6_Determinants.pdf

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DETERMINANTS

2.1 Basic Concepts

The eliminant of the variables from the equations

(i) a₁x + b₁y = 0, a₂x + b₂y = 0 (ii) a₁x + b₁y + c₁z = 0, a₂x + b₂y + c₂z = 0 is is 2 2 1 1 b a b a = a₁b₂ – a₂b₁ = D a₃x + b₃y + c₃z = 0, 3 3 3 2 2 2 1 1 1 c b a c b a c b a = D

Expansion w.r.t. first row.

a₁ 3 3 2 2 c b c b – b₁ 3 3 2 2 c a c a + c₁ 3 3 2 2 b a b a = a₁ (b₂c₃ – b₃c₂) – b₁(a₂c₃ – a₃c₂) + c₁ (a₂b₃ – a₃b₂)

Expansion w.r.t. first column.

a₁ 3 3 2 2 c b c b _{– a} 2 _b31 _c31 c b _{+ a} 3 _b21 _c21 c b _{= a} 1 (b2c3 – b3c2) – a2 (b1c3 – b3c1) + a3 (b1c2 – b2c1)

If you compare term by term you will observe that the two expansions are same. 2.2 Properties Of Determinant

1. The value of determinant is not altered by changing rows into columns and columns into rows. 2. If any two adjacent rows or two adjacent columns of a determinate are interchanged, the

determinant retains its absolute value but changes its sign.

3. If any line of a determinant D be passed over p parallel lines the resultant determinate is (–1)p _D.

4. If any two rows or two columns of a determinant are identical, then the determinant vanishes. 5. If each constituent in any row or in any column be multiplied by the same factor then the

determinant is multiplied by that factor.

6. If each constituent in any row or column consists of r terms then the determinant can be expressed as the sum of r determinants.

7. If to or from each constituent of a row (or column) of a determinant are added or subtracted the equi-multiples of the corresponding constituent of any other row (or column) the determinant remains unaltered.

(a) The value of a determinant is obtained by multiplying the elements of any row (or column) with the corresponding cofactors and adding the resulting products.

(b) If we multiply the elements of any row (or column) with the corresponding cofactors of any other row (or column) and add them, then the result is zero.

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(c) If D’ is the determinant formed by replacing the elements of a determinant D by their corresponding cofactors, then D’ = D2_.

2.3 Special Determinants : (1) Symmetric determinant. _gh b_{f c}f g h a = abc + 2fgh – af2 _{– bg}2 _{– ch}2

In this the elements a_ij = a_ji or the elements situated at equal distance from the diagonal are equal both in magnitude and sign.

(2) Skew symmetric determinant of odd order.

0 a cb 0 a c b 0 -- - _{= 0.}

All the diagonal elements are zero and a_ij = – a_ji or the elements situated at equal distance from the diagonal are equal in magnitude but opposite in sign. Its value is zero.

(3) Circulant _cb _ac _ba

c b a

= – (a3 _{+ b}3 _{+ c}3 _{– 3abc)}

The three rows or columns are the cyclic arrangement of the letters a, b, c, i.e. a, b, c ; b, c, a and c, a, b respectively.

(4) Factors of three important determinants :

(i) 2 2 2 _{b c} a c b a1 1 1 _{= (a – b) (b – c) (c – a)} (ii) 3 3 3 _{b c} a c b a 1 1 1 =(a – b) (b – c) (c – a) (a + b + c) (iii) 3 3 3 2 2 2 c b a c b a 1 1 1 = (a – b) (b – c) (c – a) (ab + bc + ca)

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Ex.1 Prove that the determinant -sinθx sinθ cosθ-x 1

cosθ 1 x is independent of q.

Sol. We shall expand this determinant, using Sarrus Method.

–sin –x 1 –sin –x

x x

cos 1 x cos 1

q q

q q

sinq cosq sinq

Thus, D = (–x3 _{+ sin q cos q – sin q cos q) – (–x cos}2 _{q + x – x sin}2 _{q) = –x}3

Hence, the determinant is independent of q.

Ex.2 Evaluate : (i) 11 x + yx yy 1 x x + y (ii) x y x + y y x + y x x + y x y

Sol. (i) Let D =

y x x 1 y y x 1 x y 1 + +

Expanding it along the first column, we have

D = 1 x_x+y _x+y_y – 1 _xx _xy_+y + 1 _xx_{+y y}y

= [(x + y)2 – xy] – [x (x + y) – xy] + [xy – xy – y2] = x2 + y2 + xy – x2 – y2 = xy (ii) Let D = y x y x x y x y y x y x + + + = x x+_xy x_y – y _xy_+y x_y + (x + y) _xy_+y x+_xy

= x (xy + y2 – x2) – y (y2 – x2 – xy) + (x + y) (xy – x2 – y2 – 2xy)

= x2y + xy2 – x3 – y3 + x2y + xy2 + x2y – x3 – xy2 – 2x2y + xy2 – x2y – y3 – 2xy2 = – 2x3 – 2y3 = –2 (x3 + y3)

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Ex.3 Solve the equation x + αx x + αx xx

x x x + α = 0, (a ¹ 0). Sol. Let D = a + a + a + x x x x x x x x x = (x + a) x_x+a _xx_+a – x _xx _x+x_a + x _xx x+_xa = (x + a) [(x + a)2 – x2] – x [x (x + a) – x2] + x [x2 – x (x + a)] = (x + a)3 – x2 (x + a) – ax2 – ax2 = (x + a)3 – x3 – 3ax2 = x3 + 3ax2 + 3a2x + a3 – x3 – 3ax2 = 3a2x + a3

Equating it to zero, we have

3a2x + a3 = 0 Þ x = –

3 a

Ex.4 If a, b and c are real numbers, and

D = b + c c + a a + bc + a a + b b + c a + b b + c c + a = 0

Show that either a + b + c = 0 or a = b = c

Sol. Here, D = a c c b ) c b a ( 2(a b c) a b b c 2 b a a c ) c b a ( 2 + + + + + + + + + + + + (C₁ ® C₁ + C₂ + C₃) = 2 (a + b + c) a c c b 1 a b b c 1 c a a b 1 + + + + + + = 2 (a + b + c) b c a b 0 b c c a 0 c a a b 1 -- -- + + (R₂ ® R₂ – R₁; R₃ ® R₃ – R₁) = 2 (a + b + c) × 1 _bb_--_ac _cc_--_ba = 2 (a + b + c) [(b – c) (c – b) – (b – a) (c – a)] = 2 (a + b + c) [bc – c2 – b2 + bc – bc + ac + ab – a2] = 2 (a + b + c) (ab + bc + ca – a2 – b2 – c2)

Equating D to zero, we have

2 (a + b + c) (ab + bc + ca – a2 – b2 – c2) = 0 Þ Either a + b + c = 0 or ab + bc + ca = a2 + b2 + c2

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Ex.5 Show that 2 2 2 α α β + γ β β γ + α γ γ α + β = (b – g) (g – a) (a – b) (a + b + g) Sol. Let D = b + a g g a + g b b g + b a a 2 2 2 = b + a g g b -g g -b g -b a -b b -a b -a 2 2 2 2 2 (R₁ ® R₁ – R₂; R₂ ® R₂ – R₃) = (a – b) (b – g) b + a g g -g + b+b -a 2 1 1 1 1 = (a – b) (b – g) b + a g g + b + a -g + b+b -a 2 1 0 1 0 (C₁ ® C₁ + C₃) = (a – b) (b – g) (a + b + g) a_b++b_{g -}-₁1 = (a – b) (b – g) (a + b + g) (–a –b + b + g) = (a – b) (b – g) (g – a) (a + b + g)

Ex.6 Show that -b + a3a -a + b -a + c3b -b + c

-c + a -c + b 3c = 3 (a + b + c) (ab + bc + ca)

Sol. Using C₁ ® C₁ + C₂ + C₃, we have D = c 3 b c c b a b c 3b b c a b c a b a c a + -+ + + - + + + - + - + + = (a + b + c) c 3 b c 1 3b b c 1 a b a c 1 + - - + + -+ -= (a + b + c) c 2 a c a 0 a 2b a b 0 a b a c 1 + - -++ - + (R₂ ® R₂ – R₁; R₃ ® R₃ – R₁) = (a + b + c) × 1. a_a+_-2_cb _aa₊-₂b_c = (a + b + c) [(a + 2b) (a + 2c) – (a – b) (a – c)] = (a + b + c) (3ab + 3bc + 3ca) = 3 (a + b + c) (ab + bc + ca)

Ex.7 Prove that 12 3 + 2p 4 + 3p + 2q1 + p 1 + p + q 3 6 + 3p 10 + 6p + 3q = 1 Sol. Let D = q 3 p 6 10 p 3 6 3 3 2p 4 3p 2q 2 1 p 1 p q 1 + + + + + ++ + + = p 3 7 3 0 p 2 1 0 1 p 1 p q 1 ++ + + + (R₂ ® R₂ – R₁; R₃ ® R₃ – R₁) = ₃1 ₇2₊+₃p_p = (7 + 3p) – 3(2 + P) = 1

Ex.8 If a, b, c are in AP, prove that : x + 2 x + 3 x + 2ax + 3 x + 4 x + 2b x + 4 x + 5 x + 2c = 0

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Sol. Let D = c 2 x 5 x 4 x 3 x 4 x 2b x 2 x 3 x 2a x + + ++ + + + + + = ) a c ( 2 2 21 1 2(b a) a 2 x 3 x 2 x -+ + + (R₂ ® R₂ – R₁; R₃ ® R₃ – R₁) = ) a b ( 4 ) a c ( 2 0 0 ) a b ( 2 1 12 x 3 x 2a x -- -+ + + (R₃ ® R₃ – 2r₂) = [2 (c – a) – 4 (b – a)] x₁+2 x+₁3 = [2c – 2a – 4b + 4a] (x + 2 – x – 3) = [2c + 2a – 4b] (–1) = (4b – 2a – 2c) (....1) Since, a, b, c are in AP,

2b = a + c

\ 4b = 2a + 2c

Hence by (1), D = 0

Ex.9 Prove that

2 2 2 2 2 2 2 2 2 (b + c) a a b (c + a) b c c (a + b) = 2abc (a + b + c)3

Sol. Replacing C₂ by C₂ – C₁ and C₃ by C₃ – C₁, we have

D = 2 2 2 2 2 2 2 2 2 2 2 c ) b a ( 0 c 0 b ) a c ( b ) c b ( a ) c b ( a ) c b ( -+ -+ + -+ -+ = (a + b + c)2 c b a 0 c 0 b a c b c b a c b a bc 2 c b 2 2 2 2 -+ -+ -+ + Replacing R₁ by R₁ – R₂ – R₃, we have D = (a + b + c)2 c b a 0 c 0 b a c b b 2 c 2 bc 2 2 2 -+ -+ -Replacing C₂ by C₃ + b 1 _C 1 and C3 by C3 + _c 1 _C 1, we have D = (a + b + c)2 b a b c c c b a c b 0 0 bc 2 2 2 2 2 + + = 2bc (a + b + c)2 [(a + c) (a + b) – bc] = 2bc (a + b + c)2 (a2 + ab + bc – bc) = 2abc (a + b + c)3

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Ex.10 Find the values of h if area of a triangle is 4 sq. units and vertices are (i) (k, 0) (4, 0) (0, 2)

Sol. (i) The area of the triangle =

2 1 1 2 0 0 1 4 0 1 k = 2 1 [k (0 – 2) – 0(4 – 0) + 1(8 – 0)] = 2 1 (–2k + 8) = (4 – k) sq units According to the question,

4 – k = 4 or 4 – k = –4

Þ k = 0 k = 8

Hence, the possible values of k are 0 and 8. (ii) Area of the triangle =

2 1 1 k 0 4 1 02 0 1 = 2 1 _{[–2 (4 – k) – 0 (0 – 0) + 1 (0 – 0)]} = (k – 4) sq units According to the question,

k – 4 = 4 or k – 4 = –4

Þ k = 8 k = 0

Hence, the possible values of k are 8 and 0.

Ex.11 Using determinants, find the equation of the line AB, joining

(i) A (1, 2) and B (3, 6) (ii) A (3, 1) and B (9, 3)

Sol. Let P (x, y) be any point on AB. Then, area of the triangle ABP is zero, since the points A, B and P are collinear. So, (i) 2 1 1 y x 6 1 3 2 1 1 = 0 Þ 2 1 [1 (6 – y) – 2(3 – x) + 1(3y – 6x)] = 0 Þ 6 – y – 6 + 2x + 3y – 6x = 0 Þ y = 2x

Hence, the equation of the line AB is y = 2x. (ii) 2 1 1 y x 3 1 9 1 1 3 = 0 Þ 2 1 [3 (3 – y) – 1(9 – x) + 1(9y – 3x)] = 0 Þ 9 – 3y – 9 + x + 9y – 3x = 0 Þ 2x = 6y Þ x = 3y

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Q.1 If A = _êëé₂2 ₁5_úûù and B = _êëé₂4 -₅3_úûù, verify that |AB| = |A| |B|

Q.2 Without expanding evaluate the determinant

b a ab b a a c ca a c c b bc c b 2 2 2 2 2 2 + + +

Q.3 Without expanding evaluate the determinant

ab c c 1 ac b b 1 bc a a 1 2 2 2 --

-Q.4 Without expanding evaluate the determinant

1 ) a a ( ) a a ( 1 ) a a ( ) a a ( 1 ) a a ( ) a a ( 2 z z 2 z 2 2 y y 2 y y 2 x x 2 x x -+ -+

-Q.5 Without expanding evaluate the determinant

) sin( cos sin ) sin( cos

sin cos sin( ) sin d + g g g b b+d b a a+d a Q.6 Prove that c 1 1 1 1 b 1 1a 1 1 1 + + + = abc 1 ÷ ø ö ç è æ _{+ + +} c 1 b 1 a 1 1 = abc + ab + bc + ca Q.7 Prove that b 2 a c a c b c 2a b c 2c a b b a + + + + + + = 2(a + b + c)3 Q.8 Prove that ab c ) b a ( ca b ) a c ( bc a ) c b ( 2 2 2 2 2 2 + + + = (a2 _{+ b}2 _{+ c}2_{) (a + b + c) (b – c) (c – a) (a – b)} Q.9 Show that 2 2 2 2 2 2 b a 1 a 2 b 2 a 2 b a 1 ab 2 ab 2 ab 2 b a 1 -+ -+ = (1 + a2 _{+ b}2₎2 Q.10 Solve : 8 x 3 3 3 3x 8 3 3 8 3 3 x 3 = 0 Q.11 If x ¹ y ¹ z and 3 2 3 2 3 2 z 1 z z y 1 y y x 1 x x + + +

= 0, then prove that 1 + xyz = 0

Q.12 If a, b, c are all positive and are pth, qth and rth terms respectively of a G.P. then prove that

1 r c logb q 1 log 1 p a log = 0

Q.13 Find the area of a triangle with vertices : (–3, 5), (3, –6), (7, 2)

Q.14 Find the value of l so that the points given below are collinear (l, 2 – 2l), (–l + 1, 2l) and (–4, –l, 6 – 2l)

Q.15 Using determinants, find the area of the triangle, whose vertices are (–2, 4), (2, –6) and (5, 4). Are the given points collinear ?

UNSOLVED PROBLEMS

EXERCISE – I

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Q.1 Prove 3 3 3 z z 1 y y 1 x x 1 = (x – y) (y – z) (z – x) (x + y + z). [C.B.S.E. 2000]

Q.2 Prove using properties of determinants

b a ab b a a c ca a c c b bc c b 2 2 2 2 2 2 + + + = 0. [C.B.S.E. 2001]

Q.3 Prove using properties of determinants [C.B.S.E. 2002]

(a) 2 2 2 2 2 2 x z x z 1 z y z y 1 y x y x 1 + + + + + + = (x – y) (y – z) (z – x). (b) x 3 x 8 y 8 x 105x 4y 4x 2x x x y x + + + = x3_.

Q.4 Using the properties of determinants, show that [C.B.S.E. 2002]

(a) 2 9 17 21 4 3543 3 6 = 0. (b) 12 9 1 3 4 1 9 12 9

-Q.5 Using the properties of determinants, show that [C.B.S.E. 2003]

(a) z a y x z y a xx y z a + + + = a2_{(a + x + y + z). (b)} 0 cb c a bc 0 b a ac ab 0 2 2 2 2 2 2 .

Q.6 Using the properties of determinants, prove that _yz z_xx _xx_y

y z z y + + + = 4xyz [C.B.S.E. 2004]

Q.7 Using the properties of determinants, solve for x

x a x a x a x a x a x ax a a x a x + -- + -- - -+ = 0. [C.B.S.E. 2005] Q.8 Show that c x 4 x 3 x 2 x 3 x b xx 1 x 2 x a + + + + +

++ + + = 0 where a, b and c are in A.P.. [C.B.S.E. 2005]

Q.9 Using the properties of determinants, prove that [C.B.S.E. 2005]

(a) 3 3 3 _{b c} a c b a1 1 1 = (a – b) (b – c) (c – a) (a + b + c) (b) y x x z z y z y x z y x 2 2 2 + + + = (x – y) (y – z) (z – x) (x + y + z).

Q.10 If a, b and c are in A.P., show that

c x 4 x 3 x 2 x 3 x b xx 1 x 2 x a + + + + + ++ + + _{= 0 [C.B.S.E. 2006]}

BOARD PROBLES

EXERCISE – II

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Q.11 Using the properties of determinants, prove that [C.B.S.E. 2006]

(a) y x x z z yq r r p p q b a a c c b + + ++ + + + + + = 2 z y x r q p b c a (b) _aa _cb _b3b_c c_3cb c a b a a 3 --

-- - + - + _{= 3(a + b + c) (ab + bc + ca)}

Q.12 Using the properties of determinants, prove that [C.B.S.E. 2007]

(a) ) b a ( ab ab 1 ca ca(c a) 1 ) c b ( bc bc 1 + + + = 0. (b) xy z z zx y y yz x x 2 2 2 = (x – y) (y – z) (z – x) (xy + yz + zx)

Q.13 If x, y, z are different and

3 2 3 2 3 2 z 1 z z y 1 y y x 1 x x + + +

= 0, show that xyz = –1. [C.B.S.E. 2008]

Q.14 If a, b and c are all positive and distinct, show that D =

b a c c a b b c a

has a negative value.

[C.B.S.E. 2008] Q.15 Solve for x : 8 x 3 3 3 3x 8 3 3 8 3 3 x 3 = 0. [C.B.S.E. 2008]

Q.16 Using the properties of determinants. [C.B.S.E. 2008]

(a) 3 2 3 2 3 2 az 1 z z ay 1 y y ax 1 x x + + + = (1 + axyz) (x – y) (y – z) (z – x) (b) 2 2 2 2 2 2 b a 1 a 2 b 2 a 2 b a 1 ab 2 b 2 ab 2 b a 1 -+ -+ = (1 + a2 _{+ b}2₎3 (c) 1 x x x 1 x x x 1 2 2 2 = (1 – x3₎2 _(d) 3 2 3 2 3 2 c ab c 1 b ca b 1 a bc a 1 + + + = – (a – b) (b – c) (c – a) (a2 _{+ b}2 _{+ c}2_).

Q.17 Using properties of determinants, prove the following : [C.B.S.E. 2009] (a) 2 2 2 2 2 2 b a 1 a 2 b 2 a 2 b a 1 ab 2 b 2 ab 2 b a 1 -+ -+ = (1 + a2 _{+ b}2₎3_(b) 1 c cb ca bc 1 b ba ac ab 1 a 2 2 2 + + + = 1 + a2 _{+ b}2 _{+ c}2 (c) q 3 p 6 1 p 3 6 3 q 2 p 3 1 p 2 3 2 1 p 1 p q 1 + + + + + ++ + + = 1 (d) b a a c c b b b c c a aa b c + + +- - - = a3 + b3 + c3 – 3abc

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Q.18 Using properties of determinants, prove the following : [C.B.S.E. 2010]

(a) 2 2 2 ) b a ( bc ac bc ) c a ( ab ca ab ) c b ( + + + = 2abc (a + b + c)3 (b) b 2 a c a c b c 2a b c 2c a b b a + + + + + + = 2 (a + b + c)3 _{[C.B.S.E. 2010]}

Q.19 Using properties of determinants, solve the following for x : [C.B.S.E. 2011]

64 x 3 27 x 2 8 x 16 x 3 9 x 2 4 x 4 x 3 3 x 2 2 x = 0

Q.20 Using properties of determinants, show that b_c c_ca _ab_b

a a c b + + + [C.B.S.E. 2012]

Q.21 Using properties of determinants prove the following : [C.B.S.E. 2013]

x y 2 x y x y x x y 2 x y 2 x y x x + + + + + + = y2_{(x + y)}

ANSWER KEY

EXERCISE – 1 (UNSOLVED PROBLEMS)

2. 0 3. 0 4. 0 5. 0 10. x = 3 11_, 3 11_, 3 2 13. 46 sq. unit 14. l = 2 1 , –1 15. 35 sq. units, No

EXERCISE – 2 (BOARD PROBLEMS)

4. (b) 576 5. (b) 2a3b3c3 7. 0 or 3a 15.

3 2 _or

3 11