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Determinants

1. Determinants

Determinants

Introduction

|A| =

11 12 13 21 22 23 31 32 33

a a a a a a a a a ⇒ Number of rows = Number of columns (i.e., defined only corresponding to square matrices) ⇒ The number of rows (or columns) of a determinant is called order of determinant 1 2 1 2

a a b b is a determinant of order 2 or we can say that it is a

second order determinant Number of rows = 2 Number of columns = 2 Note: Determinant is a special case of matrix

Expansion of a Determinant

1 1 1 2 2 2 3 3 3

a b c a b c a b c

= a 1 b 2 c 3 + b 1 c 2 a 3 + c 1 a 2 b 3 – a 3 b 2 c 1 – b 3 c 2 a 1 – c 3 a 2 b 1

Minors and Cofactors Minors: Minor Mij corresponding to element aij of determinant |[aij]| is the determinant corresponding to submatrix obtained by eliminating ith^ row and jth^ column.

Note: Minor of an element of a determinant of order n (n ≥ 2) is a determinant of order n – 1

Cofactor: Cofactor Aij corresponding to element aij is defined as Aij = (–1)i+j^ Mij where Mij is minor of aij

Note: Minors and cofactors of the elements a 11 , a 21 in the determinant 11 12 13 21 22 23 31 32 33

a a a a a a a a a

3. Determinants

Q.

Evaluate 2 4 − 1 2

Sol. 2(2) – 4(–1)

Q.

Evaluate x x 1 x 1 x

−

Sol. = x(x) – (x – 1) (x + 1)

= x^2 – (x^2 – 1) = 1

Q.

Evaluate

0 sin cos sin 0 sin cos sin 0

α − α ∆ = − α β α − β

Sol. Expansion about R^1

∆ = 0 0 sin sin 0

β − β

• sinα sin^ sin cos 0

− α β α

• cosα sin^0 cos sin

− α α − β ∆ = 0 – sinα(–cosα sinβ) – cosα (sinα sinβ) ∆ = sinα cosα sinβ – sinα cosα sinβ ∆ = 0

Q.

Find value of x for which^3 x^3 x 1 4 1 =

Sol. 3 – x

(^2) = 3 – 8 ⇒ x (^2) = 8 ⇒ x = ± 2 2

Determinants 4.

Q.

Evaluate

2 3 2 1 2 3 2 1 3

−

− −

by expanding it along the second row.

Sol. = – 1^

Q.

Show that the value of the determinant

tan A P tan B P tan C P tan A Q tan B Q tan C Q tan A R tan B R tan C R

vanishes for all values of A, B, C, P, Q & R where A + B + C + P + Q + R = 0

Sol. After expanding the determinant

tan(A + P) tan(B + Q) tan(C + R) + tan(B + P) tan(C + Q) tan(A + R) + tan(C + P) tan(A + Q) tan(B + R) – tan(A + R) tan(B + Q) tan(C + P) – tan(B + R) tan(C + Q) tan(A + P)

• tan(C + R) tan(A + Q) tan(B + P) …(1) Now ∵ α + β + γ = 0 then tanαtanβtanγ = tanα + tanβ + tanγ hence from (1) {tan(A + P) + tan(B + Q) + tan(C + R)} + {tan(B + P) + tan(C + Q) + tan(A + R)} + {tan(C + P) + tan(A + Q) + tan(B + R)} – {tan(A + R) + tan(B + Q) + tan(C + P)}
• {tan(B + R) + tan(C + Q) + tan(A + P)} – {tan(C + R)

• tan(A + Q) + tan(B + P)} = 0 Hence proved.

Note: If elements of a row (or column) are multiplied with cofactors of any other row (or column) then their sum is zero.

Determinants 6.

Properties of Determinants

Property- The value of a determinant remains unaltered, if the rows and columns are inter-changed

D =

1 1 1 1 2 3 2 2 2 1 2 3 3 3 3 1 2 3

a b c a a a a b c b b b D' a b c c c c

Note: (i) D & D' are transpose of each other. (ii) If D' = – D then it is Skew Symmetric (iii) The value of a skew symmetric determinant of odd order is zero. Property- If any two rows (or columns) of a determinant be interchanged, the value of determinant is changed in sign only.

Let D =

1 1 1 2 2 2 3 3 3

a b c a b c a b c

2 2 2 1 1 1 3 3 3

a b c D' a b c a b c

Then D' = – D Property- If any two rows (or columns) of a determinant are identical (all corresponding elements are same) then value of determinant is zero.

Let D =

1 1 1 1 1 1 3 3 3

a b c a b c a b c

then it can be verified that D = 0

Property- If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.

e.g., If

1 1 1 2 2 2 3 3 3

a b c D a b c a b c

= and D' =

1 1 1 2 2 2 3 3 3

ka kb kc a b c a b c

then D' = kD

Property- If all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants. 1 1 1 2 2 2 3 3 3

a x b y c z a b c a b c

1 1 1 2 2 2 3 3 3

a b c a b c a b c

3 3 3

x y z a b c a b c

7. Determinants

Property- If, to each element of any row (or column) of a determinant, the equimultiples of corresponding elements of other row (or column) are added, then value of determinant remains the same i.e., the value of determinant remains same if we apply the operation Ri → Ri + kRj or Ci → Ci + kCj

Let

1 1 1 2 2 2 3 3 3

a b c D a b c a b c

= and

D' =

1 2 1 2 1 2 2 2 2 3 1 3 1 3 1

a ma b mb c mc a b c a na b nb c nc

then D' = D

Note: While applying this property atleast one row (or column) must remains unchanged

Property-

(i) If ∆r =

f 1 ( )r f 2 ( )r f 3 ( )r

a b c d e f

where f 1 (r), f 2 (r), f 3 (r) are functions of r and a,

b, c, d, e, f are constants. Then

n n n 1 2 3 n (^) r 1 r 1 r 1 r r 1

f r f r f r

a b c d e f

= = =

=

∆ =

∑ ∑ ∑

∑

(ii) Also for ∆(x) =

f 1 ( x) f 2 ( x ) f 3 ( x)

a b c d e f

where f 1 (x), f 2 (x), f 3 (x) are functions

of x and a, b, c, d, e, f are constants. We have

q q q 1 2 3 q p p p

p

f x dx f x dx f x dx

x dx a b c d e f

∆ =

∫ ∫ ∫

∫

9. Determinants

Special Determinants (i) Symmetric Determinant

a h g h b f g f c

= abc + 2fgh –af^2 – bg^2 – ch^2

(ii) Skew symmetric Determinant

0 a b a 0 c 0 b c 0

− = − −

(iii) Cyclic Determinants

(a) 2 2 2

a b c a b c

= (a – b) (b – c) (c – a)

Q.

Evaluate

Sol. R^1 →^ R^1 – 6R^3

0 0 0 1 3 4 17 3 6

Q.

Show that

a b c a 2x b 2y c 2z 0 x y z

Sol. In LHS apply R

2 →^ R 2 – 2R 3

now

a b c a b c 0 x y z

=

Determinants 10.

Proof: In LHS C 1 → C 1 – C 2 , C 2 → C 2 – C 3

2 2 2 2 2

a b b c c a b b c c

Now C 1 → ( 1 )

C ,

a − b C^2 →^ ( )

C 2

b −c

(a – b) (b – c) 2

1 1 c a + b b +c c Expansion about R 1 ⇒ (a – b) (b – c) {0 – 0 + 1 (b + c – a – b)} ⇒ (a – b) (b – c) (c – a) Hence proved

(b) 3 3 3

a b c a b c

= (a – b) (b – c) (c – a) (a + b + c)

Proof: In LHS C 1 → C 1 – C 2 , C 2 → C 2 – C 3

3 3 3 3 3

a b b c c a b b c c

now C 1 → ( 1 )

C

a − b, C^2 →^ ( )

C 2

b −c

(a – b) (b – c) 2 2 2 2 3

1 1 c a + b + ab b + c +bc c Expansion about R 1 ⇒ (a – b) (b – c) {b^2 + c^2 + bc – a^2 – b^2 – ab} ⇒ (a – b) (b – c) {(c^2 – a^2 ) + b(c – a)} ⇒ (a – b) (b – c) (c – a) (c + a + b) Hence proved.

(c)^2 2 3 3 3

a b c a b c

= (a – b) (b – c) (c – a) (ab + bc + ca)

Proof: In LHS C 1 → C 1 – C 2 , C 2 → C 2 – C 3

Determinants 12.

Proof: C 1 → C 1 + C 2 + C 3

a b c b c b c a c a c a b a b

now C 1 → C 1 /(a + b + c)

(a + b + c)

1 b c 1 c a 1 a b Expansion about C 1 ⇒ (a + b + c) {(cb – a^2 ) – (b^2 – ac) + (ab – c^2 )} ⇒ – (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) ≤ 0 ⇒ – (a^3 + b^3 + c^3 – 3abc) ≤ 0 Hence proved.

Q.

Without expanding evaluate the determinant

Sol. apply C^1 →^ C^1 – (C^2 + 8C^3 )

0 1 5 0 7 9 0 0 5 3

=

Q.

Without expanding show that

2 2 2 2 2 2

b c bc b c c a ca c a 0 a b ab a b

Sol. R^1 →^ aR^1 , R^2 →^ bR^2 , R^3 →^ cR^3

2 2 2 2 2 2

ab c abc ab ac (^1) bc a abc bc ab abc ca b abc ac bc

C 1 → C 1 /(abc), C 2 → C 2 /(abc)

⇒ abc

bc 1 ab ac ca 1 bc ab ab 1 ac bc

13. Determinants

C 3 → C 3 + C 1

⇒ abc

bc 1 ab bc ca ca 1 ab bc ca ab 1 ab bc ca

C 3 → C 3 /(ab + bc + ca)

⇒ abc(ab + bc+ ca)

bc 1 1 ca 1 1 ab 1 1 = 0 Hence proved.

Q. Without expanding evaluate the determinant

( ) ( )

( ) ( )

( ) ( )

x x 2 x x^2

y y 2 y y^2

z z 2 z z^2

a a a a 1

a a a a 1

a a a a 1

− −

− −

− −

where

a, b, c and x, y, z ∈ R

Sol.

2x 2x 2x 2x 2y 2y 2y 2y 2z 2z 2z 2z

a a 2 a a 2 1 a a 2 a a 2 1 a a 2 a a 2 1

− − − − − −

C 1 → C 1 – C 2

2x 2x 2y 2y 2z 2z

4 a a 2 1 4 a a 2 1 4 a a 2 1

− − −

• −
• −
• − C 1 → C 1 / 2x 2x 2y 2y 2z 2z

1 a a 2 1 4 1 a a 2 1 0 1 a a 2 1

− − −

15. Determinants

Q.

Prove that

a b c 2a 2a 2b b c a 2b 2c 2c c a b

= (a + b + c)^3

Sol. C^1 →^ C^1 – C^2 , C^2 →^ C^2 – C^3

a b c 0 2a a b c a b c 2b 0 a b c c a b

C 1 → C 1 /(a + b + c), C 2 → C 2 /(a + b + c)

(a + b + c)^2

1 0 2a 1 1 2b 0 1 c a b

− − − − Expansion about R 1 ⇒ (a + b + c)^2 [–1 (–c + a + b – 2b) + 2a(1)] ⇒ (a + b + c)^3 Hence proved

Q.

Evaluate

2 2 2 2 2 2

1 a b 2ab 2b 2ab 1 a b 2a 2b 2a 1 a b

Sol. R^1 →^ R^1 /2b, R^2 →^ R^2 /2a, R^3 →^ R^3 /

2 2

2 2

2 2

1 a b (^) a 1 2b 8ab b 1 a^ b 1 2a b a 1 a^ b 2

− −^ −

C 1 → 2bC 1 , C 2 → 2aC 2 , C 3 → 2C 3

2 2 2 2 2 2 2 2 2 2

1 a b 2a 2 (^1) · 8ab 2b 1 a b 2 8ab 2b 2a 1 a b

R 1 → R 1 + R 2 , R 2 → R 2 – R 3

Determinants 16.

2 2 2 2 2 2 2 2 2 2 2 2

1 a b 1 a b 0 0 1 a b 1 a b 2b 2a 1 a b

R 1 → R 1 /(1 + a^2 + b^2 ), R 2 → R 2 /(1 + a^2 + b^2 )

⇒ (1 + a^2 + b^2 )^2 2 2 2 2

2b −2a 1 − a −b Expansion about R 1 ⇒ (1 + a^2 + b^2 )^2 [(1 – a^2 – b^2 + 2a^2 ) – (–2b^2 )] ⇒ (1 + a^2 + b^2 )^3

Q.

If

r 1 r 1 r 1 r n n n

2 2.3 4. x y z 2 1 3 1 5 1

− − − ∆ = − − −

show that

n r r = 1

∑∆ =^ constant

Sol.

n n n r 1 r 1 r 1 n (^) r 1 r 1 r 1 r r 1 n n n

2 2 3 4 5

x y z 2 1 3 1 5 1

− − − = = = =

∆ = − − −

n r 1 ( n ) n

r 1

−

∑ −

n r 1 ( n^ ) ( n )

r 1

−

∑ −

( )

n n n r 1 r 1

−

∑ −

n^ n^ n^ n r r 1 n^ n^ n

x y z 0 = 2 1 3 1 5 1

∑

Determinants 18.

R 3 → R 3 – R 2 , R 2 → R 2 – R 1

D = n!(n + 1)! (n + 2)!

1 n 1 n 2 n 1 0 1 2 n 2 0 1 2 n 3

expansion about C 1

D = n!(n + 1)! (n + 2)! {2(n + 3) – 2(n + 2)} D = (n!)^3 (n + 1) (n + 2) (n + 1) × 2

3

D n!

= (n^3 + 4n^2 + 5n +2) × 2

3

D (^4) n!

− = 2n(n^2 + 4n + 5) which is divisible by n

Q.

Without expanding, prove that

2 2 2

x x x 1 x 2 2x 3x 1 3x 3x 3 x 2x 3 2x 1 2x 1

= xA + B where A and

B are determinants of 3 × 3 square matrices not involving x.

Sol. Apply R^2 →^ R^2 – (R^1 + R^3 )

2

2

x x x 1 x 2 4 0 0 x 2x 3 2x 1 2x 1

now R 1 → R 1 + x^2 4

R 2 , R 3 → R 3 +

2 2

x (^) R 4 x x 1 x 2 4 0 0 2x 3 2x 1 2x 1

now R 3 → R 3 – 2R 1

x x 1 x 2 4 0 0 3 3 3

• − − −

19. Determinants

x x x 0 1 2 4 0 0 4 0 0 3 3 3 3 3 3

x 4 0 0 4 0 0 3 3 3 3 3 3

= xA + B

Q. If (x^1 – x^2 )

(^2) + (y 1 – y 2 ) (^2) = a 2 (x 2 – x 3 )^2 + (y 2 – y 3 )^2 = b^2 and (x 3 – x 1 )^2 + (y 3 – y 1 )^2 = c^2 then prove that 2 1 1 2 2 3 3

x y 1 4 x y 1 x y 1

= (a + b + c) (b + c – a) (c + a – b) (a + b – c)

Sol. Let three points in xy plane P(x^1 , y^1 ), Q(x^2 , y^2 ) and R(x^3 , y^3 ) hence

given that PQ^2 = a^2 ⇒ PQ = a QR^2 = b^2 ⇒ QR = b RP^2 = c^2 ⇒ RP = c now area of ∆PQR

∆ = S S ( − a) ( S − b) ( S −c)

⇒ ∆^2 = S(S – a) (S – b) (S – c) 2 (^2 1 ) 2 2 3 3

x y 1 (^1) x y 1 a b c b c a c a b a b c (^2) x y 1 2 2 2 2

  (^) = (^) + +   (^) + −   (^) + −   (^) + −              

2 1 1 2 2 3 3

x y 1 4 x y 1 x y 1

= (a + b + c) (b + c – a) (c + a – b) (a + b – c) Hence proved.