known as its determinant denoted by det  A or ∣A∣
If A=[ a ] then ∣A∣=a .
Determinant of a square matrix of order 2:
a b
If A=
then ∣A∣=ad – bc
c d
[ ]
∣
∣
2
3 = 2×4 – 3×– 3=89=17
–3 4
Determinant of a square matrix matrix of order 3:
a b c
A=
Let
x y z then we can find its determinant in six possible ways
u v w
Expansion Along First Row:
∣A∣= a y z – b x z  c x y
v w
u w
u v
Expansion Along Second Row:
∣A∣=−x b c  y a c − z a b
v w
u w
u v
Expansion Along Third Row:
∣A∣=u b c −v a c w a b
y z
x z
x y
Expansion along first column:
∣A∣=a y z – x b c u b c
v w
v w
y z
Expansion along second column:
∣A∣=−b x z  y a c – v a c
u w
u w
x z
Expansion along third column:
∣A∣=c x y – z a b  w a b
u v
u v
x y
Note: We expand the determinant along a row or column which has maximum number of zeros.
2 −3 1
Example-1: Evaluate the determinant 3 −2 4 by expanding it along first row.
1 2 5
2 −3 1
−2 4 −−3 3 4 1 3 −2
Solution: 3 −2 4 =2
2 5
1 5
1 2
1 2 5
2 −3 1
⇒ 3 −2 4 =2−10−83 15−462=−36338=5
1 2 5
For example
[ ]
∣ ∣ ∣ ∣ ∣ ∣
∣ ∣ ∣ ∣ ∣ ∣
∣ ∣ ∣ ∣ ∣ ∣
∣ ∣ ∣ ∣ ∣ ∣
∣ ∣ ∣ ∣ ∣ ∣
∣ ∣ ∣ ∣ ∣ ∣
∣
∣
∣
∣
∣
∣ ∣ ∣ ∣ ∣ ∣
∣
2
Example-2: Find the value of the determinant
∣
∣
2
3 −1
−1 2 −4
2 −1 3
by expanding it along the second row.
Solution:
2
3 −1
3 −1 2 2 −1 −−4 2 3 =9−12 624 −2−6
−1 2 −4 =−−1
−1 3
2 3
2 −1
2 −1 3
∣
∣
∣
∣
∣ ∣ ∣ ∣ ∣
∣
2
3 −1
⇒ −1 2 −4 =816−32=−8
2 −1 3
Example-3: Find the value of the determinant
∣
∣∣
∣
∣
3
2 −1
−2 1
1
4 −2 3
by expanding along it third row.
3
2 −1
2 −1 −−2 3 −1  3 3 2
−2 1
1 =4
1 1
−2 1
−2 1
4 −2 3
3
2 −1
⇒ −2 1
1 =42123−23 34=12221=35
4 −2 3
2 1 1
Example-4: Find the value of the determinant 3 −2 3 by expanding along first column.
1 2 1
2 1 1
−2 3 −3 1 1 1 1 1
Solution: 3 −2 3 =2
2 1
2 1
−2 3
1 2 1
2 1 1
3 −2 3 =2−2−6−31−232=−1635=−8
1 2 1
2 −1 3
Example-5: Find the value of the determinant 4 2 −1 by expanding it along second column.
5 2
1
2 −1 3
4 −1 2 2 3 −2 2 3
Solution: 4 2 −1 =−−1
5 1
5 1
4 −1
5 2
1
2 −1 3
⇒ 4 2 −1 =452 2−15−2 −2−12=9−2628=11
5 2
1
3 −1 −1
Example-6: Find the value of the determinant 2 2 −1 by expanding it along the third column.
2 3 −1
Solution:
∣
∣ ∣
∣ ∣ ∣
∣
∣
∣ ∣∣
∣ ∣
∣ ∣ ∣ ∣ ∣
∣
∣
∣
∣
∣
∣ ∣ ∣ ∣ ∣ ∣
∣
∣
Solution:
∣
∣
∣
∣
3 −1 −1
2 2 −−1 3 −1 −1 3 −1
2 2 −1 =−1
2 3
2 3
2 2
2 3 −1
∣ ∣ ∣ ∣ ∣ ∣
3
∣
∣
3 −1 −1
2 2 −1 =−6−492−62=−211−8=1
2 3 −1
Exercise-4.1
∣ ∣
∣ ∣
∣ ∣
∣ ∣
∣ ∣
3 1 2
2 −1 3 by expanding it along the first row. Also find
4 1 2
its value by expanding it along first column.
3 −1 1
2. Find the value of the determinant −2 1 1 by expanding it along the second row. Also
4
2 1
find its value by expanding it along second column.
3 4 0
3. Find the value of the determinant −1 2 3 by expanding it along the third row. Also find
2 4 5
its value by expanding it along third column.
a 0 1
4. Find the value of the determinant
1 2 1
−1 1 2
1. Find the value of the determinant
2 0 1
5. Find the value of the determinant −1 0 2
1 2 1
cos  −sin 
6. Find the value of the determinant
sin  cos 
3 x=3 2
7. Find the values of x for which
x 1 4 1
∣
∣
∣ ∣∣ ∣
8. Find the value of thew determinant
9. Find the value of the determinant
∣
∣
x 2 −x1 x−1
x1
x1
0
sin  −cos 
−sin 
0
sin 
cos  −sin 
0
∣
∣
Properties of Determinant
Property-1: Let A=[ aij ] be a square matrix of order n then ∣A∣=∣AT∣
∣
a
For example x
u
b
y
v
∣∣
c
a
z = b
w
c
x
y
z
∣
u
v
w
Property-2: Let A=[ aij ] be a square matrix of order n n≥2 and B be matrix obtained from A by
interchanging any two rows(columns) of A , then ∣A∣=– ∣B∣
4
∣
a
For example x
u
b
y
v
∣ ∣
c
x
z =– a
w
u
y
b
v
∣ ∣
z
c
w
a
or x
u
∣ ∣
b
y
v
c
b
z =– y
w
v
∣
a
x
u
c
z
w
Property-3: If any two rows(columns) of a square matrix A=[ aij ] of order n are identical, then its
determinant is zero i.e. ∣A∣=0
For example
∣
a
x
a
b
y
b
∣ ∣
c
a
z = 0 or x
c
u
b
y
v
∣
a
x =0
u
Property-4: Let A=[ aij ] be a square matrix of order n , and let B be the matrix obtained from A
by multiplying each element of a row(column) of A by a scalar  then ∣B∣=k ∣A∣
ka kb kc
a b c
ka b c
a b c
=
k
=
k
For example x
y
z
x y z or kx y z
x y z
u
v
w
u v w
ku v w
u v w
∣
∣∣
∣ ∣
∣∣
∣
Property-5: Let A be a square matrix such that each element of a row(column) of A is expressed as
the sum of two or more terms. Then, the determinant of A can be expressed as the sum of
determinants of two or more matrices of of the same order.
∣
∣
a  a'
For example
x
u
or
a
x  x'
u  u'
b  b'
y
v
b
y
v
∣∣
c  c'
a
=
z
x
w
u
∣∣
c
a
z = x
w
u
b
y
v
b
y
v
∣∣
c
a'
z  x'
w
u'
∣∣
c
a'

z
x
w
u
b
y
v
b'
y
v
c'
z
w
∣
∣
c
z
w
Property-6: Let A be any square matrix and B be a matrix obtained from A by adding to a
row(column) of A a scalar multiple of another row(column) of A , then ∣B∣=∣A∣
∣ ∣∣
∣ ∣∣
∣
a
For example x
u
b
y
v
c
a  ku
=
z
x
w
u
b  kv
y
v
a
x
u
b
y
v
c
a  kc
=
z
x  kz
w
u  kw
b
y
v
Example1. Prove that
1
1
a
b
bc ca
c  kw
z
w
∣
or
∣
∣
c
z
w
1
c =0
ab
5
∣
∣
1
1
1
Solution: Let = a
b
c
bc ca ab
Applying R2  R2 R3 we get
1
1
1
= abc abc abc
bc
ca
ab
Taking abc  common from R2 we get
1
1
1
=abc  1
1
1
bc ca ab
⇒=0
∣
∣
∣
∣
Example-2: Prove that
∣
 .̈ R1 and R 2 are identical
∣
a−b b−c c−a
b−c c−a a−b =0
c−a a−b b−c
∣
∣
a−b b−c c−a
= b−c c−a a−b
c−a a−b b−c
Applying R1  R1C 2R3
0
0
0
= b−c c−a a−b
c−a a−b b−c
⇒ =0
Solution: Let
∣
∣
( .̈ all elements in R1 are zero)
∣
∣
1
1
1
ab
bc
ca =0
c  ab a bc  b ca 
Example-3 Prove that
∣
∣
1
1
1
ab
bc
ca
c ab abc bca
Applying R2  R 2R3 we get
1
1
1
= abbcca abbcca abbcca
c ab 
a bc
b ca
Solution: Let =
∣
∣
∣
⇒ =abbcca
⇒=0
∣
1
1
1
1
1
1
c ab a bc bca
(As R1 and R2 are identical)
6
Example-4: Prove that
∣
∣
1 1 1
a b c =a−bb−c c−a 
2
2
2
a b c
∣
∣
1 1 1
Solution: Let = a b c
a2 b2 c2
Applying C 1  C 1 – C 2 and C 2  C 2−C 3 we get
0
0
1
= a−b
b−c c
2
2
2
2
2
a −b b −c c
Taking a – b and b – c common from C 1 and C 2 respectively we get
∣
∣
∣
∣
0
0
1
=a−b b−c  1
1
c
ab bc c2
Expanding along R1 we get
=a – b b – c  {bc  – ab }=a−b b−cc−a
∣ ∣
∣ ∣
1 1 1
Example-5 Prove that a b c =a – bb – c c – aabc 
a 3 b3 c3
1 1 1
Solution: Let = a b c
a3 b3 c3
Applying C 1 C 1−C 2 and C 2 C 2 −C 3 we get
∣
∣
0
0
1
= a−b
b−c c
3
3
3
3
3
a −b b −c c
∣
∣
0
0
1
⇒=a – bb – c
1
1
c
2
2
2
2
a abb b bcc c3
Applying C 1  C 1 – C 2 we get
0
0
1
=a−b b−c 
0
1
c
2
2
2
2
a ab−bc−c b bcc c 2
∣
∣
∣
∣
0
0
1
=a−b b−c 
0
1
c
2
2
a−c abc  b bcc c 2
∣
∣
0
0
1
=a−bb−ca−cabc  0
1
c
2
2
1 b bcc c 2
(Taking out a−c  abc from C 1 )
7
∣
∣
0
1
2
2
1 b bcc
=a – b b – c a−c abc×−1
=a−bb−cc−aabc 
=a−bb−ca−cabc 1⋅
∣
(Expanding along R3 )
∣
1a
1
1
1 1 1
Example 6: Prove that 1
1b
1 =abc  1   =abcbccaab
a b c
1
1
1c
∣


∣
1a
1
1
=
Solution: Let
1
1b
1
1
1
1c
Taking a , b , c common from R1, R2 and R3 we get
∣ ∣
∣
1
1
1
1
a
a
a
1
1
=abc 1
1
b
b
b
1
1
1
1
c
c
c
Applying R1  R 2R3 we get
1 1 1
1 1 1
1  
1  
a b c
a b c
1
1
=abc
1
b
b
1
1
c
c

1 1 1
1  
a b c
1
b
1
1
c
∣

1 1 1
Taking 1   common from R1 we get
a b c
∣ ∣
∣ ∣
1
1
1
1
1
1
1 1 1
1
=abc 1   b
b
b
a b c
1
1
1
1
c
c
c
Applying C 1  C 1 – C 2 and C 2  C 2 – C 3 we get


0



∣
0
1
1
b
1 1 1 –1 1
=abc 1  
a b c
1
0 – 1 1
c
Expanding along R1 we get
1 1 1
1 1 1
=abc 1   1⋅ – 1 1 =abc 1   =abcbccaab
a b c
0 –1
a b c
∣ 

8
∣
∣
1a 2 – b2
2 ab
– 2b
2
2
Example-7 Prove that
=1a 2b 23
2 ab
1 – a b
2a
2b
–2a
1 – a2 – b2
∣
∣
1a2 – b2
2 ab
– 2b
2
2
Solution: Let = 2 ab
1 – a b
2a
2b
– 2a
1 – a2 – b2
Applying C 1 C 1−bC 3 and C 2  C 2a C 3 we get
∣
∣
1a 2b 2
0
– 2b
2
2
=
0
1a b
2a
2
2
2
2
b 1a b  – a1a b  1 – a 2 – b 2
Taking 1a 2b2  common from both C 1 and C 2 we get
1 0
– 2b
=1a2 b 22 0 1
2a
b – a 1 – a2 – b2
Applying R3  R3 – bR 1aR2 we get
1 0
– 2b
=1a2 b 22 0 1
2a
0 0 1a2b2
Expanding along R3 we get
=1a2 b 22 1a 2b 2⋅ 1 0
0 1
2
2 3
⇒ =1a b 
∣
∣
∣
∣
∣ ∣
∣
∣
∣
∣
bc 2
a2
a2
3
Example 8: Prove
b2
 ca2
b 2 =2 abc  abc
c2
c2
ab2
bc2
a2
a2
Solution: Let = b 2
ca2
b2
c2
c2
ab2
Applying C 1  C 1 – C 3 and C 2 C 2 – C 3 we get
∣
∣
bc2 – a 2
0
a2
=
0
ca2 – b 2
b2
c 2 – ab2 c 2 – ab2  ab2
Taking abc  common from C 1 and C 3 we get
∣
∣
∣
bc – a
0
a2
=abc2
0
ca – b
b2
c – a – b c – a – b ab2
Applying R3  R3 –  R1R 2 we get
∣
bc – a
0
a2
=abc2
0
ca – b b 2
–2b
– 2a
2 ab
9
∣
∣
abac – a2
0
a2
0
bcab – b 2 b2
– 2 ab
– 2 ab
2 ab
Applying C 1  C 1C 3 and C 2 C 2C 3
abc 2
⇒ =
ab
abc 2
=
ab 
∣
2
2
C 1 a C 1 and C 2  b C 2 
∣
abac
a
a
2
b
bcab b 2
0
0
2 ab
∣
∣
abc 2
abbc
a2
=
2 ab⋅
(By Expanding along R3 )
2
ab
b
bcab
Taking a common from R1 and b common from R2 we get
a
=2 ab abc 2 bc
b
ca
2
⇒ =2ab abc {bcbac2 ac−ab }
⇒=2abc abc3
∣
∣
Example 9: Prove that
∣
1
Solution: Let = a
bc
C

C
–
Applying 1
1 C2
∣
∣
1 1 1
a b c =a – bb – cc – a
bc ca ab
1
b
ca
and
∣
∣
1
c = a – bb – cc – a
ab
C 2 C 2 – C 3 we get
0
0
1
= a – b
b –c
c
– c a – b – a b – c  ab
∣
∣
∣
0
0
1
⇒ =a – b b – c  1
1
c
– c – a ab
Expanding along R1 we get
1
=a – b b – c c – a1⋅ 1
–c –a
⇒=a – bb – cc – a 
∣
(Taking a – b common from C 1 and b – c from C 2 )
∣
∣
∣
y z
x
Example-10 Prove that
y
z x
z
z
∣
x
y =4 xyz
x y
∣
yz
x
x
y
z x
y
z
z
x y
Applying C 1  C 1−C 3 and C 2 C 2 −C 3 we get
yz −x
0
x
=
0
z x− y
y
z −x− y z− x− y x  y
Solution: Let =
∣
∣
10
Applying R3  R3 –  R1R 2 we get
yz −x
0
x
=
0
z x− y y
−2y
−2x
0
Expanding along R1 we get
= yz – x 2 xyx  2 y  z x – y 
⇒=2 xy  yz – xz x – y =4 xyz
∣
∣
∣
∣∣ ∣
bc ca
Example-11 Prove that qr r p
y z z x
ab
a b c
=2
pq
p q r
xy
x y z
∣
∣
bc ca ab
Solution: Let = qr r  p pq
yz z x x y
Applying C 1 C 2 C 3 we get
2 abc ca ab
= 2  pqr  r p pq
2  x yz  z x x y
∣
∣
∣
∣
abc ca ab
⇒=2 pqr r  p pq
x yz z x x y
Applying C 2 C 2 – C 1 and C 3 C 3−C 1 we get
abc −b −c
=2 pqr −q −r
x yz − y −z
Applying C 1 C 1C 2 C3 we get
a −b −c
=2 p −q −r
x − y −z
a b c
⇒= p q r
x y z
∣
∣
∣ ∣
∣ ∣
Example-12: Prove that
∣
∣
2
a 1
ab
ac
2
2
2
2
ab
b 1
bc =1a b c
ca
cb
c 21
∣
∣
a 21
ab
ac
2
Solution: Let = ab
b 1
bc
2
ca
cb
c 1
∣
2
2
2
∣
a a 1
a b
a c
1
2
2
2
⇒=
ab
bb 1
b c
abc
2
2
2
c a
c b
c c 1
( R1 
1
1
1
R1 , R2  R2 and R3  R3 )
a
b
c
11
∣
∣
a21
a2
a2
abc
⇒=
b2
b 21
b2
abc
c2
c2
c 21
∣
∣
a21
a2
a2
⇒= b 2
b2 1
b2
c2
c2
c 21
Applying R1  R1 R2R3 we get
∣
∣
1a2b2 1a2 b2 1a 2b 2
=
b2
b 21
b2
c2
c2
c 21
∣
∣
∣
1
1
1
2
2
=1a b  b 2 b 21
(By taking 1a2b2 coomon from R1 )
b2
c2
c2
c 21
Applying C 1  C 1 – C 2 and C 2 C 2 – C 3 we get
0
0
1
2
2
=1a b  – 1 1
b2
0 – 1 c 21
Expanding along R1 we get
=1a2 b 2 – 1 1 =1a 2b 2 
0 –1
∣
∣
∣
Example-13: Prove that
∣
a –b–c
2b
2c
2a
b– c–a
2c
∣
∣
2a
=abc3
2b
c –a–b
∣
a–b–c
2a
2a
2b
b–c–a
2b
2c
2c
c –a – b
Applying R1  R1R 2R3 we get
abc abc abc
= 2 b
b–c–a
2b
2c
2c
c–a–b
Solution: Let =
∣
∣
∣
∣
1
1
1
⇒=abc 2 b b – c – a
2b
2c
2c
c–a–b
Applying C 1  C 1 – C 2 and C 2 C 2 – C 3 we get
0
0
1
=abc abc –  abc
2b
0
abc
c−a−b
∣
∣
∣
∣
0 0
1
=abc3 1 – 1
2b
0 1 c–a–b
(Taking abc common from C 1 and C 2 )
12
∣ ∣
=abc3 1 – 1
0 1
⇒=abc3
(Expandibg along R1 )
∣
∣
sin  cos  cos 
Example 14: Prove that sin  cos  cos  =0
sin  cos  cos 
∣
∣
sin  cos  cos 
Solution: Let = sin  cos  cos 
sin  cos  cos  
∣
sin 
⇒= sin 
sin 
By applying
∣
cos  cos  cos  – sin sin 
cos  cos  cos  – sin sin 
cos  cos  cos  – sin sin 
C 3 C 3sin  C 1 – cos  C 2
∣
∣
sin  cos  0
⇒= sin  cos  0
sin  cos  0
⇒=0
∣
∣
a2
bc
acc 2
2 2 2
Example-15: Prove that a2ab
b2
ac =4 a b c
ab
b2 bc
c2
∣
a2
Solution: Let = a 2ab
ab
∣
bc
acc 2
b2
ac
2
b bc
c2
∣
∣
a
c
ac
⇒=a b c ab
b
a
b
bc
c
Applying R1  R1R 2R3 we get
2 ab 2 bc  2 ca
=abc ab
b
a
b
bc
c
∣
∣
∣
∣
∣
ab bc ca
=2 abc ab
b
a
b
bc
c
Applying R1  R1 – R 2 and we get
0
c
c
=2 abc ab
b
a
b
bc c
C
C
Applying 2
2 – C 3 we get
∣
13
∣
∣
0
0
c
=2 abc ab b – a a
b
b
c
=2 abc {c abb – b ab }=4 a b c
2
2
2
2 2
Example-16 If a , b , c are positive and unequal, show that the value of the determinant
a b c
= b c a
c a b
is negative.
Solution: Applying C 1  C 1C 2C 3 we get
abc b c
1 b c
= abc c a =abc  1 c a
abc a b
1 a b
∣ ∣
∣
∣
∣
0 b–c c–a
⇒=abc 0 c – a a – b
1
a
b
∣ ∣
∣
 Applying R 1  R 1−R2 and R 2  R2− R3
⇒=abc {b−c a−b− c−a2 }
⇒=abc ba – b 2 – cabc – c2 – a22 ca
⇒=abc – a 2 – b 2 – c 2abbcca=–  abca 2b2c 2 – ab – bc – ca
−1
2
2
2
⇒=
abc2 a 2 b 2 c – 2 ab – 2 bc – 2 ca
2
−1
⇒=
abc {a – b2b – c 2  c – a2 }
2
As a , b , c are positive and unequal therefore abc0 and a – b2b – c2c – a2 0
⇒0
∣
∣
a x a – x a – x
Example 17: Solve a – x ax a – x =0
a – x a – x ax
∣
∣
ax a – x a – x
Solution: Let = a – x ax a – x
a – x a – x a x
∣
∣
3a – x a – x a – x
Applying C 1  C 1C 2C 3 we get = 3 a – x a x a – x
3 a – x a – x ax
∣
∣
1 a– x a – x
⇒=3 a – x  1 a x a – x
1 a – x ax
Applying R2  R 2 – R1 and R3  R3 – R1 we get
1 a– x a – x
=3 a – x  0 2 x
0
0
0
2x
∣
∣
14
∣
∣
⇒=3 a – x ×1× 2 x 0
(Expanding along C 1 )
0 2x
⇒=3 a – x  4 x 2
Thus =0 ⇒3 a – x 4 x 2=0⇒ x=0,3 a
∣
2
x
Example 18: Prove that = y
z
∣
2
x
Solution: We have = y
z
x2
y2
z2
1
1
1  pxyz 1
1
1
1
⇒ = – 1
1
x2
y2
z2
1
⇒ = 1
1
x2
1
2

pxyz
y
1
2
z
1
x
y
z
1  px
3
1  py
1  pz 3
3
x
2
y
z2
1  px
3
1  py
1  pz 3
∣ ∣ ∣
∣ ∣ ∣
∣ ∣ ∣
∣ ∣
∣
x
⇒= y
z
3
x
2
y
z2
x
y
z
∣
∣
{As each element of C 3 is sum of two elements}
x
y
z
x2
y2
z2
x2
y2
z2
x
1
y  pxyz 1
z
1
∣
x2
y2
z2
x
y
z
∣
∣
x2
2
y
z2
1
⇒ =  1  pxyz  1
1
x
y
z
1
⇒ = 1  pxyz  0
0
x
y – x
z – x
2
x
2
2
y – x
z2 – x2
∣
1
⇒ = 1  pxyz  y – x z – x  0
0
∣
∣
x
1
1
(Applying R2  R 2 – R3 and R3  R3 – R1 )
x2
yx
zx
∣
∣
⇒=1 pxyz  y – x z – x 1 y x
(Expanding along C 1 )
1 z x
⇒=1 pxyz  y – z z – x  z x – y – x=1 pxyz  x – y  y – z  z – x 
∣∣
∣
∣
1 1 1
a b c
2
2
2
Example-19: Show that a b c = a2 b 2 c 2 =a – bb – c c – aabbcca
bc ca ab a3 b 3 c 3
∣
∣
a
b
c
2
2
=
Solution: Let
a
b
c2
bc ca ab
1
1
1
Applying C 1  C 1 , C 2  C 2 and C 3  C 3 we get
a
b
c
15
∣
a2
1
=
a3
abc
abc
b2
b3
abc
∣
∣∣
c2
c3
abc
∣
a2 b 2 c2
a2
abc 3
⇒ =
a b 3 c3 = a3
abc
1
1 1
1
Applying R1 ⇔ R3 we get
∣
b2
b3
1
c2
c3
1
∣
∣
1
1 1
 = – a3 b 3 c3
a2 b 2 c 2
Applying R2 ⇔ R3 we get
1
1 1
2
 = a b2 c2
a3 b 3 c 3
Applying C 1  C 1 – C 2 and C 2 C 2 – C 3 we get
0
0
1
0
 = a2 – b2 b 2 – c 2 c 2 =
a – ba  b
3
3
3
3
3
a –b
b –c
c
 a – ba 2  ab  b 2
∣
∣
∣
∣∣
∣
0
b – cb  c
b – cb 2  bc  c 2 
1
c2
c3
∣
∣
0
0
1
⇒ = a – b b – c
ab
bc
c2
a 2  ab  b 2 b2  bc  c 2 c 3
Applying C 2 C 2 – C 1 we get
0
0
1
 = a – bb – c 
ab
c– a
c2
a2ab  b 2 c – aa  b  c c3
∣
∣
∣
∣
0
0
1
⇒ = a – b b – cc – a
ab
1
c2
a2  ab  b2 a  b  c c3
Expanding along R1 we get
2
2
=a – b b – c c – a { ababc  – a abb  }
2
2
2
2
⇒=a – bb – cc – a  a abacabb bc – a – ab−b 
⇒=a – bb – cc – a  abbcca
∣
a
Example-20: Prove that 2 a
3a
∣
ab
3a  2 b
6a  3b
∣
a  b c
3
4 a  3 b  2 c =a
10 a  6 b  3 c
∣
a
ab
abc
Solution: Let = 2 a 3 a  2 b 4 a  3 b  2 c
3 a 6 a  3 b 10 a  6 b  3 c
Since each element of the second column is sum of two elements we get
16
∣
a
 = 2a
3a
a
3a
6a
∣
∣
a
⇒ = 2 a
3a
∣∣
∣
∣ ∣
∣
∣
∣∣ ∣∣ ∣
∣ ∣ ∣ ∣ ∣
ab c
a
4a  3b  2c  2 a
10 a  6 b  3 c
3a
a
3a
6a
b
2b
3b
a bc
1

ab⋅
4 a  3b  2c
2
10 a  6 b  3 c
3
a b  c
4a  3b  2c
10 a  6 b  3 c
1
2
3
ab c
4a  3b  2c
10 a  6 b  3 c
a
a
abc
⇒ = 2 a 3 a 4 a  3b  2 c ab ⋅0 [ .̈ C 2 and C 3 are identical in second determinant)
3 a 6 a 10 a  6 b  3 c
As each element of C 3 is sum of three elements therefore
a
a
a
a
a
b
a
a
c
= 2 a 3 a 4 a  2 a 3 a 3 b  2 a 3 a 2 c
3 a 6 a 10 a
3a 6 a 6b
3a 6a 3c
∣
∣
1
⇒=a 2
3
C
As 2 and
thus we get
3
∣
1 1
1 1 1
1 1 1
2
2
3 4 a b2 3 3 a c 2 3 2
6 10
3 6 6
3 6 3
C 3 are identical in second determinant and C1 and C3 are identical in third determinant
∣
∣ ∣
1
=a3 2
3
1
3
6
1
2
2
4  a b ⋅0 a c ⋅0
10
1 0 0
⇒=a 3 2 1 2 [Applying C 2  C 2 – C 1 and C 3 C 3 – C 1 ]
3 3 7
3
[Expanding along R1 ]
⇒=a 7 – 6
3
⇒ =a
Exercise 4.2
Using Properties of determinants, prove the following
x a xa
1. y b y  b = 0
z c z c
2.
3.
4.
∣
∣
∣
1
1
1
∣
0
–a
b
∣
∣
bc
ca
ab
–a
ba
ca
a b  c
b c  a = 0
c  a  b
a
0
c
2
∣
–b
– c =0
0
ab
2
– b
cb
∣
ac
bc =0
– c2
17
5.
6.
7.
8.
9.
∣
∣
b2 c2
c2 a2
a2 b 2
x
y
z
∣
bc
ca
ab
b c
ca =0
ab
∣
x2
2
y
2
z
yz
zx =  x – y y – z  z – x xy  yz  zx 
xy
∣
∣
∣
∣
x4
2x
2x
2x
x4
2x
2x
= 5 x  4  4 – x 2
2x
2x  4
yk
y
y
y
yk
y
y
= k 2 3y  k 
y
yk
∣
1
1
1
∣
a 2 – bc
b 2 – ca = 0
c 2 – ab
a
b
c
∣
∣
∣
bc
10. b
c
a –b–c
11. 2 b
2c
∣
∣
x
y  z 2x
x
y
= 2  x  y  z 3
y
zx2y
∣
∣
∣
bc
15. c  a
ab
2
∣
x
4x
8x
a–b
b–c
c –a
 y  z
16. xy
xz
b 2  c2
17. ba
ca
∣
2a
= a  b  c 3
2b
c –a–b
x2
2 3
x = 1 – x 
1
x
1
x2
xy
14. 5 x  4 y
10 x  8 y
∣
∣
a
= 4 abc
b
a b
2a
b– c–a
2c
x  y  2z
12. z
z
1
13. x 2
x
∣
a
c a
c
x
3
2x = x
3x
∣
a
3
3
3
b = 3 abc – a – b – c
c
xy
2
 x  z
yz
ab
c2  a2
cb
∣
zx
= 2 xyz  x  y  z 3
yz
2
 x  y
∣
ac
= 4 a2 b 2 c 2
bc
a2  b 2
18
∣
∣
3a
18. – b  a
– ca
1
19. 2
3
– ab
3b
– cb
∣
1 p
32p
63p
20. Prove that
21. Evaluate
∣
– a c
– b  c = 3 a  b  c ab  bc  ca
3c
1 pq
4  3 p  2q = 1
10  6 p  3 q
∣
x
– sin 
cos 
∣
cos  cos 
– sin 
sin  cos 
sin 
– x
1
∣
∣
∣
x –2
24. Solve x – 4
x –8
∣
cos  sin 
cos 
sin sin 
x a
22. Solve the equation x
x
x
x≠
y
≠z
23. If
and y
z
cos 
1
x
x2
2
y
z2
2x – 3
2x – 9
2 x – 27
is independent of 
– sin 
0
cos 
∣
∣
x
xa
x
x
= 0 , a≠0
x
xa
∣
1  x3
3
1  y = 0 , then prove that xyz=– 1
1  z3
∣
∣
3x – 4
3 x – 16 =0
3 x – 64
2y  4
25. If a , b , c are in A.P, find value of 3 y  5
4y6
5y 7
6y 8
7y 9
∣
8ya
9yb
10 y  c
Area of Triangle
Result : Area of triangle whose vertices are  x 1, y1, z 1 , x 2, y 2, z 2 ; x 3, y3, z 3 is given by
x
y1 1
1 1
 = ± x2 y 2 1
2
x3 y3 1
Collinearity of three points: Three points  x 1, y1, z 1 , x 2, y 2, z 2 ; x 3, y3, z 3 will be collinear if and only
∣
∣
∣
∣
x1 y 1 1
x
y2 1 = 0
if
2
x 3 y3 1
Example-1: Find the area of the triangle whose vertices are 1,2 ,2 ;3 , 3; – 1
1 1 2 1
1
=±
Solution:
2 3 1 =± {131−22−31−2−9 }
2
2
3 −1 1
1
5
⇒=± 42 – 11= square units
2
2
∣ ∣
Example-2: Show that the points 1,2; 4,−1; 5.−2 are collinear.
19
∣
1
Solution: 4
5
2
–1
–2
∣
1
1 =1  – 12 – 2 4 – 51 – 85=12 – 3=0
1
Example-3: If points a , 0 ,0, b  , x , y  are collinear then show that
x y
 =1
a b
∣ ∣
a
a
,
0
,0,
b

,
x
,
y

Solution: As points
are collinear 0
x
⇒ ab – y  – bx=0 ⇒ ab – ay – bx=0 ⇒ aybx =ab
Divide both sides by ab we get
x y
 =1
a b
0
b
y
1
1 =0
1
Exercise 4.3
1. Find the area of the triangle whose vertices are
(i) 2,3 , 4, – 1 ,0,2 (ii) 3, – 1 , 2,4 ,9,2 (iii) 4, – 1 , 4,1 ,7,3
2. Show that the points 3,1 , 4,0 ,6, – 2 are collinear.
3. Show that the points 3,1 ,2, – 1 ,1, – 3 are collinear.
4. If the points a , 0 ,0, b  ,1,1 are collinear then show ab=ab
5. Using determinants prove that the points a , b , a ' , b '  and a – a ' , b – b '  are collinear if
ab ' =ba '
6. Find the value of  such that 2,3 , ,1 , – 1,2 are collinear.
7. Using determinants find the equation of the line joining the points 2,3 and  – 3,1
Minors and Co-Factors
Minors: Let A=[ aij ] be a square matrix of order n . Then we define minor of a ij as a determinant of
order n – 1 obtained from A be deleting its i th row and jth column. Minor of a ij is denoted by M ij
Co-Factor: Let A=[ aij ] be a square matrix of order n . Then we define co-factor of a ij as  –i j M ij
. We denote the co-factor of a ij as Aij or C ij
Thus Aij = – 1i j M ij
M ij if i j is even
or Aij =
−M ij if i j is odd
Result: If A=[ aij ] be a square matrix of order n
{
n
∣A∣=∑ aij Aij (Expansion along i th row)
j=1
n
∣A∣=∑ aij C ij (Expanding along j th column)
j=1
To be more specific let A=[ aij ] be a square matrix of order 3.
Then ∣A∣=a11 A11a12 A12a13 A13 (Expansion along first column)
∣A∣=a21 A21 a 22 A22a 23 A23
(Expanding along second row)
∣A∣=a31 A31a32 A32 a 33 A33
(Expanding along third row)
∣A∣=a11 A11 a21 A21a31 C31
(Expansion along first column)
∣A∣=a12 A12a 22 A22 a32 A32
(Expanding along second column)
∣A∣=a13 A13 a23 A23 a 33 A33
(Expanding along third column)
Note: If we multiply the elements of a row(column) by the co-factors of elements of different
row(column) and then add the value of such expression is zero.
20
For example for a square matrix of order 3
a 11 A21 a 12 A 22a13 A23=0
[
2 3 1
Example-1: Find the minor and co-factor of each element of the matrix – 1 4 2
2 5 1
∣
∣
∣
∣
∣
∣
∣
∣
∣
]
∣
M 11= 4 2 = 4−10=−6 ; A11 =– 6
5 1
M 12 = −1 2 =−1−4=−5, A12 =5
2 1
M 13= −1 4 =−5−8=−13, A13 =−13
2 5
M 21 = 3 1 =3−5=−2, A 21=2
5 1
2 1
M 22 =
= 2−2=0, A22=0
2 1
M 23 = 2 3 =10−9=1, A23 =−1
3 5
M 31= 3 1 =6−4=2, A31 =2
4 2
2 1
M 32=
=41=5, A32 =−5
−1 2
M 33= 2 3 =83=11, A33=11
−1 4
2 3 4
Example-2: Find the determinant of the matrix – 1 2 3 and verify the relation
3 2 5
∣A∣=a11 A11a12 A12a13 A13
Solution:
∣
∣
∣
∣
∣
∣
∣
∣
[
]
∣ ∣ ∣ ∣ ∣ ∣
∣ ∣
∣ ∣
∣ ∣
2 3 – 3 – 1 3  4 – 1 2 =210 – 6 – 3 – 5 – 94 – 2 – 6=842 – 32=18
Solution: ∣A∣=2
2 5
3 5
3 2
A11 =−111 2 3 =10−6=4 , A12 =−112 −1 3 =−−5−9=14
2 5
3 5
13 −1
2
A13 =−1
=−2−6=−8
3 2
Thus a11 A11a12 A12 a 13 A13=2 43144−8=842−32=18
Thus ∣A∣=a11 A11a12 A12a13 A13
[
]
2 3 1
1 2 4 , verify that a 21 A11 a 22 A12a23 A13=0
–1 2 1
2 4 =2 – 8=– 6
2 1
Example-3: For the matrix
∣ ∣
A11 =−111
Solution:
12
A12 =−1
∣ ∣
1 4
=−14=−5
−1 2
21
∣ ∣
1 2
=22=4
−1 2
Thus a 21 A11 a22 A12a23 A13 =1−62−54 4=−6−1016=0
13
A13 =−1
Exercise-4.4
∣ ∣
2 3
1 4
2 −1 3
2. Write the minors and cofactors of each element of the determinant −2 4 −1
5
1
2
nd
3. Using co-factors of the elements of the 2 column find the value fo the determinant
2 –1 0
3 2 –1
2 1
4
1. Write the minors and co-factors of each element the determinant
∣
∣
∣
∣
∣
∣
3 –1 1
4. Using co-factors of elements of 3rd row find the value of the determinant – 1 0 2
3
5 1
5. For the determinant
∣
∣
∣
∣
2 −1 3
4 1 5
1 0 2
show that a 11 A21 a 12 A 22a13 A23=0
–1 2 5
6. For the determinant 1 2 – 1 , show that a 13 A12 a 23 A 22a33 A23 =0
2 5 7
Adjoint of a Matrix
Definition: : The adjoint of a square matrix A=[ aij ]n×n is defined as the transpose of the matrix
[ Aij ]n× n , where Aij is the co-factor of the element a ij . Adjoint of the matrix A is denoted by
adj A
[ ]
2 3
4 1
Solution: A11 =1, AA12 =−4, A21 =−3, A22=2
1 –4
1 –3
=
Hence adj A=transposeof
–3 2
–4 2
a b adj A= d – b
Note: For a 2×2 matrix A=
c d
–c a
2×2
Thus the rule to write the adjoint of
matrix is that
(i) interchange the position of diagonal elements and
(ii) change the sign of the non-diagonal elements
Result: ∣adj A∣=∣A∣n−1
Result: A adj A=∣A∣I
Example-1: Find the adjoint of the matrix
[
][
[ ]
]
[
]
22
[
]
2 1 3
Example-2: Find the adjoint of the matrix – 1 4 1
2 5 1
Solution: A11 =−1, A12 =3, A13 =−13 ; A21 =14, A22 =−4, A23 =−8 ;
– 1 14 – 11
adj A= 3
– 4 –5
– 13 – 8
9
[
A31 =−11, A32 =−5, A33=9
]
[
]
2 3 –1
Example 3: Find the adjoint of the matrix A= 2 1 – 1 and verify that A adjA=∣A∣I . Also
4 1 2
n−1
verify that ∣adj A∣=∣A∣
Solution: A11 =3, A12=−8, A13=−2 ; A21 =−7, A22=8, A23 =10 ; A31=−2, A32 =0, A33 =−4
[
]
][
3 –7 – 2
Thus adj A= – 8 8
0
– 2 10 – 4
and ∣A∣=233−8−−2=6−242=−16
2 3 −1 3 −7 −2
– 16
0
0
Now A adj A= 2 1 −1 −8 8
0 = 0
– 16
0 =– 16 I =∣A∣I
4 1 2 −2 10 −4
0
0
– 16
Now
3 −7 −2
2
∣adj A∣= −8 8
0 =3−32−0−−732−0−2 −8016=−96224128=256=−16
−2 10 −4
Thus ∣adj A∣=∣A∣2=∣A∣3−1=∣A∣n−1 (Here n=3 )
[
∣
][
]
∣
Exercise 4.5
2
1. Find the adjoint of the matrix
–1
3
2. Find the adjoint of the matrix – 3
1
[
[
3. Find the adjoint of the matrix
]
[
3
and verify the result ∣adj A∣=∣A∣n−1
–3
[
–1 0
n−1
3 1 and verify the result ∣adj A∣=∣A∣
1 2
2
1
3
4. Find the adjoint of the matrix 2
4
[ ]
]
3
and verify the result A adj A=∣A∣I
5
–1 2
1 2 and verify the result A adj A=∣A∣I
3 1
]
2 3 1
A=
5. For the matrix
1 2 4 show that
3 5 5
]
A adj A=0
Inverse of a Matrix
Singular Matrix: A square matrix A is said to be singular if ∣A∣=0
Non-Singular Matrix: A square matrix A is said to be non-singular if ∣A∣≠0
Result: If A and B are nonsingular matrices of the same order, then AB and BA are also non
23
singular matrices of the same order.
Result: ∣AB∣=∣A∣∣B∣ where A and B are square matrices of the same order.
Invertible Matrix: A square matrix A is invertible if and only if A is nonsingular.
1
A – 1= adj  A
and
∣A∣
2 3
Example-1 Show that the matrix
is invertible and find its inverse.
1 3
2 3
Solution: Let A=
. Then
1 3
∣A∣=6−3=3≠0 ⇒ A is nonsingular. Thus A is invertible.
A11 =3, A12 =– 1, A21=– 3, A22 =2 ⇒adj A= 3 −3
−1 2
1 −1
1
1 3 −3
−1
= 1 2
Thus A = adj A=
∣A∣
−
3 −1 2
3 3
[ ]
[ ]
[
][
[
[
∣
2
Example-2: Show that A= 1
2
∣
[
3
4
–1
]
]
]
–1
2 is invertible and find A – 1
3
2 3 −1
∣
∣
A
=
Solution:
1 4
2 =2 122 – 33 – 4 – 1 – 1 – 8=2839=40
2 −1 3
A11 =14, A12 =1, A13 =– 9, A 21=– 8, A22 =8, A23 =8, A31 =10, A32 =−5, A33 =5
14 −8 10
8 −5
Thus adj A= 1
−9 8
5
1
−1
..˙ A = adj  A
∣A∣
]
[ ]
7
20
1 14 −8 10
−1
1
⇒A =
1
8 −5 =
40
40
−9 8
5
9
–
40
[
]
[
]
[
]
3
1 –2
and B=
, then verify that  AB −1=B−1 A−1
–4
–1 3
2 3
1 – 2 = –1
5
Solution: We have AB=
1 – 4 –1 3
5 – 14
−1
∣
∣
AB
=14−25=−11≠0
Since
,  AB  exists and is given by
1
1 – 14 – 5
1 14 5
−1
 AB =
adj AB=−
=
∣AB∣
11 – 5 – 1 11 5 1
Further ∣A∣=−11≠0 and ∣B∣=1≠0 . Therefore, A−1 and B – 1 both exist and are given by
1 −4 −3
−1
−1
3 2
A =−
,B =
11 −1 2
1 1
Example-3:: If A=
2
1
1 1
5 4
1 −1
5
8
1
1
5
8
–
[
][
[
[
]
][
]
] [ ]
[ ]
24
B−1 A−1=−
Therefore
−1
−1
[ ][
] [ ]
1 3 2 −4 −3
1 14 5
=
11 1 1 −1 2
11 5 1
−1
Hence  AB =B A
[ ]
[ ][ ] [
[ ][
2 3
then show that A2 – 7 A – 2 I =0 and hence find A – 1 .
4 5
2
2 3 2 3 = 16 21 7A=7 2 3 = 14 21
2 0
Solution: A =
,
, 2I=
4 5 4 5 28 37
4 5
28 35
0 2
2
16 21
14 21 2 0
0 0
−
−
=
Thus A −7A−2I=
28 37
28 35 0 2
0 0
2
Thus A – 7 A – 2 I =0
Pre-Multiply by A−1 we get
–1 2
–1
–1
–1
A A – 7 A A – 2 A A=0 ⇒ A – 7 I – 2 A =0
−1
2 3
7 0
–5 3
⇒ 2A =A−7I=
−
=
4 5
0 7
4 –2
1
⇒ A−1= – 5 3
2 4 –2
5 3
–
−1
⇒A = 2 2
2 –1
Example-4: If A=
] [ ][
][ ][ ]
[ ][ ][
]
[
]
[ ]
]
[ ]
[ ] [ ]
[ ][ ] [ ]
2 1 1
3 –1
A=
B=
Example-5: If
1 2 1 and
–1 3
1 1 2
– 1 –1
–1
– 1 then find AB and hence find
3
A−1
2 1 1 3 –1 –1 4 0 0
Solution: AB= 1 2 1 – 1 3 – 1 = 0 4 0
1 1 2 –1 –1 3
0 0 4
Thus AB=4 I
1
–1
Pre-Multiply by A – 1 AB=4 A – 1 I ⇒ IB=4 A– 1 ⇒ B=4 A– 1 ⇒ A = B
4
[ ]
3
4
−1
1
Thus A = –
4
1
–
4
1
4
3
4
1
–
4
–
1
4
1
–
4
3
4
–
Example:6 Find the inverse of the matrix
[
cos  −sin 
Solution: Let A= sin  cos 
0
0
2
2
∣A∣=cos sin =1
A11 =cos  , A12 =−sin  , A13 =0 ;
[
cos −sin  0
sin  cos  0
0
0
1
]
]
0
0
1
(By expanding along third row)
A21 =sin  , A22 =cos  , A23 =0 ;
A31 =0, A32 =0, A33 =1
25
[
cos  sin 
adj A= −sin  cos 
0
0
1
−1
Thus A = adj A
∣A∣
cos  sin 
⇒ A−1= −sin  cos 
0
0
[
[
0
0
1
0
0
1
]
]
]
1
−1
2 4
then show that A = A
16
3 –2
∣
∣
Solution: We have A =−4−12=−16≠0 Thus A−1 exists.
1
1 −2 −4
1 2 4
1
A−1= adj  A=−
=
= A
∣A∣
16 −3 2
16 3 −2 16
1
tan x
T −1
cos 2 x −sin 2 x
Example -8: If A=
then show that A A =
sin 2 x cos 2 x
−tan x
1
2
−1
Solution: ∣A∣=1tan x≠0 , thus A exists and is given by
1
1
1
−tan x
A−1= adj A=
2
∣A∣
1
1tan x tan x
−tan x
AT = 1
tan x
1
Example: -7: If A=
[
] [
]
[
]
[
[
[
Thus
]
]
AT A=
[
][
[
]
][
[
]
2 3
Example-9: For the matrix A=
4 –1
–1
Hence find A
2 3
Solution: We have A=
4 −1
2
2 3 2 3 = 16
Therefore A =
4 −1 4 −1
4
2
As A aAbI =0
16 3  2 a 3 a  b 0
Therefore
4 13
4 a −a 0 b
[
][
[
[
∣
∣
1
1
1−tan 2 x −2tan x
1
−tan x
1
−tan x =
1
tan x
1
1−tan2 x
1tan2 x tan x
1tan2 x 2tan x
1−tan 2 x
2tan x
−
2
1tan x
1tan 2 x
T −1
⇒A A =
= cos 2 x −sin 2 x
2
sin 2 x cos 2 x
2 tan x
1−tan x
2
2
1tan x
1tan x
[
]
][
]
][
]
, find the values of a ad b such that
3
13
2
A aAbI =0 .
]
][ ][ ]
][ ]
=0 0
0 0
33 a = 0 0
⇒ 162 ab
44 a
13−ab 0 0
⇒162 ab=0, 33 a=0, 44 a=0, 13−ab=0
⇒ a=−1 and b=−14
Thus A2a Ab I =0 ⇒ A2− A−14I=0
26
Pre-Multiply by A−1 we get
A−1 A2−A−1 A−14 A−1 I =0
⇒ A−I −14A−1=0
⇒14A −1= A−I
⇒ A−1=
{[
] [ ]}
[
][ ]
1 2 3
−1 0
14 4 −1
0 1
[ ][ ]
1
3
1
3
14
14
14
14
⇒A =
=
4
2
2
1
−
−
14
14
7
7
−1
Example-10: Find a 2×2 matrix
[
]
B such that
B
1 −2
6 0
=
1 4
0 6
[ ]
6 0
A= 1 −2 and C=
1 4
0 6
−1
∣
∣
Then, A =42=6≠0 . Thus A exits and is given by
1
1 4 2
A−1= adj A=
∣A∣
6 −1 1
1 −2
6 0
=
⇒ B A=C
Now B
1 4
0 6
Post-Multiply both side by A−1 we get B A A−1=C A−1 or B I =C A−1 or
1
1
B=C A−1= 6 0 4 2 = 24 12
6 0 6 −1 1 6 −6 6
4 2
Thus B=
−1 1
2 3
1 –2
2 1
A
=
Example:11 Find a matrix A such that
1 2
3 1
1 3
2 3 ,C = 1 – 2 ,D = 2 1
Solution: Let B =
1 2
3
1
1 3
Thus we have BAC=D
Pre-Multiply by B – 1 and post multiply by C – 1 we get
–1
–1
–1
–1
B BAC C =B DC
⇒ IAI=B– 1 DC – 1
–1
–1
⇒ A=B D C
∣B∣=1,∣C∣=7
1
–1
–3 = 2
–3
adj B = 2
Thus B =
∣B∣
–1
2
–1
2
1
1 1 2
–1
and C = adjC=
∣C∣
7 –3 1
1 2
–3 2 1
1
2
Thus A =
7 –1
2 1 3 –3 1
1 1 –7
1 2
Thus A=
7 0 5
–3 1
Solution: Let
[
[
]
][ ]
[ ][
[ ]
] [
[ ] [
[
[
]
[ ][ ][ ]
] [ ]
[
] [
[ ]
][ ][ ]
][ ]
]
27
[
]
1 22 – 5
=
Thus A=
7 – 15 5
[ ]
22
7
15
–
7
5
7
5
7
–
Exercise 4.6
[ ]
2 3 4
1. Show that 1 2 1 is non singular.
1 1 4
3 1 2
2. Show that the matrix 2 1 3 is singular
5 2 5
[ ]
[
]
[ ]
a– b b–c c–a
3. Show that the matrix b – c c – a a – b is singular.
c–a a –b b–c
2  3
4. Find the value of  such that the matrix 1 2 3 is singular.
0 1 2
5. Find the inverse of the matrix
6. Find the inverse of the matrix
7. Find the inverse of the matrix
8. Find the inverse of the matrix
9. Find the inverse of the matrix
[ ]
[ ]
[
]
[ ]
2 3
1 –2
3 –1
–2 4
cos  – sin 
sin  cos 
a b
c d
2 3 –1
1 2 5
2 0 –1
[
[
]
]
3 –1 – 2
10. Find the inverse of the matrix 2 1
3
4 1
2
2 3
4 1
11. If A=
and B=
then verify that AB – 1=B – 1 A – 1
4 –1
2 –1
[
12. If A =
[
[
]
1
2
–1
2
13. If A =
4
[
]
2
1
2
]
[
2
0
B
=
and
3
2
1
2
–1
1
–1
]
2
1 then verify that AB – 1=B – 1 A – 1
1
]
3
then prove that A2 – A – 14I=0 and hence find A^{-1}
–1
28
14. If A =
[ ]
2
3
1
then find a and b such that A 2aAbI=0 and hence find A – 1
2
[
1
0
0
15. Find the inverse of the matrix 0 cos  sin 
0 sin  – cos 
[
[
]
]
2 –1 1
16. For the matrix A= – 1 2 – 1 , show that A3 – 6A29A – 4 I=0 . Hence find A– 1
1 –1 2
]
1 1
1
17. For the matrix A= 1 2 – 3 show that A 3 – 6A25A11 I=0
2 –1 3
[
]
2 3
18. Show that the matrix A=
is zero of the polynomial f  x=x2 – 6x11 and hence find
–1 4
A– 1
3 2 X= 1 −2
19. Find the matrix X for which
7 5
1 3
3 –1
1 2
=
20. Find the matrix X for which X
2
4
3 4
3 1 A –1 2 = 1 3
21. Find the matrix A for which
2 2
3
4
−1 0
a
b
A
=
22. Find the inverse of the matrix
1  bc and show that aA – 1= A 2bc1I – aA
c
a
[ ] [ ]
[ ] [ ]
[ ][ ][ ]
[
[
]
]
2
–3
, compute A– 1 and show that 2A – 1=9I – A
–4
7
4 5
24. If A =
, then show that A−3I=2I 3A – 1
2 1
5 0 4
1 3 3
–1
25. Given A = 2 3 2 , B = 1 4 3 . Compute AB – 1
1 2 1
1 3 4
23. Given A =
[ ]
[
26. Show that
27. If A =
[
] [
1
– tan

tan
2
[
1 –8
4
9
1
[
[
3
28. If A = 2
0
–1
29. If A = – 1
0

2
1
1
4
–8
][
]
1
tan

– tan
2
1

2
–1
]
[
= cos 
sin 
– sin 
cos 
]
]
4
7 , prove that A– 1=A T
4
]
]
–3
–3
–1
4
4 , prove that A – 1=A 3
1
2
1
1
0
1 , show that A2=A – 1
0
29
System of Linear Equations
a1 xb1 y=c1
Consider a system of linear equations
.
a2 xb2 y=c 2
{
[ ][ ] [ ]
a1 b1 x
c
= 1
a2 b2 y
c2
Using matrix notations this can be written as
[]
or AX=B where A=
[ ]
a 1 b1
,
a 2 b2
[]
c
x
and B= 1
c2
y
If ∣A∣≠0 then A – 1 exists
Pre-Multiplying AX=B by A– 1 we get A– 1 AX=A – 1 B ⇒ I X=A – 1 B
–1
⇒ X=A B
X=
Consider the system of equations
This can be written as
or
A X =B where
{
a1 xb 1 yc1 z=d 1
a 2 x b 2 yc 2 z =d 2
a3 xb 3 yc3 z=d 3
[ ][ ] [ ]
[ ] []
a1 b1 c 1 x
d1
a2 b2 c 2 y = d 2
a3 b3 c 3 z
d3
a1 b1 c 1
A= a 2 b 2 c 2
a 3 b3 c 3
If ∣A∣≠0 then A−1 exists.
Pre-Multiplying A X =B by
,
x
X= y
z
and
A−1 we get A−1 A X = A−1 B ⇒ I X = A−1 B ⇒ X = A−1 B
Note: The system of equations A X =B will have
(i) unique solution if ∣A∣≠0 and the solution is given by
(ii) infinitely many solutions if adj A B=0
(iii) no solution if adj A B≠0
Example-1: Solve the system of linear equations
[
[
][ ] [ ]
2 4 x = 10
3 −1 y
8
] [] [ ]
A= 2 4 , X = x , B= 10
3 −1
y
8
∣ ∣
2 4 =−2−12=−14≠0
Now ∣A∣=
, thus
3 −1
A11 =−1, A12 =−3, A21=−4 , A22 =2
−1 −4
Thus adj A=
−3 2
1
1 −1 −4
−1
..˙ A = adj  A=−
∣A∣
14 −3 2
[
X = A−1 B
y=10
{23x4
x− y =8
Solution: The given system of equations can be written as
or A X =B where
[]
d1
B= d 2
d3
A−1 exists.
]
[
]
30
[
][ ]
[ ] [] [] []
1 −1 −4 10
14 −3 2 8
1 −42
3
x
3
=
⇒ =
Thus X =−
14 −14
1
y 1
Hence solution of the system of equations is x=3, y=1
Now
X = A−1 B
⇒ X =−
{
3 x−2 y z=3
x2 y −z=5
x− y z=1
Solution: The above system of equations can be written as A X =B where
3 −2 1
x
3
A= 1 2 −1 , X = y , B= 5
1 −1 1
z
1
Example:2 Solve the system of equations
[
∣
] [] []
∣
3 −2 1
∣A∣= 1 2 −1 =3 2−1−−2111−1−2=4≠0 . Thus A−1 exists.
1 −1 1
Now
A11 =1, A12 =−2, A13 =−3, A21=1, A22 =2, A23=1, A31 =0, A32 =4, A33 =8
1 1 0
Thus adj A= −2 2 4
−3 1 8
1
−1
Using A = adj  A we get
∣A∣
1 1 0
1
A−1= −2 2 4
4
−3 1 8
Using X = A−1 B we get
x
2
2
1 1 1 0 3 1 8
X = −2 2 4 5 = 8 = 2
⇒ y=2
4
4
−3 1 8 1
4
1
z
1
Thus solution of the given system of equations is x=2, y=2, z=1
1 2 1
Example-3: If A= 1 0 3 then find A−1 and hence solve the system of equations
2 −3 0
x2 yz =7, x3 z =11, 2x−3y=1
1 2 1
∣
∣
A
=
Solution:
1 0 3 =109−2 0−61−3−0=912−3=18≠0
2 −3 0
−1
Thus A exists.
A11 =9, A12 =6, A13 =−3 ; A21 =−3, A22 =−2, A23 =7, A31 =6, A32=−2, A33=−2
[
]
[
]
[ ][ ] [ ] [ ] [ ] [ ]
[ ]
∣ ∣
[
9 −3 6
Thus adj A= 6 −2 −2
−3 7 −2
]
31
[
9 −3 6
1
1
⇒ A = adj  A=
6 −2 −2
∣A∣
18
−3 7 −2
−1
]
Now given system of equations can be expressed as
[
] [] [ ]
][ ] [ ] [ ]
[
1 2 1
x
7
A= 1 0 3 , X = y and B= 11
1
2 −3 0
z
9 −3 6 7
36
2
1
1
−1
..˙ X = A B=
6 −2 −2 11 =
18 = 1
18
18
−3 7 −2 2
54
3
A X =B where
[
][ ] [ ]
1 2 1 x
7
1 0 3 y = 11
2 −3 0 z
1
or
.
[][]
x
2
..˙ y = 1
z
3
Hence solution of the system of equations is
[
x=2, y=1, z =3
]
1 −1 1
A=
2 1 −3 , find A−1 and hence solve the system of equations
Example-5: If
1 1
1
x2 y z=4,−x yz=0, x−3yz=2
1 −1 1
A=
Solution:
2 1 −3
1 1
1
[
∣
]
∣
1 −1 1
..˙ ∣A∣= 2 1 −3 =1 1312312−1=10≠0
1 1
1
So, A is invertible.
A11 =5, A12 =−5, A13=1, A21 =2, A22=0, A23=−2, A31 =2, A32 =5, A33 =3
4
2 2
..˙ adj A= −5 0 5
1 −2 3
4
2 2
1
1
⇒ A−1= adj  A=
−5 0 5
∣A∣
10
1 −2 3
1
2 1 x
4
Now the given system of equations is expressible as −1 1 1 y = 0
1 −3 1 z
2
x
4
T
or, A X =B where X = y , B= 0
z
2
T
Now, ∣A ∣=∣A∣=10≠0 . So, the system of equations is consistent with a unique solution.
X = AT −1 B= A−1T B
[
]
[
]
[
][ ] [ ]
[] []
32
[] [
x
2 2
1 4
y=
−5 0 5
10
z
1 −2 3
T
][ ]
4
0
2
[]
9
5
x
4
−5
1
4
18
1
1
2
⇒ y=
2 0 −2 0 =
4 =
10
10
5
z
2 5
3 2
14
7
5
9
2
7
Hence x= y= z = is the solution of the given system of linear equations.
5,
5,
5
2 1 1
3 –1 –1
Example-6: Find the product of the matrices 1 2 1 and – 1 3 – 1 and hence solve the
1 1 2
–1 –1 3
system of equations 2 x y z=4, x2 yz =4, x y2 z=4
2 1 1
3 −1 −1
Solution:Let A= 1 2 1 and B= −1 −3 −1
1 1 2
−1 −1 3
[] [
][ ] [ ]
[ ] [
[ ] [ ]
[ ][ ] [ ]
]
2 1 1 3 −1 −1
4 0 0
A B= 1 2 1 −1 −3 −1 = 0 4 0 =4 I
1 1 2 −1 −1 3
0 0 4
1
−1
−1
−1
−1
−1
A−1 we get A A B=4 A I ⇒ I B=4 A ⇒ 4A =B ⇒ A = B
4
3 −1 −1
1
−1
Thus A = −1 3 −1
4
−1 −1 3
2 1 1 x
4
=
Given system of equations can be written as 1 2 1 y
4 or A X =C where
1 1 2 z
4
Pre-Multiplying by
[
]
[ ][ ] [ ]
[] []
[
[] []
x
4
X = y ,C = 4
z
4
][ ] [ ]
4
3 −1 −1
4
1
1
..˙ X = A−1 C= −1 3 −1 4 = 4
4
4 4
−1 −1 3
4
4
x
1
⇒ y=1
z
1
Hence the solution of the given system of equations is x=1, y=1, z=1
Example-7: Solve the system of equations 2 x− y z=4, x2 y−z =3,3 x y=7 if consistent.
2 −1 1 x
4
=
3
Solution: The given system of equations can be expressed as 1 2 −1 y
3 1
0 z
7
[
][ ] [ ]
33
[
] [] []
2 −1 1
x
4
or A X =B where A= 1 2 −1 , X = y , B= 3
3 1
0
z
7
∣A∣=2011 031 1−6=23−5 =0 ⇒ A−1 does not exist.
Thus system of equations does not have unique solution.
A11 =1, A12 =−3, A13=−5, A21 =1, A22 =−3, A23=−5, A31=−1, A32 =3, A33=5
1
1 −1
Thus adj A= −3 −3 3
−5 −5 5
[
]
][ ] [ ]
[
1
1 −1 4
0
⇒adj A B= −3 −3 3 3 = 0
−5 −5 5 7
0
Thus system of equations has infinitely many solutions.
To find the solutions let us put z=k in the first two equations. Thus we get
2 x− yk =4, x2 y−k =3
or 2 x− y=4−k , x2y=3k This can be written as
2 −1 x
4−k
=
1 2 y
3k
[
][ ] [ ]
[][ ][ ]
[ ] [ ][ ]
[] [ ]
−1
x
2 −1
=
y
1 2
x 1 2 1
⇒ =
y 5 −1 2
x 1 11−k
⇒ =
y 5 23k
⇒
4−k
3k
4−k
3k
[ ]
11−k
⇒ x = 5
y
23 k
5
[]
Thus solution of the given system of equations is
x=
11−k
23 k
, y=
z=k where k is any real
5
5,
number.
Example-8: Show that the system of equations 2 x y z=4 , x y z=1, 3 x2 y2 z =4 is
inconsistent.
2 1 1 x
4
Solution: The given system of equations can be written as 1 1 1 y = 1
3 2 2 z
4
2 1 1
x
4
or A X =B where A= 1 1 1 , X = y , B= 1
3 2 2
z
4
∣A∣=22−2−1 2−31 2−3=0 . Thus A is singular matrix. Thus A−1 does not exist.
Now A11 =0, A12=1, A13 =−1, A21=0, A22 =1, A23 =−1, A31 =0, A32 =−1, A33=1
0
0
0
adj
A=
1
1
−1
Thus
−1 −1 1
[ ][ ] [ ]
[ ] [] []
[
]
34
[
][ ] [ ]
0
0
0 4
0
⇒adj A B= 1
1 −1 1 = 1
−1 −1 1 4 −1
As ∣A∣=0 and adj A B≠0 , therefore the system of equations do not have any solution.
Thus the given system of equations is inconsistent.
Homogeneous system of equations: A system of equations is said to be homogeneous if it can be
written as A X =B
Trivial Solution: A solution in which each variable is zero is known as trivial solution. Every system
homogeneous linear equations always has trivial solution. A homogeneous system of linear equations
will have trivial solution only if ∣A∣=0
Non-trivial Solution: A homogeneous system of linear equations AX=B will have non-trivial
solution if ∣A∣≠0
Example-9: Solve the following homogeneous system of linear equations
2x− y z=0, x y− z=0, x− y=0
Solution:The given system of equations can be written as A X =0 where
2 −1 1
x
A= 1 1 −1 , X = y
1 −1 0
z
Now ∣A∣=20−11 011−1−1=−21−2=−3≠0
Thus the given system of equations has trivial solution.
Thus solution is x=0, y=0, x=0
[
] []
Exercise 4.7
Solve the following system of linear equations by matrix method
5 x7 y2
1.
4 x6 y3=0
2.
3 x4 y−5=0
x− y3=0
3.
x yz =3
2 x− y z=−1
2 x y−3z=−9
4.
6 x−12 y25 z =4
4 x15 y−20 z=3
2 x 18 y15 z =10
5.
3 x4 y7 z =16
2 x− y3 z =19
x 2 y−3z=25
35
6.
2 3 3
−  =10
x y z
1 1 1
  =10
x y z
3 1 2
−  =13
x y z
7.
3 x4 y2 z=8
2 y−3 z =3
x−2 y6 z=−2
8.
2 x6 y=2
3 x−z=−8
2 x− yz=−3
9.
8 x4 y3 z =18
2 x y z=5
x2 y z=5
2 x yz=4
10. x2 yz=4
x y2 z=4
[
[
]
[
]
1 −1 0
2
2 −4
A= 2 3 4 and B= −4 2 −4 are two square matrices, find
0 1 2
2 −1 5
solve the system of equations x− y=3, 2 x3 y4 z=17, y2 z=7
2 −3 5
A=
12. If
3 2 −4 find A−1 and hence solve the system of equations
1 1 −2
2 x−3 y5 z=11, 3 x2 y−4 z=−5, x y2 z=−3
1 2
5
−1
13. Find A if A= 1 −1 −1 . Hence solve the system of equations
2 3 −1
x2 y5 z =10, x− y−z=−2, 2 x3 y− z=−11
4
5 2
−1
14. Find A
if A= −5 −4 2 and hence solve the system of equations
−2 2 8
4 x−5 y−2 z=2,5 x−4 y2 z=−2, 2 x2 y8 z=−1
15. Solve the following system of equations if consistent
2 x y z=4, x yz=1, 3 x2 y2 z=5
16. Show that the system of equations 3 x yz =1, x y−z=2, 4 x2 y=2
17. Solve the following homogeneous\system of linear equations
2 x− yz=0
2 x− y− z=0
x yz=0
(i) 3 x2 y− z=0 (ii)
x y z=0 (iii) x− y−5 z=0
x4 y3 z=0
3 x2 y− z=0
x2 y4 z =0
11. If
AB and hence
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[
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x y−z =0
(iv)
x−2 yz =0 (v)
3 x6 y−5 z=0
2 x yz =0
2 x− y2 z=0
xx2 y z=0 (vi) 5 x3 y−z =0
x y2z=0
x5 y−5 z =0