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  • Maths Class 12

Maths Class 12 Determinants R.D Sharma Question Answer

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Page 6.11 Ex. 6.1

Q1.

Answer :

(i)
M11=-1M21= 20Cij= -1i+jMijC11= -11+1-1 = -1C21 = -11+220 = -20D = -1×5-20×0=-5

(ii)
M11= 3M21= 4Cij = -1i+jMijC11=-11+1M11= 3C21=-12+1M21=-4 = -4D=3 × -1 – 4 × 2 = -3 – 8 = -11

(iii)
M11= -1252 =-2 – 10=-12M21 = -3252 = -6 – 10 = -16M31 = -32-12 = -6 + 2 = -4C11= -11+1M11= -12C21= -12+1M21 = –16 = 16C31 = -13+1M31= -4D=1-12 + 38 – 6 + 220 + 3=-12 + 6 + 46 = 40

(iv)
M11 = bcacab= ab2 – c2a = ab2 – c2M21 = abccab = a2b – c2b = ba2 – c2M31 = abcbca = a2c – b2c = ca2 – b2C11=-11+1M11 =ab2-c2C21=-12+1M21 =-ba2-c2C31=-13+1M31 = ca2-b2D=1.ab2 – c2 – aab – ca + b.cc – b=ab2 – ac2 – a2b + a2c + c2b – b2c=a2c – b + b2a – c + c2b – a

(v)
M11= 5071 = 5 – 0 = 5M21= 2671 = 2 – 42 = -40M31= 2650 = 0 – 30 = -30C11= -11+1M11 = 5C21=-12+1M21 =–40C31=-13+1M31 =-30D=05 – 0 -21 – 0 + 67 – 15 = -2 – 48 =-50

(vi)
M11= bffc =bc – f2M21 = hgfc = hc – fgM31 = hgbf = hf – gbC11= -11+1M11 = bc – f2C21=-12+1M11=-hc – fg = fg – hcC31=-13+1M11= hf – gbD = abc – f2 – hhc – fg + gfh – bg = abc – af2 – h2c + fgh + fgh – bg2 = abc + 2hfg – af2 – bg2 – ch2

(vii)

M11= 00 – 5 -10 + 1 – 25 – 1 = -1 – 8= -9M21=-10 – 5 + 1(5 – 1) = 5 + 4 = 9M31=-10 + 10 + 1(0 + 1) = -10 + 1 = -9M41=-1(1 – 2) + 10 – 1 = 1 – 1 = 0C11=-11+1M11= -9C21=-12+1M21= -1 × 9C31=-13+1M31=-9C41=-14+1M41= 0D=201-21-11-150 + 1-31-21-1 1250 -1-30111-12-15=-18 – 27 + 15 = 30

Q2.

Answer :

(i)
∆ = x(5x + 1) + 7x = 5×2 + x + 7x=5×2 + 8x

(ii)
∆ = cos2θ – – sin2θ = cos2θ + sin2θ = 1

(iii)
∆=cos15°cos75°- sin15°sin75°=cos15°cos75°-sin(90°- 75°)sin(90° – 15°) ∵ sin90° – θ = cosθ=cos15°cos75° – cos75°cos15°=cos15°cos75° – cos15°cos75°= 0

(iv)
∆=a2 – i2b2 – i2d2 – c2=a2 – i2b2 – i2d2 + c2=a2 + c2 – i2b2 + d2 ∵ i2 =-1=a2 + c2 + b2 + d2

Q3.

Answer :

Let A=23713175152012= 2 204 – 100 -3 156 – 75 + 7 260 – 255⇒A=2104 – 381 + 75⇒A=208 – 243 + 35⇒A=243 – 243 = 0∴ 23713175152012 = 0⇒237131751520122=02 = 0 ∵ det A2 = det A2

Q4.

Answer :

Let ∆=sin10°-cos10°sin80°cos80°⇒∆ = sin10°cos80° + cos10°sin80°=sin10°cos(90°-10°) + cos10°sin(90° – 10°) ∵ cosθ = sin90 – θ⇒∆ = sin10°sin10° + cos10°cos10° = sin210° + cos210° ∵ sin2θ + cos2θ = 1⇒∆ = 1

Q5.

Answer :

Let ∆ = 23-571-2-341

First method

∆=-11+1 21 + 8 + -11+2 37 – 6 + -11+3 -528 + 3= 21 + 8 – 37 – 6 – 528 + 3= 18 – 3 – 155=-140

Second method is the Sarus Method, where we adjoin the first two columns to the right to get

23-52371-271-341-34=2 × 1 × 1 + 3 × -2 × -3 – 5 × 7 × 4 – -5 × 1 × -3 + 2 × -2 × 4 + 3 × 7 × 1=2 + 18 – 140 – 15 – 16 + 21=-120 – 20=-140

Q6.

Answer :

Let ∆ = 0sinα-cosα-sinα0sinβcosα-sinβ0
∆=-11+1 0 0+sin2β + -11+2 sinα0 – sinβcosα + -11+3 -cosαsinαsinβ – 0 Expanding along R1=00 + sin2β – sinα0 – sinβcosα – cosαsinαsinβ – 0=sinαsinβcosα – sinαsinβcosα=0

Q7.

Answer :

Given: ∆=cosαcosβcosαsinβ-sinα-sinβcosβ0sinαcosβsinαsinβcosα
⇒∆=-11+1 cosα cosβcosα cosβ – 0 + -11+2 cosα sinβ-sinβ cosα – 0 + -11+3 -sinα-sin2β sinα – sinα cos2β Expanding along R1=cosα cosβcosα cosβ – 0 – cosα sinβ-sinβ cosα – 0 – sinα-sin2β sinα – sinα cos2β=cos2α cos2β + cos2α sin2β + sin2α sin2β + sin2α cos2β=cos2αcos2β + sin2β + sin2αsin2β + cos2β ⇒∆ = cos2α + sin2α ∵ sin2θ + cos2θ = 1⇒∆ = 1 ∵ sin2θ + cos2θ = 1

 

Page 6.12 Ex. 6.1

Q8.

Answer :

Consider LHSAB=25214-325=8 + 10-6 + 258 + 2-6 + 5 = 181910-1AB = -18 – 190 = -208Consider RHSA =2 – 10 =-8B=20 – -6 = 26AB=-8 × 26 =-208∴ LHS= RHS

Q9.

Answer :

A=101012004⇒3A=3030360012 Multiplying each element of A by 3⇒3A = -11+1 336 – 0 + -11+2 00 – 0 + -11+3 30 – 0 = 336 – 0 – 00 – 0 + 30 – 0 Expanding along R1=3 × 36 = 108 …1⇒A = -11+1 14 – 0 + -11+2 00 – 0 + -11+3 10 – 0 = 14 – 0 – 00 – 0 + 10 – 0 = 4 Expanding along R1⇒27A = 27 × 4 = 108 …2∴ 3A = 27 A From eqs. (1) and (2)

Q10.

Answer :

(i)
Given: 2451 = 2x46x⇒2 – 20 = 2×2 – 24⇒-18 = 2×2 – 24⇒2×2 = 6⇒x2= 3⇒x = ±3

(ii)
Given: 2345 = x32x5⇒10 – 12 = 5x – 6x⇒-2 = -x⇒ x = 2

(iii)
Given: 3xx1 = 3241⇒3 – x2 = 3 – 8⇒-x2 = -8⇒x2 = 8⇒x = ±22

(iv)
Given: 3×724 = 10⇒12x – 14 = 10⇒12x = 24⇒x = 2

Q11.

Answer :

Given: x2x1021314 = 28⇒x28 – 1 – x0 – 3 + 10 – 6⇒8×2 – x2 + 3x – 6 = 28⇒7×2 + 3x – 6 = 28⇒7×2 + 3x – 34 = 0⇒7x + 17 x – 2 = 0⇒x = 2

Integral value of x is 2. Thus, x = -177 is not an integer.

Q12.

Answer :

Matrix A will be singular if

A = 0
⇒x – 1x – 12 – 1 -1x – 1 – 1 + 11 -x – 1 = 0⇒x – 1×2 – 2x -1x – 2 + 12 – x = 0⇒x3 – 2×2 – x2 + 2x – x + 2 – x + 2 = 0⇒x3 – 3×2 + 4 = 0⇒x – 22 x + 1 = 0⇒x = 2 or x = -1

 

Page 6.48 Ex. 6.2

Q1.

Answer :

(i) ∆=1352610311138= 1 6101138 – 32103138 + 5263111=1228 – 110 – 376 – 310 + 522 – 186=1(118) – 3( -234) + 5(-164)=118 + 702 – 820=0(ii)∆=671921391314812426= 67338 – 336 -191014 – 1134 + 21936 – 1053= 67(2) – 19( -120) + 21( -117)= 134 + 2280 – 2457=-43(iii) ∆=ahghbfgfc=abffc -hhfgc +ghbgf=a(bc – f2) -h(hc – fg) + g(hf – gb)=abc – af2 – h2c + fgh + fgh – g2b=abc + 2fgh – af2 – ch2 – bg2(iv) ∆=1-324-12352=1-1252 -(-3)4232 + 24-135=1-2 – 10 + 38 – 6 + 220 + 3=(-12) + 6 + 46= 40(v)∆= 149491691625=19161625 -4416925 + 949916=1225 – 256 -4100 – 144 + 964 – 81=1(-31) – 4(-44) + 9(-17)=-31 + 176 – 153=-8(vi) ∆=6-322-12-10 52=6(-2 – 10) – (-3)(4 + 20) + 2(10 – 10)=-72 + 72 + 0=-72 + 72=0(vii) ∆=13927392719271327139=192712713139 – 332719132739 + 939192732719 – 27392792712713= 19(9 – 9) – 27(243 – 3) + 1(81 – 1) -33(9 – 9) -27(81 – 81) + 1(27- 27) + 93(243 – 3) – 9(81 – 81) + 1(9 – 729) – 27(81 – 1) – 9(27 – 27) + 27(9 – 729)= 10 – 6480 + 80 – 30 – 0 + 0 + 9720 – 0 – 720 – 2780 – 0 – 19440= – 6400 + 522720= 516320(viii) ∆=2-101-301-211-112-150=201-21-11-150 – (-1)-31-21-11250 + 0-30-21112-10 -1 -30111-12-15=20 – 1(0 + 1) – 2(5 – 1) + 1( -3(0 – 5) – 1(0 – 2) – 2(5 + 2) + 0 – 1-3(5 – 1)-0 + 1(-1 – 2)=2(-1 – 8) + (15 + 2 – 14) – (-12 – 3)=-18 + 3 + 15= 0

Q2.

Answer :

(i) ∆=82712351643=027035043 Applying C1→C1-4C2⇒∆=0

(ii) ∆=6-322-12-1052=0-320-12052 Applying C1→C1+2C2⇒∆=0

(iii) ∆=23713175152012=2371317513175 Applying R3→R3-R1⇒∆=0

(iv) ∆=1aa2bc1bb2ac1cc2ab=1a3abc1b3abc1c3abc Applying R1→aR1, R2→bR2 and R3→cR3=abc1a311b311c31⇒∆=0

(v) ∆=a+b2a+b3a+b2a+b3a+b4a+b4a+b5a+b6a+b=aaa2a2a2a4a+b5a+b6a+b Applying R1→R2-R1 and R2→R3-R2=2aaaaaa4a+b5a+b6a+b=0

(vi) ∆=1aa2-bc1bb2-ac1cc2-ab=0a-ba2-bc-b2+ac0b-cb2-ac-c2+ab1cc2-ab Applying R1→R1-R2, R2→R2-R3=0a-ba-ba+b+ca-b0b-cb-cb+c+ab-c1cc2-ab=a-bb-c01a+b+c01a+b+c1cc2-ab⇒∆=0

(vii) ∆=491639742623=116774223 Applying C1→C1-8C3⇒∆=0

(viii) ∆=0xy-x0z-y-z0=xyzxyz0xy-x0z-y-z0=1xyz0xzyz-xy0zy-yx-zx0=1xyz-2xy02yz-xy0zy-yx-zx0 Applying R1→R1+R2+R3=1xyz000-xy0zy-yx-zx0=0 Applying R1→R1-2R2

(ix)∆=143673543172=116774332=0 Applying C2→C2-7C3

x)∆=12223242223242523242526242526272=14916491625916253616253649=1491649162557911791113 Applying R3→R3-R2 and R4→R4-R3=14916491625791113791113=0 Applying R3→2+R3

xii) ∆=abca+2xb+2yc+2zxyz=a+2xb+2yc+2za+2xb+2yc+2zxyz Applying R1→R1+2R3=000a+2xb+2yc+2zxyz=0 Applying R1→R1-R2

Q3.

Answer :

∆=ab+ca2bc+ab2ca+bc2 When a=b, the first two rows become identical. Hence, a-b is a factor. Similarly, when b=c the second and third rows become identical. So, b-c is also a factor. Also, when c=a, the third and first rows become identical. Hence, c-a is also a factor. The product of diagonal elements, a(c + a) c2 is 4. So, the other factor should be a linear in a, b and c. It should also remain unaltered when any two letters are changed. Let this factor be λ(a+b+c).Here, λ is a constant. To find this, we havea=0, b=1, c=2032121214= λ(a – b)(b – c)(c – a)(a + b + c)032121214 = λ(0 – 1)(1 – 2)(2 – 1)(0 + 1 + 2)⇒-6 = 6λ⇒λ = -1Thus, ab + ca2bc + ab2ca + bc2 =-((a + b + c))(a – b)(b – c)(c – a)

 

Page 6.49 Ex. 6.2

Q4.

Answer :

∆=1abc1bca1cab When a=b, the first two rows become identical. Hence, a-b is a factor. Similarly, when b=c and c=a, the second and third and third and first rows become identical. Hence, b-c and c-a are also factors. The degree of product of the diagonal elements is 3. Hence, there are no other factors. 1abc1bca1cab=λ(a-b)(b-c)(c-a) Where λ is a constant102110120=2λ Putting a=0, b=1 and c=2 to find λ⇒2=2λ⇒λ=1Hence,1abc1bca1cab =(a-b)(b-c)(c-a)

Q5.

Answer :

∆=x + λxxxx + λxxxx + λ= λ 0x-λ λx0-λ x + λ ApplyingC1→C1 -C2, C2→C2-C3= λ 0x -λ 0 2x +λ 0-λ x + λ Applying R1→R2+R3= λ02x +λ -λx + λ + x-λ 0 0-λ= λ[λ(2x + λ)] + xλ2= λ2 (2x + λ + λ2x)= 3λ2x + λ3= λ2 (3x + λ )

Q6.

Answer :

∆=abccabbca=a(a2 – bc) – b(ca – b2) + c(c2 – ba)=a3 – abc – bca + b3 + c3 – abc=a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Q7

Answer :

∆=sinαcosαcos(α+δ)sinβcosβcos(β+δ)sinγcosγcos(γ+δ)=sinα (cosβ cos (γ+δ)-cosγ cos (β+δ)) + cosα (sinβ cos (γ+δ) – sinγcos (β+δ)) + cos (α+δ)(sinβcosγ-cosβsinγ)=0

Q8.

Answer :

∆=abca-bb-cc-ab+cc+aa+b=abca-bb-cc-aa+b+cc+a+ba+b+c Applying R3→R3+R2=(a+b+c)abca-bb-cc-a111 Taking (a+b+c) common=(a+b+c)abcbca111 Applying R2→R1-R2=(a+b+c)a-bb-ccb-cc-aa001 C1→C1-C2 and C2→C2-C3=(a+b+c)-1a-bc-a-b-c2=(a+b+c)-ac-bc-a2+ab-b2-c2+2bc=(a+b+c)a2+b2+c2-ab-bc-ca=a3+b3+c3-3abc

Q9.

Answer :
Let LHS =Δ = b + c a – b ac + a b – c b a + b c – a c Δ=b + c b – c b c – a c – a – b c + a b a + b c + a c + a b – c a + b c – a Expanding =b + c bc – c2 – bc + ab – a – bc2 + ac – ab – b2 + ac2 – a2 – ab + ac – b2 + bc=bc2 – c3 + abc – ac2 – a2c + a2b + ab2+ bc2+ abc – ab2 – b3+ ac2 – a3 – a2c – ab2 + abc⇒Δ=3abc-a3-b3-c3 Simplyfying =RHS

Q10.

Answer :

Let Δ=a+b b+c c+a b+c c+a a+bc+a a+b b+c Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we getΔ =a b c b c ac a b + b c a c a ba b c =a b c b c ac a b + -1a c b b a cc b a Applying C1↔ C3 in second determinant to get negative value of the deteminant=a b c b c ac a b + -1-1 a b c b c ac a b Applying C2↔C3= 2 a b c b c ac a b = RHS

Q11.

Answer :

Let LHS= Δ=a+b+2c a b c b+c+2a b c a c+a +2b ⇒∆ = 2a+2b+2c a b 2a+2b+2c b+c+2a b 2a+2b+2c a c+a +2b Applying C1→ C1+C2+C 3 = 2 a+b+c 1 a b 1 b+c+2a b1 a c+a +2b Taking out 2(a+b+c) common from C1∆ =2 a+b+c 1 a b 0 b+c+a 00 -b-c-a c+a +b Applying R2→ R2-R1 and R2→R2 -R3=2a+b+ca+b+ca+b+c1 a b 0 1 00 -1 1 Taking out (a+b+c) common from R2 and R3=2a+b+c3 11-0 Expanding along C1=2a+b+c3 =RHS

Q12.

Answer :

Let LHS=∆=a-b-c 2a 2a 2b b-c-a 2b 2c 2c c-a-b⇒∆= a+b+c a+b+c a+b+c 2b b-c-a 2b 2c 2c c-a-b Applying R1→R1+R2+R3=a+b+c 1 1 1 2b b-c-a 2b 2c 2c c-a-b=a+b+c 0 1 1 b +c+a b-c-a 2b 0 2c c-a-b Applying C1→ C1-C2= a+b+c a+b+c× 1 1 2c c-a-b Expanding along C1=a+b+c3=RHS

Q13.

Answer :

Let LHS =∆= 1 b+c b2+c21 c+a c2+a21 a+b a2+b2⇒∆=0 b+c-c+a b2+c2-c2+a20 c+a – a+b c2+a2-a2+b21 a+b a2+b2 Applying R1→ R1-R2 and R2→ R2-R3 = 0 b-a b2-a20 c-b c2-b21 a+b a2+b2= -12 0 a-b a2-b20 b-c b2-c21 a+b a2+b2 Taking out -1 common from R1 and R2=a-bb-c 0 1 a+b0 1 b+c1 a+b a2+b2 = a-bb-c1×1 a+b1 b+c Expanding along C1=a-bb-cc-a=RHS

Q14.

Answer :

Let LHS=∆= a a+b a+2ba+2b a a+ba+b a+2b a Δ =3a+3b 3a+3b 3a+3b a+2b a a+ba+b a+2b a Applying R1→R1+R2+R3=3 a+b1 1 1 a+2b a a+ba+b a+2b a Taking out 3 a+b common from R1= 3 a+b 0 0 12b -b a+b-b 2b a Applying C1→ C1-C2 and C2→C2-C3 = 3 a+b b2 0 0 1 2 -1 a+b-1 2 a Taking out b common from C1 and C2= 3 a+b b2×3=9a+b b2=RHS

Q15.

Answer :

Let LHS =∆=1 a bc1 b ca 1 c ab=1abca a2 abcb b2 bca c c2 abc Applying R1→a R1, R2→b R2 and R3→c R3 and then dividing it by abc =abcabca a2 1b b2 1c c2 1 Taking out abc common from C3= -1 1 a2 a1 b2 b1 c2 c Interchanging C3 and C1 to get-ve value of original determinant=-1-11 a a21 b b21 c c2 Applying C2↔C3= 1 a a21 b b21 c c=RHS

Q16.

Answer :

Let Δ1 =z x yz2 x2 y2 z4 x4 y4, Δ2=x y zx2 y2 z2 x4 y4 z4, Δ3=x2 y2 z2x4 y4 z4 x y z and Δ4=xyzx-y y-z z-x x+y+zNow,Δ1 =z x yz2 x2 y2 z4 x4 y4 Using the property that if two rows ( or columns ) of a determinant are interchanged, the value of the determinant becomes negetive, we get⇒ Δ1 = -1 x z yx2 z2 y2x4 z4 y4 ∵ C1↔ C2 =-1-1x y zx2 y2 z2x4 y4 z4 ∵ C2 ↔C3= x y zx2 y2 z2x4 y4 z4 = Δ2 …(1)= -1 x2 y2 z2x y zx4 y4 z4 Applying R1↔ R2= -1 -1 x2 y2 z2x4 y4 z4x y z Applying R2 ↔ R3 = x2 y2 z2x4 y4 z4x y z =Δ3 …(2)Thus,Δ1 = Δ2 = Δ3 From eqs. (1) and (2)
∆2 = x y zx2 y2 z2 x4 y4 z4= xyz 1 1 1x y z x3 y3 z3 Taking out common factor x from C1 , y from C2 and z from C3=xyz 0 0 1 x-y y-z z x3-y3 y3 -z3 z3 Applying C→ C1 -C2 and C2→ C2 -C3=xyz x-y y-z 0 0 1 1 1 z x2+2xy+y2 y2 +2yz+z2 z3 ∵ a3-b3=a-ba2+ab+b2 Taking out common factor x-y from C1 and y-z from C2=xyz x-y y-z1× 1 1 x2+xy+y2 y2 +yz+z2 Expanding along R1 =xyz x-y y-zy2 +yz+z2-x2-xy-y2=xyz x-y y-zyz-xy +z2-x2=xyz x-y y-zyz-x+z-xz+x=xyz x-y y-zz-xy+x+z=xyz x-y y-zz-xx+y+z=∆4 Thus, ∆1 =∆2 =∆3=∆4

 

Page 6.50 Ex. 6.2

Q17.

Answer :

Let LHS =Δ=b+c2 a2 bcc+a2 b2 caa+b2 c2 ab= b+c2-c+a2 a2- b2 bc-cac+a2-a+b2 b2 -c2 ca-aba+b2 c2 ab Applying R1 →R1-R2 and R2→R2-R3= b-ab+2c+a a+b a-b cb-a c-b b+2a+c b-c b+c ac-b a+b2 c2 ab=a-bb-c-b+2c+a a+b -c -b+2a+c b+c -a a+b2 c2 ab Applying x2-y2=x+y x-y and taking out a-b common from R1 and b-c from R2= a-bb-c -2b+c+a a+b -c -2b+a+c b+c -a a+b2 -c2 c2 ab Applying C1 →C1-C2= a-bb-c -2b+c+a a+b -c -2b+a+c b+c -a a+b+c a+b-c c2 ab Applying x2-y2=x+y x-y in C1= a-bb-ca+b+c -2 a+b -c -2 b+c -a a+b-c c2 ab Taking out a+b+c common from C1= a-bb-ca+b+c-2 a+b -c 0 c-a c-aa+b-c c2 ab Applying R2→R2-R1= a-bb-ca+b+cc-a-2 a+b -c 0 1 1a+b-c c2 ab Taking out c-a common from R2= a-bb-ca+b+cc-a-2 a+b +c -c 0 0 1a+b-c c2-ab ab Applying C2→ C2-C3= a-bb-ca+b+cc-a -1-2 a+b +c a+b-c c2-ab Expanding along R2= – a-bb-ca+b+cc-a-2c2+2ab -a2-b2-2ab+c2= – a-bb-ca+b+cc-a-a2-b2-c2= a-bb-ca+b+cc-aa2+b2+c2=RHS

Q18.

Answer :

Let LHS= Δ = a+1a+2 a+2 1a+2a+3 a+3 1 a+3a+4 a+4 1= a+1a+2-a+2a+3 a+2 -a+3 0a+2a+3-a+3a+4 a+3-a+4 0 a+3a+4 a+4 1 Applying C1→ C1-C2 and C2→C2-C3=-2a+2 -1 0-2a+3 -1 0a+3a+4 a+4 1=1×-2a+2 -1 -2a+3 -1 Expanding along C3 =4+2a-2a-6=-2=RHS

Hence proved.

Q19.

Answer :

Let LHS =Δ=a2 a2-b-c2 bcb2 b2-c-a2 cac2 c2- a-b2 ab

⇒∆=a2 -b-c2 bcb2 -c-a2 cac2 -a-b2 ab Applying C2→C2-C1=-1a2 b-c2 bcb2 c-a2 cac2 a-b2 ab=-a2 b2 +c2 bcb2 c2 + a2 cac2 a2 +b2 ab Applying C2→C2-2C1
=-a2 +b2 +c2 b2 +c2 bcb2+ c2 + a2 c2 + a2 cac2 +a2 +b2 a2 +b2 ab Applying C1→C1+C2=-a2 +b2 +c21 b2 +c2 bc1 c2 + a2 ca1 a2 +b2 ab
=-a2+b2 +c2 1 b2 +c2 bc0 c2 + a2 -b2 +c2 ca-bc0 a2 +b2 -b2 +c2 ab-bc Applying R2→R2-R1 and R3→R3-R1=a2+b2 +c2 1 b2 +c2 bc0 a2- b2 ca-b0 a2-c2 b a-c=-a2+b2 +c2 a-b a-c1 b2 +c2 bc0 a+b c0 a+c b Taking a-b common from R2 and a-c common from R3=a2+b2 +c2 a-b c-a×1× a+b ca+c b ∵ c-a=-a-c Expanding along C1=a2+b2 +c2 a-b c-a ab+b2-ac-c2
=a2+b2 +c2 a-bc-aab-c+b+cb-c=a-bc-ab-ca+b+ca2+b2 +c2

= RHS

Hence proved.

Q20.

Answer :

Let LHS =Δ=1 a2+bc a31 b2+ca b31 c2+ab c3⇒Δ =0 a2+bc-b2+ca a3-b30 b2+ca-c2+ab b3-c31 c2+ab c Applying R1→R1-R2 and R2→R2- R3

=0 a2-b2-ca+ bc a3-b30 b2-c2-ab+ ca b3-c31 c2+ab c3=0 a-b a+b-c a-ba2+ab+b20 b-cb+ c-a b-cb2+bc+a21 c2+ab c3

=a-bb-c0 a+b-c a2+ab+b20 b+c-a b2 +bc+c21 c2+ab c3 Taking out a-b common from R1 and b-c from R2= a-bb-c0 a+b-c a2+ab+b20 b+c-a- a+b-c b2 +bc+c2-a2+ab+b21 c2+ab c3 Applying R 2→R2 -R1

=a-bb-c0 a+b-c a2+ab+b20 2 c-a bc-a+c2-a21 c2+ab c3=a-bb-cc-a 0 a+b-c a2+ab+b20 2 a+b+c1 c2+ab c3=a-bb-cc-a×1× a+b-c a2+ab+b2 2 a+b+c Expanding along C1
=a-bb-cc-a× a+b2-c2 – 2a2+2ab+2b2 =a-bb-cc-aa+b2-c2 -a+b2-a2+b2=-a-bb-cc-aa2+b2+c2=RHS

Hence proved.

Q21.

Answer :

Let LHS =Δ=a2 bc ac+c2a2+ab b2 acab b2+bc c2Δ=abc a c a+ca+b b ab b+c c Taking out a, b and c common from C1, C2 and C3

=abc a c 0a+b b -2bb b+c -2b Applying C3→C3-C2-C1 =abc-2b a c 0a+b b 1b b+c 1 Taking (-2b) common from C3=abc-2b a c 0a -c 0b b+c 1 Applying R2→ R2-R3 =abc-2b×1 a c a -c Expanding along C3=abc-2b-2ac=4a2b2c2=RHS

Q22.

Answer :

Let LHS =Δ=x+4 x xx x+4 xx x x+4=3x+4 3x+4 3x+4x x+4 xx x x+4 Applying R1→ R1+R2+R3 =3x+4 1 1 1x x+4 x x x x+4 Taking out 3x+4 common from R1=3x+4 1 0 0x 4 0x 0 4 Applying C2→C2-C1 and C3→C3-C1=3x+4 42 Expanding along R1=163x+4 =RHS

Q23.

Answer :

Let LHS= Δ=1 1+p 1+p+q2 3+2p 4+3p+2q3 6+3p 10+6p+3q =1 1 1+p2 3 4+3p3 6 10+6p+1 p q2 2p 2q3 3p 3q =1 1 12 3 43 6 10+1 1 p2 3 3p3 6 6p+pq 1 1 12 2 23 3 3 Taking out pq common from last determinant=1 1 12 3 43 6 10+p1 1 12 3 33 6 6+0 Taking out p common from second determinant =1 1 12 3 43 6 10+0 ∵ Value of determinant with two identical columns is zero=1 0 02 1 23 3 7 Applying C2→C2-C1 and C3→C3-C1=1×1237 Expanding along R1=7-6=1 =RHS

Q24.

Answer :

Let LHS =Δ=a b-c c-ba-c b c-aa-b b-a cΔ=a 0 c-b+aa-c b+c-a 0a-b b+c-a c+a-b Applying C2→C2+C3 and C3→C1+C3 = b+c-ac+a-ba 0 1a-c 1 0a-b 1 1 Taking out common factor from C2 and C3= b+c-ac+a-ba ×1 01 1+1× a-c 1a-b 1 Expanding along R1=a+b-c b+c-ac+a-b=RHS

Q25.

Answer :

Let LHS=∆=a2 2ab b2 b2 a2 2ab 2ab b2 a2= a2a2 2ab b2 a2 -2ab b2 2ab 2ab a2 +b2 b2 a2 2ab b2 Expanding=a2a4-2ab3 -2ab b2a2-4a2b2+b2b4-2a3b=a6-2a3b3-2a3b3+8a3b3 +b6-2a3b3=a6+2a3b3 +b6=a32+2a3b3 +b32=a3+b32=RHS

Hence proved.

Q26.

Answer :

Let LHS = Δ=a2+1 ab acab b2+1 bcca cb c2+1=abc a+1a b ca b+1b ca b c+1c Taking out a, b and c common from R1 , R2 and R3= abc a+1a b c-1 a 1b 0 -1 a 0 1c Applying R2→R2-R1 and R3→R3-R1 = abc 1abca2+1 b2 c2-1 1 0 -1 0 1 Applying C1 →aC1, C2→bC2 and C3→cC3 =a2+1 b2 c2-1 1 0 -1 0 1= -1 b2 c2 1 0 + 1 a2+1 b2 -1 1 Expanding along R3=-1 -c2 +a2+1+ b2=a2+1+ b2+c2=a2+ b2+c2+1=RHS

Q27.

Answer :

Let LHS=Δ= 1 a a2a2 1 aa a2 1Δ= 1 +a2+a 1 +a2+a 1 +a2+aa2 1 aa a2 1 Applyng R1→R1+R2+R2= 1 +a2+a 1 1 1 a2 1 aa a2 1 Applying C2→C2-C1 and C3→C3-C1= 1 +a2+a 1 0 0 a2 1-a2 a-a2a a2-a 1-a= 1 +a2+a 1 0 0 a2 1-a1+a a1-aa aa-1 1-a= 1 +a2+aa-1a-1 1 0 0a2 -1+a -aa a -1 Taking out (a-1) common from C2 and C3= a3-1a-1 1 0 0a -1+a -aa a -1 ∵ 1 +a2+aa-1=a3-1= a3-1a-11+a +a2=a3-1a3-1=a3-12= RHS

Hence proved.

Q28.

Answer :

Let LHS =∆=a+b+c -c -b-c a+b+c -a-b -a a+b+c= a -c -bb a+b+c -ac -a a+b+c Applying C1→C1+C2+C3= a+b a+b -a+b b+c b+c b+c c -a a+b+c Applying R1→R1+R2 and R2→R2+R3=a+bb+c 1 1 -1 1 1 1 c -a a+b+c Taking out common factor from R 1 and R2= a+bb+c 0 0 -2 1 1 1 c -a a+b+c Applying R1 →R1- R2=a+bb+c -2-a-c Expanding along R1=2 a+bb+c c+a =RHS

Hence proved.

 

Page 6.51 Ex. 6.2

Q29.

Answer :

∆=b+caabc+abcca+b=0-2c-2bbc+abcca+b Applying R1→R1-(R2+R3)=0-2c-2bbc+a-b0c0a+b-c Applying C2→C2-C1 and C3→C3-C1 =0c+a-b00a+b-c-(-2c)b0c a+b-c-2bbc+a-bc0 Expanding along R1 =2c[b(a+b-c)-0]-2b[0-c(c+a-b)]=2bc[a+b-c]-2bc[b-c-a]=2bc[(a+b-c)-(b-c-a)]=4abc

Hence proved.

Q30.

Answer :
∆=b2+c2abacbac2+a2bccacba2+b2

a(b2+c2)a2ba2cb2ab(c2+a2)b2cc2ac2bc(a2+b2) Multiplying the three rows by a, b and c
=abcabcb2+c2a2a2b2c2+a2b2c2c2a2+b2 Taking out a, b and c common from the three columns=2(b2+c2)2(a2+c2)2(a2+b2)b2c2+a2b2c2c2a2+b2 Applying R1→R1+R2+R3=2b2+c2a2+c2a2+b2-c20-a2-b2-a20 Taking out 2 common from the three columns and then applying R2→R2-R1 and R3→R3-R1 =20c2b2-c20-a2-b2-a20 Applying R1→R1+R2+R3=2{[-c2(-a2b2)]+[b2(c2a2)]} Expanding along R1=4a2b2c2

Q31.

Answer :

∆=0 b2a c2aa2b0c2ba2c b2c0=1abc0b3ac3aa3b0c3ba3cb3c0 Multiplying the three columns by a, b and c =abcabc0b3c3a30c3a3b30 Taking out a, b and c common from the three rows =b3a3c3a30+c3a30a3b3=2a3b3c3 Expanding along R1

Q32.

Answer :

∆=a2+b2cccab2+c2aabbc2+a2b=1abca2+b2c2c2a2b2+c2a2b2b2c2+a2 Multiplying R1, R2 and R3 by c,a and b and then dividing by abc=1abca2+b2 c2-a2-b2 c2-a2-b2a2b2+c2-a20b20 c2+a2-b2 Applying C2→C2-C1 and C3→C3-C1=1abc0 -2b2 -2a2a2b2+c2-a20b20 c2+a2-b2 Applying R1→R1-R2-R3=1abc[-a2-2b2-2a20c2+a2-b2+b2-2b2-2a2b2+c2-a20 Expanding along C1=1abc-a2-2b2(c2+a2-b2)+b20+2a2b2+c2-a2=1abc-a2-2b2c2-2b2a2+2b4+b22a2b2+2a2c2-2a4=1abc2a2b2c2+2a4b2-2a2b4+2a2b4+2a2b2c2-2a4b2=1abc4a2b2c2=4abc

Hence proved.

Q33.

Answer :

∆=-bcb2+bcc2+bca2+ac-acc2+aca2+abb2+ab-ab=1abc-abcab2+abcac2+abca2b+abc-abcc2b+abca2c+abcb2c+abc-abc Applying R1→aR1, R2→bR2 and R3→cR3 and then dividing by abc=abcabc-bcab+acac+abab+bc-accb+abac+bcbc+ac-ab Taking out a, b and c common from the three columnsab+bc+caab+bc+caab+bc+caab+bc-accb+abac+bcbc+ac-ab Applying R1→R1+R2+R3=(ab+bc+ca)111ab+bc-accb+abac+bcbc+ac-ab=(ab+bc+ca)0010-(ab+bc+ac)cb+abac+bc+abbc+ac+ab-ab Applying C1→C1-C3 and C2→C2-C3=(ab+bc+ca)0-(ab+bc+ac)ac+bc+abbc+ac+ab=(ab+bc+ca)(ab+bc+ac)2=(ab+bc+ca)3

Hence proved.

Q34.

Answer :

∆=x+42x2x2xx+42x2x2xx+4=5x+45x+45x+42xx+42x2x2xx+4 Applying R1→R1+R2+R3=5x+41112xx+42x2x2xx+4 Take out 5x+4 common from R1=5x+41002×4-x02x04-x Applying C2→C2-C1 and C3→C3-C1=5x+4(4-x)2 Expanding along R1

Hence proved.

Q35.

Answer :

∆=b+caabc+abcca+b=b+caab+c2c+aa+2bcca+b Applying R2→R2+R3=2b+caa0cbcca+b Applying R2→R1-R2 and taking out 2 common from R2=2b+caa0cbc0a Applying R3→R2-R3=2ba00cbc0a Applying R1→R1-R3=4abc Expanding

Q36.

Answer :

∆=1xx2x21xxx21=11xx21 – xx2xx1 + x2x21xx2 Expanding=(1 – x3) – x(x2 – x2) + x2(x4 – x)=1 – x3 + x6 – x3=1 – 2×3 + x6=(1 – x3)2

Q37.

Answer :

∆=-a(b2+c2-a2)2b32c32a3-b(c2+a2-b2)2c32a32b3-c(a2+b2-c2)=abc-b2-c2+a22b22c22a2-c2-a2+b22c22a22b2-a2-b2+c2 Taking out a, b and c common from C1, C2 and C3=abca2+b2+c22b22c2a2+b2+c2-c2-a2+b22c2a2+b2+c22b2-a2-b2+c2 Applying C1→C1+C2+C3=abc(a2+b2+c2)12b22c21-c2-a2+b22c212b2-a2-b2+c2 Taking out a2+b2+c common from C1=abc(a2+b2+c2)12b22c20-c2-a2-b2000-a2-b2-c2 Applying R2→R2-R1 and R3→R3-R1=abc(a2+b2+c2)3 Expanding

Hence proved.

Q38.

Answer :

∆=1+a1111+a1111+a=1+a 1+a111+a – 11111+a +111+a11 Expanding=(1+a)(1+a)2 -1-1(1 + a – 1) + (1 – 1 – a)=(1+a)[1 + a2 + 2a – 1] -a -a=1 + a + a2 + a3 + 2a + 2a2 – 2a= a3 + 3a2

Q39.

Answer :

xyzpqrabc R2↔R3=-xyzabcpqr R1↔R2=abcxyzpqrybqxapzcr =yxzbacqpr C1↔C2=-xyzabcpqr R1↔R2=abcxyzpqr

Hence proved.

 

Page 6.52 Ex. 6.2

Q40.

Answer :

Given: a, b, c are in A.P.

2b=a+c

∆=x+1x+2x+ax+2x+3x+bx+3x+4x+c Applying R2=2R2∆=12x+1x+2x+a2x+42x+62x+2bx+3x+4x+c ∆=12x+1x+2x+a000x+3x+4x+c ∵ 2b=a+c Applying R2→R2-R1+R3∆=0

Q41.

Answer :

Given:α, β, γ areinA.P.

Now,
2β=α+γ

∆=x-3x-4x-αx-2x-3x-βx-1x-2x-γ ∆=12x-3x-4x-α2x-42x-62x-2βx-1x-2x-γ Applying R2→2R2∆=12x-3 x-4 x-α00 -2β+α+γx-1 x-2x-γ ∵ 2β=α+γ Applying R2→R2-R1+R3∆=12x-3 x-4 x-α000x-1x-2x-γ∆=0

Q42.

Answer :

Let ∆=y+z x yz+x z x x+y y z⇒ ∆=2x+y+z x+y+z x+y+z z+x z x x+y y z Applying R1→ R1 +R2+ R3=x+y+z 2 1 1 z+x z x x+y y z= x+y+z 0 1 10 z x x-z y z Applying C1→C1-C2-C3= x+y+z x-z×11zx Expanding along C1= x+y+z x-z2

Q43.

Answer :

Let ∆ = x111x111x=x-11-x01x111x Applying R1→R1-R2=x-11-101x111x =x-11001x+211x+2x Applying C2→C1+C2+C3= x-1x+210011111x Expanding along R1=x-1x+2x-1∆=x-12x+2

Q44.

Answer :

Let Δ =b+c c+a a+bc+a a+b b+ca+b b+c c+a =2a+b+c 2a+b+c 2a+b+c c+a a+b b+c a+b b+c c+a Applying R1→R1+R2+R3=2a+b+c 1 1 1c+a a+b b+ca+b b+c c+a =2a+b+c 1 0 0c+a b-c b-aa+b c-a c-b Applying C2→C2-C1 and C3→C3-C1=2a+b+c1b-cb-ac-ac-b=2a+b+cb-cc-b-b-ac-a=-2a+b+ca2+b2+c2-ab-bc-ca=-a+b+c2a2+2b2+2c2-2ab-2bc-2ca=-a+b+ca-b2+b-c2+c-a2But Δ=0 Given⇒-a+b+ca-b2+b-c2+c-a2=0⇒Either a+b+c=0 or a-b2+b-c2+c-a2=0⇒a+b+c=0 or a=b=cHence proved.

Q45.

Answer :

Let ∆=x-6-12-3xx-3-32xx+2=x-6-12-3xx-3-3-x2x+6x+3 Applying R3→R3-R1=x+3x-6-12-3xx-3-121 =x+3x-23x-6-x+22-3xx-3-121 Applying R1→R1-R2=x+3x-213-12-3xx-3-121 =x+3x-21302-3xx-1-120 Applying C3→C3+C1=x+3x-2x-11302-3×1-120 =x+3x-2x-1-113-12 Expanding along C3=-5x+3x-2x-1x=2,-3, 1

Q46.

Answer :

(i)

Let ∆=x+abcax+bcabx+c=x+a+b+cbcx+a+b+cx+bcx+a+b+cbx+c Applying C1→C1+C2+C3=x+a+b+c1bc1x+bc1bx+c =x+a+b+c1bc0 x01bx+c Applying R2→R2-R1=x+a+b+c1bc0 x000x Applying R3→R3-R1∆=x+a+b+cx2-0=0 Given⇒x2=0 or x+a+b+c=0⇒x=0 or x=-a+b+c

(ii)
Let ∆=x+axxxx+axxxx+a=3x+axx3x+ax+ax3x+axx+a Applying C1→C1+C2+C3=3x+a1xx1x+ax1xx+a=3x+a1xx0a01xx+a Applying R2→R2-R1=3x+a1xx0a000a Applying R3→R3-R1∆=3x+aa2-0=0x=-a3

(iii)

Let ∆=3x-83333x-83333x-8=3x-2333x-23x-833x-233x-8 Applying C1=C1+C2+C3=3x-213313x-83133x-8 =3x-213303x-110133x-8 Applying R2→R2-R1=3x-213303x-110003x-11 Applying R3→R3-R1∆=3x-23x-112=0x=23,113,113

(iv)
Let ∆=1xx21aa21bb2=1xx20x-ax2-a21bb2 Applying R2→R1-R2=1xx20x-ax2-a20x-bx2-b2 Applying R3→R1-R3=x-ax-b1xx201x+a01x+b ∆=x-ax-bx+b-x-a=0x=a,b

(v)

Let ∆=x+1352x+2523x+4=x+935x+9x+25x+93x+4 Applying C1=C1+C2+C3=x+91351x+2513x+4 =x+91350x-1013x+4 Applying R2→R2-R1=x+91350x-1000x-1 Applying R3→R3-R1∆=x+9x-12=0x=-9, 1, 1

(vi)
Let ∆=1xx31bb31cc3=1xx30b-xb3-x31cc3 Applying R2→R2-R1=1xx30b-xb3-x30c-xc3-x3 Applying R3→R3-R1=1xx30x-bx3-b30x-cx3-c3=x-bx-c1xx201x2+xb+b201x2+xc+c2 ∆=x-bx-cxc-b-b2+c2=0x=b, c, -b+c
(vii)

Let Δ=15-2x 11-3x 7-x 11 17 14 10 16 13=0⇒15-2x-14+2x 11-3x 7-x 11-28 17 14 10-26 16 13=0 Applying C1→C1-2C3⇒ 1 11-3x 7-x-17 17 14 -16 16 13 =0⇒ 12-3x 4-2x 7-x 0 3 14 0 3 13 =0 Applying C1→C1+C2 and C2→C2 -C3⇒ 12-3x 3×13-3×14 =0⇒12-3x-3=0⇒12-3x =0⇒3x =12⇒ x=4

(viii)
Let ∆=11xp+1p+1p+x3x+1x+2=11xppp3x+1x+2 Applying R2→R2-R1=p11x1113x+1x+2 =p11x1112x2 Applying R3→R3-R1=p01x0112-xx2 Applying C1→C1-C2=p2-x×1×11 Expanding along C1=p2-x1-x=0x=1, 2

 

Page 6.57 Ex. 6.3

Q1.

Answer :

(i)
∆=12381-4215-11 ∆=12381-7-605-11 Applying R2→R2-R1∆=123 81-7-602-90 Applying R3→R3-R1∆=12-7-62-9∆=1263 + 12∆= 1275 = 752square units

(ii)
∆=122711111081 ∆=12271-1-601081 Applying R2→R2-R1∆=12 2 71-1-60 8 10 Applying R3→R3-R1∆=12-1-681∆=12-1 + 48∆=1247 = 472square units

(iii)
∆=12-1-81-2-31321 ∆=12-1-81-150321 Applying R2→R2-R1∆=12-1-81-1504100 Applying R3→R3-R1∆=12-15410∆=12-10-20∆=1230=15 square units

(iv)
∆=12001601431 ∆=12001600431 Applying R2→R2-R1∆=12001600430 Applying R3→R3-R1∆=126043∆=1218-0∆=1218=9 square units

Q2.

Answer :

(i) If the points (5, 5), (−5, 1) and (10, 7) are collinear, then

∆=551-5111071=0=551-10-401071 Applying R2→R2-R1= 5 51-10-40520 Applying R3→R3 – R1=-10-452=-20 + 20 = 0

Thus, these points are colinear.

(ii) If the points (1, −1), (2, 1) and (4, 5) are collinear, then

∆=1-112 114 51 = 0=1-111 204 51 Applying R2→R2-R1=1-11120360 Applying R3→R3-R1=1236 = 6 – 6 = 0

Thus, these points are collinear.

(iii) If the points (3, −2), (8, 8) and (5, 2) are collinear, then

∆=3-21881521 = 0=3-215100521 Applying R2→R2-R1=3-215100240 Applying R3→R3-R1=51024=20 – 20 = 0

Thus the points are colinear.

(iv) If the points (2, 3), (−1, −2) and (5, 8) are collinear, then

∆=231-1-21581=0=231-3-50581 Applying R2→R2-R1=231-3-50350 Applying R3→R3-R1=-3-535=-15+15=0

Thus the points are colinear.

Q3.

Answer :

If the points (a, 0), (0, b) and (1, 1) are collinear, then

a010b1111=0⇒a01-ab0111=0 Applying R2→R2-R1⇒a01-ab01-a10=0 Applying R3→R3-R1⇒∆=-ab1-a1=0⇒-a-b1-a=0⇒a+b=ab

Q4.

Answer :

ab1a’b’1a-a’b-b’1⇒∆=ab1a’-ab’-b0a-a’b-b’1 Applying R2→R2-R1⇒∆=ab1a’-ab’-b0-a’-b’0 Applying R3→R3-R1⇒∆=a’-ab’-b-a’-b’⇒∆=-b’a’-a + a’b’-b=-b’a’ + b’a + a’b’ – a’b=b’a – a’b

If the points are collinear, then ∆ = 0. So,
ab’ − a’b = 0

Thus, ab’ = a’b

Q5.

Answer :

If the points (1, −5), (−4, 5) and λ, 7 are collinear, then

1-51-451λ71=0⇒ 1-51-5100 λ71 = 0 Applying R2→R2-R1⇒1-51-5100λ-1120 = 0 Applying R3→R3-R1⇒∆=-510λ-112 = 0⇒-60-10λ-1 = 0⇒-60-10λ+10=0⇒-10λ=50⇒λ=-5

Q6.

Answer :

∆=12×412-61541=±35=12×412-x-100541=±35 Applying R2→R2-R1=12×412-x-1005-x00=±35 Applying R3→R3-R1=122-x-105-x0=±35=0 + 105 – x = ±70⇒50 – 10x = 70 or 50 – 10x = -70⇒-10x = 20 or -10x=-120⇒x=-2 or x =12

Q7.

Answer :

∆=12141231-5-31=121 4 11-10-5-31 Applying R2→R2-R1=121411-10-6-70 Applying R3→R3-R1=121-1-6-7=12-7 – 6=132square units ∵Area cannot be negative

Therefore, (1, 4), (2, 3) and (−5, −3) are not collinear because, 141231-5-31 is not equal to 0.

Q8.

Answer :

Given:
Vertices of triangle: (− 3, 5), (3, − 6) and (7, 2)

Area of the triangle=∆=12-3 513-617 21=12-3516-110721 Applying R2→R2-R1=12-351 6-11010-30 Applying R3→R3-R1=126-1110-3=12-18 + 110=46 square units

Q9.

Answer :

If the points (k, 2 − 2 k), (− k + 1, 2k) and (− 4 − k, 6 − 2k) are collinear, then

∆=k2-2k1-k+12k1-4-k6-2k1=0⇒k2-2k1-2k+14k-20-4-k6-2k1=0 Applying R2→R2-R1⇒k2-2k1-2k+14k-20-4-2k40=0 Applying R3→R3-R1⇒-2k+14k-2-4-2k4=0⇒-8k + 4 + 16k – 8 + 8k2 – 4k = 0 ⇒8k2 + 4k – 4 = 0⇒8k – 4k + 1 = 0⇒k = -1 or k = 12

Q10.

Answer :

If the points (x, −2), (5, 2), (8, 8) are collinear, then

x-21521881=0∆=x-21521881∆=x-215-x40881 Applying R2→R2-R1=x-215-x 408-x100 Applying R3→R3-R1=5-x48-x10=50 – 10x – 32 + 4x=18 – 6x∆=18 – 6x∆= 0 Given⇒18 – 6x = 0⇒x = 3

Q11.

Answer :

If the points (3, −2), (x, 2) and (8, 8) are collinear, then

3-21×21881 = 0∆=3-21×21881=3-21x-340881 Applying R2→R2-R1=3-21x-3405100 Applying R3→R3-R1=x-34510=10x – 30 – 20∆=10x – 50∆=0 Given⇒10x – 50 = 0⇒10x = 50⇒x = 5

 

Page 6.58 Ex. 6.3

Q12.

Answer :

(i)
Given: A = (1, 2) and B = (3, 6)

Let the point P be (x, y). So,
Area of triangle ABP = 0

⇒∆=12121361xy1 = 0⇒16 – y – 23 – x + 13y – 6x = 0⇒6 – y – 6 + 2x + 3y – 6x = 0⇒2y – 4x = 0⇒y = 2x

(ii)
Given: A = (3, 1) and B = (9, 3)

Let the point P be (x, y). So,
Area of triangle ABP = 0

⇒∆=12311931xy1=0⇒33-y-19-x+19y-3x=0⇒9-3y-9+x+9y-3x=0⇒-2x+6y=0⇒x=3y

Q13.

Answer :

(i) If the area of a triangle with vertices (k, 0), (4, 0) and (0, 2) is 4 square units, then Δ=12k 0 14 0 10 2 1 =12 2 × k 14 1 Expanding along C2=k-4Since area is always+ve, we take its absolute value, which is given as 4 square units.⇒( k-4 )=±4⇒(k-4)=4 or (k-4 )=-4⇒k-4=4 or k-4 =-4⇒k=8 or k=0⇒k=8, 0(ii)If the area of a triangle with vertices (-2, 0) (0, 4) and (0, k) is 4 square units, then ∆1 =12-2 0 1 0 4 1 0 k 1=12 -2×4 1k 1 Expanding along C1=-4-kSince area is always+ve, we take its absolute value, which is given as 4 square units.⇒-4-k=±4⇒-4-k=±4⇒-4-k=4 or -4-k=-4⇒k=4+4 or k =-4+4⇒ k=8 or k=0

 

Page 6.71 Ex. 6.4

Q1.

Answer :

Given: x-2y = 4 -3x + 5y = -7Using the properties of determinants, we getD= 1 -2-3 5 = 5 – 6 = -1 ≠ 0D1= 4 -2 -7 5 = 20 – 14 = 6D2 = 1 4-3 -7 = -7 + 12 = 5Using Cramer’s Rule, we getx=D1D = 6-1 = -6y= D2D = 5-1 = -5∴ x=-6 and y =-5

Q2.

Answer :

Given: 2x – y = 1 7x – 2y = -7Using Crammer’s Rule, we getD =2 -17 -2 =-4 + 7 = 3D1= 1 -1-7 -2=-2 – 7 = -9D2=2 1 7 -7 = -14 – 7 = -21Now,x=D1D = -93= -3y=D2D= -213 = -7∴ x=-3 and y=-7

Q3.

Answer :

Given: 2x – y = 17 3x + 5y = 6Using Cramers Rule, we get D= 2 -1 3 5 = 10 + 3 = 13D1=17 -1 6 5 = 85 + 6 = 91D2= 2 17 3 6 = 12 – 51 = -39Now,x= D1D = 9113 = 7y=D2D = -39 13=-3∴ x = 7 and y = -3

Q4.

Answer :

Given: 3x + y = 19 3x – y = 23Using Cramer’s Rule, we getD=3 1 3 -1= -3 – 3 = -6D1=19 123 -1 =-19 – 23 = -42D2=3 19 3 23= 3 × 23 – 3 × 19 = 3 × 4 = 12Now,x=D1D=-42-6 = 7y=D2D= 12-6 = -2∴ x=7 and y=-2

Q5.

Answer :

Given: 2x – y = -2 3x + 4y = 3Using Cramer’s Rule, we get D=2 -1 3 4= 8 + 3 = 11D1 =-2 -1 3 4 =-8 + 3 =-5D2=2 -2 3 3 = 6 + 6 = 12Now,x=D1 D = -511y=D2D = 1211∴ x =-511 and y = 1211

Q6.

Answer :

Given: 3x+ay = 4 2x+ay= 2 Using Cramer’s rule, we get D =3 a2 a=3a-2a=aD1 =4 a2 a =4a-2a =2aD2=3 4 2 2 =6-8=-2Now,x=D1 D=2aa=2y=D2D=-2a=-2a∴ x=2 and y=-2a

Q7.

Answer :

Given: 2x + 3y =10 x + 6y = 4Using Cramer’s Rule, we get D =2 31 6 =12 – 3 = 9D1 =10 34 6=60 – 12 = 48D2= 2 101 4 = 8 – 10 =-2Now,x= D1 D = 489 = 163y=D2D = -29∴ x = 163 and y = -29

Q8.

Answer :

Given: 5x + 7y = -2 4x + 6y = -3Using Cramer’s Rule, we get D= 5 7 4 6= 30 – 28 = 2D1 =-2 7-3 6=-12 + 21 = 9D2=5 -2 4 -3= -15 + 8 = -7Now,x= D1 D = 92y = D2D = -72∴ x = 92 and y = -72

Q9.

Answer :

Given: 9x + 5y = 10 3y – 2x = 8 Rearranging the second equation, the two equations can be written as 9x + 5y = 10-2x + 3y = 8Now,D = 9 5-2 3 = 27 + 10 = 37D1 = 10 5 8 3 = 30 – 40 = -10D2 = 9 10 -2 8 = 72 + 20 = 92Using Cramer’s rule, we getx= D1D=-1037 y= D2D=9237∴ x=-1037 and y=9237

Q10.

Answer :

Given: x + 2y = 1
3x + y = 4

D=1231 = -5D1=1241 = -7D2=1134 = 1Now,x = D1D = 75y = D2D = -15∴ x = 75 and y = -15

Q11.

Answer :

Given: 3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11
D=3112-4341-3= 312 – 3 – 2-3 – 1 + 43 + 4= 27 + 8 + 28= 63D1=211-1-43-111-3=212 – 3 + 1-3 – 1 – 113 + 4=18 – 4 – 77=-63D2=3212-134-11-3=33 + 33 – 2-6 + 11 + 46 + 1=108 – 10 + 28=126D3=3122-4-141-11=344 + 1 – 2-11 – 2 + 4-1 + 8=135 + 26 + 28=189Now,x=D1D=-63 63=-1y=D2D=12663=2z=D3D=18963=3∴ x=-1, y=2 and z=3

Q12.

Answer :

Given: x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1
D=1-4-12-52-321=1-5-4-(-4)(2+6)+(-1)(4-15)=1(-9)-(-4)(8)+(-1)(-11)=34D1=11-4-139-52121=11(-5-4)-(-4)(39-2)+(-1)(78+5)=11(-9)-(-4)(37)+(-1)(83)=-34D2=111-12392-311=1(39-2)-11(2+6)+(-1)(2+117)=1(37)-11(8)+(-1)(119)=-170D3=1-4112-539-321=1(-5-78)-(-4)(2+117)+11(4-15)=1(-83)-(-4)(119)+11(-11)=272Now,x=D1D=-3434=-1y=D2D=-17034=-5z=D3D=27234=8∴ x = −1, y= −5and z=8

Q13.

Answer :

Given: 6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8
D=61-313-2214=6(12+2)-1(4+4)-3(1-6)=6(14)-1(8)-3(-5)=91D1=51-353-2814=5(12+2)-1(20+16)-3(5-24)=5(14)-1(36)-3(-19)=91D2=65-315-2284=6(20+16)-5(4+4)-3(8-10)=6(36)-5(8)-3(-2)=182D3=615135218=6(24-5)-1(8-10)+5(1-6)=6(19)-1(-2)+5(-5)=91Now,x=D1D=9191=1y=D2D=18291=2z=D3D=9191=1∴ x=1, y=2 and z=1

Q14.

Answer :

These equations can be written as
x + y + 0z = 5
0x + y + z = 3
x + 0y + z = 4

D=110011101=1(1-0)-1(0-1)+0(0-1)=1(1)-1(-1)+0=2D1=510311401=5(1-0)-1(3-4)+0(0-4)=5(1)-1(-1)=6D2=150031141=1(3-4)-5(0-1)+0(0-4)=1(-1)-5(-1)=4D3=115013104=1(4-0)-1(0-3)+5(0-1)=1(4)-1(-3)+5(-1)=2Now,x=D1D=62=3y=D2D=42=2z=D3D=22=1∴ x=3, y=2 and z=1

Q15.

Answer :

These equations can be written as
0x + 2y − 3z = 0
x + 3y + 0z = − 4
3x + 4y + 0z = 3

D=02-3130340=0(0-0)-2(0-0)-3(4-9)=15D1=02-3-430340=0(0-0)-2(0-0)-3(-16-9)=75D2=00-31-40330=0(0-0)-0(0-0)-3(3+12)=-45D3=02013-4343=0(9+16)-2(3+12)-0(4-9)=-30Now,x=D1D=7515=5y=D2D=-4515=-3z=D3D=-3015=-2∴ x=5, y=-3 and z=-2

Q16.

Answer :

Given: 5x − 7y + z = 11
6x − 8y − z = 15
3x + 2y − 6z = 7

D=5-716-8-132-6=5(48+2)+7(-36+3)+1(12+24)=5(50)+7(-33)+1(36)=55D1=11-7115-8-172-6=11(48+2)+7(-90+7)+1(30+56)=11(50)+7(-83)+1(86)=55D2=5111615-137-6=5(-90+7)-11(-36+3)+1(42-45)=5(-83)-11(-33)+1(-3)=-55D3=5-7116-815327=5(-56-30)+7(42-45)+11(12+24)=5(-86)+7(-3)+11(36)=-55Now,x=D1D=5555=1y=D2D=-5555=-1z=D3D=-5555=-1∴ x=1, y=-1 and z=-1

Q17.

Answer :

Given: 2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11

D=2-3-4-25-13-15=2(25-1)+3(-10+3)-4(2-15)=2(24)+3(-7)-4(-13)=79D1=29-3-4-155-1-11-15=29(25-1)+3(-75-11)-4(15+55)=29(24)+3(-86)-4(70)=158D2=229-4-2-15-13-115=2(-75-11)-29(-10+3)-4(22+45)=2(-86)-29(-7)-4(67)=-237D3=2-329-25-153-1-11=2(-55-15)+3(22+45)+29(2-15)=2(-70)+3(67)+29(-13)=-316Now,x=D1D=15879=2y=D2D=-23779=-3z=D3D=-31679=-4∴ x=2, y=-3 and z=-4

 

Page 6.72 Ex. 6.4

Q18.

Answer :

These equations can be written as
x+ y + 0z = 1
x + 0y + z = − 6
x − y − 2z = 3
D=1101011-1-2=1(0+1)-1(-2-1)+0(-1-0)=4D1=110-6013-1-2=1(0+1)-1(12-3)+0(6-0)=-8D2=1101-6113-2=1(12-3)-1(-2-1)+0(3+6)=12D3=11110-61-13=1(0-6)-1(3+6)+1(-1-0)=-16Now,x=D1D=-84=-2y=D2D=124=3z=D3D=-164=-4∴ x=-2, y=3 and z=-4

Q19.

Answer :

These equations can be written asx + y + z = -1ax + by + cz = -da2x + b2y + x2z = -d2D=111abca2b2c2 =100aa-bb-ca2a2-b2b2-c2 Applying C2→C1-C2 , C3→C2-C3Taking (b-a) and (c-a) common from C1 and C2, respectively, we get =(a-b)(b-c)100a11a2a+bb+c=(a-b)(b-c)(c-a) …(1)D1=-111-dbc-d2b2c2 = -111dbcd2b2c2D1=-(d-b) (b-c) (c-d) Replacing a by d in eq. (1)D2=1-11a-dca2-d2c2 = -111adca2d2c2D2=-(a-d)(d-c)(c-a) Replacing b by d in eq. (1)D3=11-1ab-da2b2-d2 = -111abda2b2d2D3=-(a-b)(b-d)(d-a) Replacing c by d in eq. (1)Thus,x = D1D = -(d-b)(b-c)(c-d)(a-b)(b-c)(c-a)y = D2D = -(a-d)(d-c)(c-a)(a-b)(b-c)(c-a)z = D3D = -(a-b)(b-d)(d-a)(a-b)(b-c)(c-a)

Q20.

Answer :

D=11111-22221-223-13-31-2221-22-13-3 – 11222-2233-3 + 11-222123-1-3 – 11-2221-23-13=1-26 – 6-2-3 + 2 + 23 – 2 -116 – 6 -2-6 – 6 + 26 + 6 + 11-3 + 2 + 2-6 – 6 + 2-2 – 3 – 113 – 2 + 26 + 6 +2-2 – 3=4 – 48 – 35 – 15=-94D1=2111-6-222-51-22-3-13-32-2221-22-13-3 -1-622-5-22-33-3 +1-6-22-512-3-1-3 -1-6-22-51-2-3-13=2-26 – 6 -2-3 + 2 + 23 – 2 -1-66 – 6 -215 + 6 +2-15 – 6 +1 -6 -3 + 2 + 215 + 6 +25 + 3 -1 -63 – 2 +2-15 – 6 +25 + 3=188D2=12111-6222-5-223-33-31-622-5-22-33-3-21222-2233-3+11-622-523-3-3-11-622-5-23-331-66 – 6 -215 + 6 +2-15 – 6 -216 – 6 -2-6 – 6 +26 + 6 + 1115 + 6 + 6-6 – 6 +2-6 + 15 -11-15 – 6 -66 + 6 +2 -6 + 15=1D3=11211-2-6221-523-1-3-31-2-621-52-1-3-3-11-622-523-3-3+21-222123-1-3-11-2-621-53-1-3=1 -215 + 6 +6-3 + 2 +2-3 – 5 -1115 + 6 + 6-6 – 6 +2-6 + 15 +21 -3 + 2 + 2-6 – 6 + 2-2 – 3 -11-3 – 5 +2-6 + 15 -6-2 – 3=-141D4=11121-22-621-2-53-13-31-22-61-2-5-13-3-112-62-2-533-3+11-2-621-53-1-3-21-2221-23-1-31-26 + 15 -2-3 – 5 -63 – 2 -116 + 15 -2-6 + 15 -66 + 6 +11- 3 – 5 +2-6 + 15 -6-2 – 3 -21 -3 – 2 +2-6 + 6 +2-2 – 3=47Thus,x = D1D = 188-94 = -2y = D2D = -282-94 = 3z= D3D = -141-94 = 1.5w = D4D = 47-94 = -0.5

Q21.

Answer :

D=20-311-1020-31111102-102-311110-0-31-120-31110-11-100-31111= 2-10-1-00-1+2-3-1-310-1+10-1+20+3-11-3-1+10-1+00+3= -21D1= 10-311-1021-31111101-102-311110-0-31-121-31110-11-101-31111=1-10-1-00-1+2-3-1-310-1+10-1+21+3-11-3-1+10-1+21+3=-21D2=21-31110201111110=2102111110-1102011110+(-3)112011110-1110011111210-1+21-1-110-1+20-1-310-1-10-1+20-1-111-1-10-1=6D3=20111-1120-3111110=2-112-311110-0+11-120-31110-11-110-31111=2-10-1-10-1+2-3-1+110-1 +10-1+20+3-11-3-1+10-1+10+3=-6D4=20-311-1010-3111111= 2-101-311111-0-31-110-31111-11-100-31111= 2-11-1+1-3-1-31-3-1+10-1+10+3-11-3-1+10-1= 3So, by Cramer’s rule , we obtain x = D1D = 2121= 1y= D2D = 6-21 = -27z = D3D = -6-21 = 27w=D4D=3-21=-17Hence, x = 1, y = -27, z = 27, w= -17

Q22.

Answer :

Given: 2x − y = 5
4x − 2y = 7

D=2-14-2 = -4 + 4 = 0D1=5-17-2=-10 + 7 = -3D2=2547 = 14 – 20 = -6

Here, D1 and D2 are non-zero, but D is zero. Thus, the given system of linear equations is inconsistent.

Q23.

Answer :

Given: 3x + y = 5
− 6x − 2y = 9

D=3 1-6-2 = -6 + 6 = 0D1=5 19-2 = -10 – 9 = -19D2=35-69 = 27 + 30 = 57

Here, D1 and D2 are non-zero, but D is zero. Thus, the system of linear equations is inconsistent.

Q24.

Answer :

Given: 3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1

D=3-122131-2-1=3-1+6+1-2-3+2-4-1=0D1=3-125131-2-1=3-1+6+1-5-3+2-10-1=-15D2=33225311-1=3-5-3-3-2-3+22-5=-15D3=3-132151-21=31+10+12-5+3-4-1=-15

Here, D is zero, but D1, D2 and D3 are non-zero. Thus, the system of linear equations is inconsistent.

Q25.

Answer :

Given: 3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20

D=3-122-113653-5 – 6 + 110 – 3 + 212 + 3 = 4

Since D is non-zero, the system of linear equations is consistent and has a unique solution.

D1=6-122-112065=6-5 – 6 + 110 – 20 + 212 + 20=-66-10+64=-12D2=3622213205=310-20-610-3+240-6=-30-42+68=-4D3=3-162-123620=3-20-12+140-6+612+3=-96+34+90=28Now,x=D1D=-12 4=-3y=D2D=-44=-1z=D3D=284=7∴ x=-3, y=-1 and z=7

Q26.

Answer :

Using the equations we get D=1-1 12 1-1-1-2 2⇒12 – 2 + 14 – 1 + 1-4 + 1 = 0D1=3-1 12 1-11-2 2⇒32 – 2 + 14 + 1 + 1-4 – 1 = 0D2= 13 1 22-1-11 2⇒14 + 1 – 34 – 1 + 12 + 2 = 0D3= 1-13 2 12-1-21⇒11 + 4 + 12 + 2 + 3-4 + 1 = 0

Here,
D=D1=D2=D3=0
Thus, the system of linear equations has infinitely many solutions.

Q27.

Answer :

Using the equations, we get D = 1236 = 6 – 6 = 0D1 = 52156 = 30 – 30 = 0D2= 15315 = 15 – 15 = 0

∴ D=D1=D2

Hence, the system of linear equation has infinitely many solutions.

Q28.

Answer :

Using the equations we get D=11-11-2 136-5⇒110 – 6 -1-5 – 3 -16 + 6 = 0D1=01-10-2 106-5⇒010 – 6 -10 – 0 -10 + 0 = 0D2=10-110 130-5⇒10 – 0 -0-5 – 3 -10 – 0 = 0D3=1101-20360⇒10 – 0 -10 – 0 + 06 + 6 = 0

∴ D=D1=D2

Hence, the system of linear equations has infinitely many solutions.

Q29.

Answer :

Using the equations we getD=2 1-21-2 15-5 1⇒2-2 + 5 – 11 – 5 -2-5 + 10 = 0D1=41-2-2-2 1-2-5 1⇒4-2 + 5 -1-2 + 2 -210 – 4 = 0D2=24-21-2 15-2 1⇒2-2 + 2 -41- 5 -2-2 + 10 = 0D3=2141-2-25-5-2⇒24 – 10 -1-2 + 10 + 4-5 + 10 = 0

∴ D=D1=D2=0

Hence, the system of linear equations has infinitely many solutions.

Q30.

Answer :

Using the equations, we get D=1-1 313-3533=1(9 + 9) + 1(3 + 15) + 3(3 – 15)=18 + 18 – 36 = 0D1=6-13-43-31033=6(9 + 9) + 1(-12 + 30) + 3(-12 – 30)=108 + 18 – 126 = 0D2=1631-4-35103 =1(-12 + 30) -6(3 + 15) + 3(10 + 20)=18 – 108 + 90 = 0D3=1-1613-45310=1(30 + 12) + 1(10 + 20) + 6(3 – 15)= 42 + 30 – 72 = 0∴ D = D1= D2= D3= 0

Hence, the system of equations has infinitely many solutions.

Q31.

Answer :

Let x, y and z be the rates of commission on items A, B and C respectively. Based on the given data, we get

90x + 100y + 20z = 800130x + 50y + 40z = 90060x + 100y + 30z = 850

Dividing all the equations by 10 on both sides, we get

9x + 10y + 2z = 8013x + 5y + 4z = 906x + 10y + 3z = 85
D=910213546103 Expressing the equation as a determinant=9(15-40)-10(39-24)+2(130-30)=9(-25)-10(15)+2(100)=-175D1=80102905485103=80(15-40)-10(270-340)+2(900-425)=80(-25)-10(-70)+2(475)=-350D2=9802139046853=9(270-340)-80(39-24)+2(1105-540)=9(-70)-80(15)+2(565)=-700D3=910801359061085=9(425-900)-10(1105-540)+80(130-30)=9(-475)-10(565)+80(100)=-1925Thus,x=D1D=-350-175=2y=D2D=-700-175=4z=D3D=-1925-175=11

Therefore, the rates of commission on items A, B and C are 2, 4 and 11, respectively.

Q32.

Answer :

Let x, y and z denote the number of cars that can be produced of each type. Then,2x + 3y + 4z = 29x + y + 2z = 133x + 2y + z = 16Using Cramer’s rule, we getD=234112321=2(1 – 4) – 3(1 – 6) + 4(2 – 3)=-6 + 15 – 4= 5D1=293413121621=29(1 – 4) – 3(13 – 32) + 4(26 – 16)=-87 + 57 + 40= 10D2=229411323161=2(13 – 32) – 29(1 – 6) + 4(16 – 39)=-38 + 145 – 92= 15D3=232911133216=2(16 – 26) – 3(16 – 39) + 29(2 – 3)=-20 + 69 – 29= 20Thus,x = D1D = 105 = 2y = D2D = 155 = 3z = D3D = 205 = 4

Therefore, 2 C1 cars, 3 C2 cars and 4 C3 cars can be produced using the three types of steel.

 

Page 6.77 Ex. 6.5

Q1.

Answer :

Given: x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0

D=11-221-354-9=1(-9 + 12) – 1(-18 + 15) – 2(8 – 5)=0So, the system has infinitely many solutions. Putting z=k in the first two equations, we getx + y = 2k2x + y = 3kUsing Cramer’s rule, we getx = D1D = 2k13k11121 = -k-1 = ky = D2D = 12k23k1121 = -k-1 = k z=kClearly, these values satisfy the third equation.Thus, x = y = z = k k∈R

Q2.

Answer :

D = 2 341 112-13 = 2 (3 + 1) -3 (3 – 2) + 4(-1 -2) = 8 – 3 – 12 = -7So, the given system of equations has only the trivial solution i.e. x = 0, y = 0, z= 0

Q3.

Answer :

Given: 3x + y + z = 0
x − 4y + 3z = 0
2x + 5y − 2z = 0

D=3 1 11-4 32 5-2=0The system has infinitely many solutions. Putting z=k in the first two equations, we get3x + y = -kx – 4y = -3kSolving these equations by Cramer’s rule, we getx = D1D = -k 1-3k-43 11-4 = -7k13y=D2D=3-k1-3k3 11-4=8k13 z = k⇒ x = -7k13, y = 8k13 and z = kor x = -7k, y = 8k and z = 13kClearly, these values satisfy the third equation.Thus, x=-7ky = 8k k∈Rz = 13k

Q4.

Answer :

The given system of equations can be written as2λx – 2y + 3z = 0x + λy + 2z = 02x + 0y + λz = 0The given system of equations will have non-trivial solutions if D=0.⇒2λ-231λ220λ = 0⇒2λ(λ2) + 2(λ – 4) + 3(-2λ) =0⇒2λ3 – 4λ – 8 =0⇒λ = 2So, the given system of equations will have non-trivial solutions if λ=2.Now, we shall find solutions for λ = 2.Replacing z by k in the first two equations, we get2λx – 2y = -3kx + λy = -2kSolving these by Cramer’s rule, we getx = -3k-2-2kλ2λ-21λ = -3kλ – 4k2λ2 + 2 = -3k(2) – 4k2(2)2 + 2 = -6k – 4k10 = -ky=2λ-3k1-2k2λ-21λ = -4kλ + 3k2λ2 + 2 = -4k(2) + 3k2(2)2 + 2 = -5k10 = -k2Substituting these values of x and y in the third equation, we getLHS = 2(-k) + 0(-k2) + 2(k) = 0 = RHSThus,λ = 2, x = -k, y = -k2 and z = k k∈R

Q5.

Answer :

The three equations can be expressed as

a – 1x – y – z = 0-x + b – 1y – z = 0-x – y + c – 1z = 0

Expressing this as a determinant, we get

∆=a-1-1-1-1b-1-1-1-1c-1

If the matrix has a non-trivial solution, then

a-1-1-1-1b-1-1-1-1c-1=0

⇒a – 1b – 1c – 1 – 1 + 1-c – 1 – 1 – 11 + b – 1 = 0⇒a – 1bc – c – b + 1 – 1 + 1-c + 1 – 1 – 1b = 0⇒a – 1bc – b – c – c – b = 0⇒abc – ab – ac – bc + b + c – b – c = 0⇒ab + ac + bc = abc

Hence proved.

 

Page 6.77 (Very Short Answers)

Q1.

Answer :

Given: A is a singular matrix.

Thus, A=0

Q2.

Answer :

If a matrix A is singular, then A=0

∴ 5 – xx + 124 = 0

⇒4(5 – x) – 2(x + 1) 0⇒20 – 4x – 2x – 2⇒18 – 6x = 0⇒18 = 6x⇒x = 3

 

Page 6.78 (Very Short Answers)

Q3.

Answer :

Let Δ= 2 3 4 2x 3x 4x 5 6 8=x 2 3 4 2 3 4 5 6 8 Taking out x common from R2=0

Q4.

Answer :

Let Δ = 2 3 6 4= 2 × 4 – 6 × 3 = 8 – 18 = -10A matrix is said to be singular if its determinant is equal to zero.Since Δ=-10≠0, the given matrix is non-singular.

Q5.

Answer :

Let Δ=4200 42014202 4203 Δ=4200 14202 1 Applying C2→C2 – C1= 4200 – 4202 =-2

Q6.

Answer :

Let Δ = 101 102 103104 105 106 107 108 109 Δ = 101 1 2104 1 2107 1 2 Applying C2→C2 – C1 and C3 →C3 – C1 = 2101 1 1104 1 1107 1 1= 0 Since two columns are identitical, the value of the determinant is zero. ⇒Δ = 101 102 103104 105 106 107 108 109 = 0

Q7.

Answer :

Let ∆ = a 1 b + cb 1 c + ac 1 a + b=a + b + c 1 b + ca + b + c 1 c + aa + b + c 1 a + b Applying C1→C1 + C3=a + b + c 1 1 b + c1 1 c + a1 1 a + b=a + b + c × 0=0

Q8.

Answer :

A = 0 ii 1 ⇒A = 0-i2 =–1 = 1Also,B = 0 11 0 ⇒B = 0 – 1 = -1 So,A + B = 1 – 1 = 0

Q9.

Answer :

A=1 23 -1 B= 1 0 -1 0 AB=1 23 -1 1 0-1 0 = 1 – 20 + 03 + 10 + 0 = -1 0 4 0AB = 0 – 0 = 0

Q10.

Answer :

Let Δ=4785 47874789 4791⇒Δ=4785 24789 2 Applying C2→C2-C1=2 × 4785 14789 1=2 × 4785 – 4789 = 2 × -4 = -8⇒4785 47874789 4791 = -8

Q11.

Answer :

1 w w2 w w2 1 w2 1 w=1+w+w2 w w2 w+w2+1 w2 1 w2 +1+w 1 w Applying C1→C1+ C2+C3= 0 w w2 0 w2 1 0 1 w ∵ 1+w+w2=0, w is the imaginary cube root of unity =0

Q12.

Answer :

A=1 23-1⇒A = -1 – 6 = -7B=1 -43-2⇒B=-2 + 12 = 10 If A and B are square matrices of the same order, then AB=AB.⇒AB = A B = -7 × 10=-70

Q13.

Answer :

If A=aij is a diagonal matrix of order n, then A=a11 × a22 × a33 × … × ann.Given: a11= 1, a22 = 2 and a33= 3⇒A = 1 × 2 × 3 = 6 Applying the above property

Q14.

Answer :

A scalar matrix is a diagonal matrix, in which all the diagonal elements are equal to a given scalar number.

Given: A = ai j is 3 × 3 matrix, where a11=2 ⇒A = 2 0 00 2 00 0 2 ⇒ A = 2 0 0 0 2 0 0 0 2 =2 × 2 0 0 2 Expanding along C1=2 × 2 × 2 = 8

Q15.

Answer :

In an identity matrix, all the diagonal elements are 1 and rest of the elements are 0.
Here,

I3=1 0 00 1 00 0 1 =1 × 1 00 1 Expanding along C1= 1⇒I3= 1

Q16.

Answer :

If A = ai j is a square matrix of order n and k is a constant, thenkA = kn A Here, Number of rows = n k is a common factor from each row of k3A = 33A = 27 × 5 = 135 Given matrix is 3×3 such that A = 5Thus, 3A = 135

Q17.

Answer :

If A=ai j is a square matrix of order n, then the sum of the products of elements of a row (or a column) with their cofactors is always equal to det (A). Therefore,

∑i=1nai j Ci j=A and ∑j=1nai j Ci j=AGiven: A = a11C11 + a12C12 + a13C13 Expanding along R1Now,A=a12 C12 + a22C22 + a32C32 Expanding along R2 a12 , a22 and a32 are elements of C2

Q18.

Answer :

If A=ai j is a square matrix of order n and Ci j is a cofactor of ai j, then∑i=1nai j Ci j=A and ∑j=1nai j Ci j=AGiven: A = 5 and matrix A is of order 3×3Since a13 C13 + a23C23 + a33C33 represent expansion of A along third column, we geta13 C13 + a23C23 + a33C33 = A = 5⇒a13 C13 + a23C23 + a33C33 = 5

Q19.

Answer :

We know that in a square matrix of order n, the sum of the products of elements of a row (or a column) with the cofactors of the corresponding elements of some other row (or column ) is zero. Therefore,

A=ai j is a square matrix of order n.⇒∑j=1nai j Ck j=0 and ∑i=1nai j Ci k = 0⇒a11C21 + a12C22 + a13 C23 = 0 Since the elements are of first row and the cofactors are of elements of second row ⇒a11C21 + a12C22 + a13 C23 = 0

Q20.

Answer :

Let Δ=sin 20° -cos 20°sin 70° cos 70° =sin 20° cos 70° + cos 20°sin 70°=sin (20° + 70°) trignometric identity=sin 90° =1

Q21.

Answer :

Let A=ai j be a square matrix of order n. Here,A=AT By property of determinantsGiven: ATA = I ⇒ATA = 1Then,ATA = AT A Since the determinants are of the same order ⇒AT A =1⇒A = 1AT⇒A = 1A ∴ A = AT⇒A2= 1⇒A = ±1

Q22.

Answer :

Since A & B are square matrices of the same order, by the property of determinants we get

AB = A × BA = 3, AB = I ⇒AB = 1⇒A × B = 13 × B = 1B = 13

Q23.

Answer :

We know that if a skew symmetric matrix A is of odd order, then A=0

Since the order of the given matrix is 3, A=0.

 

Page 6.79 (Very Short Answers)

Q24.

Answer :

A = 4 Here, Order of the matrix n = 3Using properties of matrices, we getkA = knA For a square matrix of order n and constant k⇒-A = -13 A = -1 × 4 = -4

Q25.

Answer :

In a square matrix, A = AT.Since they are of same order, A AT = AAT .Given: A=2 ⇒A AT=2 × 2 = 4

Q26.

Answer :

243 156 300 81 52 100 -3 0 4=243 -81 × 3 156 – 52 × 3 300 – 100 × 3 81 52 100 -3 0 4 Applying R1→ R1-3 R2= 0 0 0 81 52 100 -3 0 4 =0

Q27.

Answer :

A=2 -3 54 -6 106 -9 15=2 -3 54 – 4 -6 + 6 10 – 106 -9 15 Applying R2→R2-2R 1=2 -3 50 0 06 -9 15 =0

Q28.

Answer :

A matrix is said to be singular if its determinant is zero. Since the given matrix is singular, we get

A= 5x 2 -10 1 ⇒A= 5x 2 -10 1 = 0⇒5x + 20 = 0 Expanding⇒x = -205 = – 4

Q29.

Answer :

A=λ Order of A is n⇒-A=-1n A=-1nλ

Q30.

Answer :
22 23 2423 24 2524 25 26 = 22×23 ×24 1 2 221 2 221 2 22 Taking out common factors from R1, R2 and R3= 22×23 ×24×2 1 1 221 1 221 1 22=0 Two rows being identical⇒22 23 2423 24 2524 25 26=0

Q31.

Answer :

Let A & B be non-singular matrices of order n.

A≠0 and B ≠0 By definition Since they are of same order, AB=ABAB=0 iff either A=0 or B=0 But it is not the case here. Thus, AB is non-zero and AB is non-singular matrix.

Q32.

Answer :

Let A be the given matrix. Then, A = 2 Order=n = 3 I = 1 I is an identity matrix3I = 3A3I = 3A = 33A A being of order 3=27 × 2 = 54⇒A3I = 54

Q33.

Answer :

K A = KnA n is the order of A⇒ 3AB = 33 AB …(1)If A and B are square matrices of the same order, then AB=A B. So,3AB=33A B From eq. (1) =27 × -1 × 3 =-81

Q34.

Answer :

a+ibc+id-c+ida-ib=a2-iab+iab-i2b2-(-c2-icd+icd+i2d2)=a2-i2b2+c2-i2d2Here, i2=-1⇒a+ibc+id-c+ida-ib=a2+b2+c2+d2

Q35.

Answer :

Given: 2 -3 56 0 41 5-7Here, a12=-3Cofactor of a12=-11+26 41-7 = -(-42 – 4) = 46

Q36.

Answer :

2x + 5 35x + 29 = 0⇒9(2x + 5) – 3(5x + 2) = 0⇒18x + 45 – 15x – 6 = 0⇒3x + 39 = 0⇒3x = -39⇒x = -393 = -13

Q37.

Answer :

x42 2x = 0⇒2×2 – 8 = 0⇒2×2 = 8⇒x2 = 82 = 4⇒x = 4 = ±2

Q38.

Answer :

2345686x9x12x= 2 3 45 6 82 3 4 Taking 2x common from R3= 0 ∵ R1 and R3 are identical

Q39.

Answer :

For any square matrix A of order n, adjA=An-1.Given: A = 2Here, order is 2.⇒adjA = 22-1 = 2

Q40.

Answer :

020234456=0(18 – 20) – 2(12 – 16) + 0(10 – 12)=8

Q41.

Answer :

6 – x 43 – x 1 is singular when its determinant is 0.⇒6 – x 43 – x1 =0⇒6 – x – 43 – x = 0⇒6 – x – 12 + 4x = 0⇒3x – 6 = 0⇒3x = 6⇒x = 63 = 2

Q42.

Answer :

KA = Kn A Here, n is the order of A.Given: A = 4⇒2A = 23 × 4 = 32

Q43.

Answer :

cos 15° sin 15°sin 75° cos 75°=cos 15° cos 75° – sin 15° sin 75°= cos (15° + 75°) ∵cos A cos B – sin A sin B = cos (A + B)=cos 90 °=0⇒cos 15° sin 15°sin 75° cos 75° = 0

 

Page 6.80 (Multiple Choice Questions)

Q1.

Answer :

(d) A + B = O

Let A= ai j and B=bi j be a square matrix of order 2.As their orders are same, A+B is defined asA + B = ai j + bi j ⇒ A + B = ai j + bi j Now, A + B = 0⇒ai j + bi j = 0⇒ai j + bi j = 0 each corrsponding term is 0⇒A + B =0

Q2.

Answer :

(d) a + bc + de + fg + h = a ceg + b dfh

a + b c + de + f g + h = a + b ce + f g + a + b de + f h=a ce g + b cf g + a d e h + b df h

Q3.

Answer :

(d) a11 C11 + a21 C21 + a13 C31

Properties of determinants state that if A is a square matrix of the order n, then Det (A) is the sum of products of elements of a row (or a column) with the corresponding cofactor of that element.

A = a11C11 + a21C21 + a31’C31    Calculating along C1

Q4.

Answer :

(b) Minor of an element can never be equal to the cofactor of the same element.

Ci j = -1i+jMi jSo, for even values of i+j, Ci j = Mi j.

Q5.

Answer :

(d) none of these

∆=x 2 xx2 x 6x x 6=x x 6 x 6 – x2 2 x x 6 + x 2 x x 6 Expanding along C1=0 – x212 – x2 + x12 – x2=x4 – 12×2 + 12x – x3 ∆=ax4 + bx3 + cx2 + dx + e Given⇒x4 – 12×2 + 12x – x3 = ax4 + bx3 + cx2 + dx + e ⇒a=1, b=-1, c=-12, d = 12, e = 0Thus, 5a + 4b + 3c + 2d + e = 5 – 4 – 36 + 24 + 0 = -11

Q6.

Answer :

(a) n

Let A = nx, B = n + 1 x, C= n + 2 x⇒C – B = x, B – A = x, C – A = 2xThus, the given determinant is a2 a 1cos A cos B cos Csin A sin B sin C=a2 cos B sin C – cos C sin B – a × cos A sin C – cos C sin A + 1 × cos A sin B – sin A cos B=a2 sin C – B – a sin C – A + sin B – A=a2 sin x – a sin 2x + sin x Independent of n

Q7.

Answer :

(a) ∆1+∆2=0

Δ2 =1 bc a1 ca b1 ab c=1abca abc a2b bca b2c cab c2 [R1, R2, R3 are multiplied by a, b and c respectively, therefore we divide by abc]=abcabc a 1 a2b 1 b2c 1 c2 Taking abc common from C2=-1 a a21 b b21 c c2 C1↔C2We know that the value of a determinant remains unchanged if its rows and columns are interchanged. So,∆2=-1 1 1a b ca2 b2 c2 =-Δ1⇒Δ1+ Δ2 = 0

 

Page 6.81 (Multiple Choice Questions)

Q8.

Answer :

(a) 4

Dk= 1 n n 2k n2+n+2 n2 +n2k-1 n2 n2+n+2= 1 n n 1 n+2 – 22k-1 n2 n2+n+2 Applying R2→R2-R3=1 n n 0 2 – 2-n2k-1 n2 n2+n+2 Applying R2→R2-R1∑k=1nDk= 1 n n 0 2 – 2-n 1 n2 n2+n+2+ 1 n n 0 2 – 2-n 3 n2 n2+n+2+…+ 1 n n 0 2 – 2-n n n2 n2+n+2∑k=1nDk=12n2+n+2+2+nn2+1n-2-n-2n+12n2+n+2+2+nn2+2n-2-n-2n+…+12n2+n+2+2+nn2+nn-2-n-2n∑k=1nDk=n2n2+n+2+2+nn2+n-2-n-2n1+3+5+7+…+n∑k=1nDk=n2n2+n+2+2+nn2+n-2-n-2nn2∑k=1nDk=2n2+4n⇒2n2+4n=48⇒n-6n-4=0⇒n=4

Q9.

Answer :

(b) 0

Let Δ=x2+ 3x x – 1 x + 3x + 1 -2x x – 4x – 3 x + 4 3x=x2 + 3x-2x x – 4 x + 4 3x – x – 1 x + 1 x – 4x – 3 3x + x + 3x + 1 -2x x – 3 x + 4=x2 + 3x-6x – x2 + 16 – x – 1 3×2 + 3x – x2 + 7x – 12 + x + 3×2 + 5x + 4 + 2×2 – 6x=-7×4 + 16×2 + 48x + 21×3 + 8×2 – 22x – 2×3 – 12 + 8×2 + x + 3×3 + 12=-7×4 + 22×3 + 32×2 + 27x + 0But x is a root of ax4 + bx3 + cx2 + dx + e.⇒e = 0

Q10.

Answer :

(a) 4

Δ=-2a a + b a + cb + a -2b b + cc + a c + b -2c Let a + b = 2C, b + c = 2A and c + a = 2B.⇒a + b + b + c + c + a = 2A + 2B + 2C⇒2a + b + c = 2A + B + C⇒a + b + c = A + B + CAlso,a = a + b + c – b + c = A + B + C – 2A = B + C – ASimilarly,b = C + A – B, c = A + B – CΔ = 2A – 2B – 2C 2C 2B 2C 2B-2C-2A 2A 2B 2A 2C-2A-2B = 8 × A – B – C C B C B – C – A A B A C – A – B taking out 2 common from R1 R2 R3=8 × A – B C + B B B – A B – C A B + A C – B C-A-B Applying C1→C1+C2 , C2→C2 + C3=8 × A – B C + B B 0 2B A + B 2B 0 C – B Applying R2→ R1+ R2, R3→ R2 +R3=8 × A – B 2B A + B 0 C – B + 2B × C + B B 2B A + B Expanding along C1=16 BA – BC – B + C + BA + B – 2B2= 32 ABC=32b + c2c + a2a + b2=4a + bb + cc + aHence, 4 is the other factor of the determinant.

Q11.

Answer :

(d) 0

When we put x = 0 in the given matrix, then it turns out to be the skew symmetric matrix of order 3 and the determinant of the skew symmetric matrix of odd order is always 0.

Q12.

Answer :

(b) α is a root of 4ax2 + 12bx + 9c = 0 or a, b, c are in G.P.
Let Δ= a b 2aα + 3b b c 2bα + 3c2aα + 3b 2bα + 3c 0= a-b b 2aα+3b b-c c 2bα+3c2aα+3b-2bα-3c 2bα+3c 0 Applying C1→C1-C2= a – b b 2aα + 3b b – c c 2bα + 3c2a – bα + 3b – c 2bα+3c 0= a-b b 2aα+3b b-c c 2bα+3c 0 0 – 2α 2aα+3b -3 2bα+3c Applying R3→R3-2α, R1-3R2=- 2α 2aα+3b -3 2bα+3c a-b b b-c c Expanding along R3=-4aα2+12bα+9cac-b2 But Δ=0 Given⇒-4aα2+12bα+9cac-b2=0⇒4aα2+12bα+9c=0 or ac-b2=0⇒α is a root of 4ax2+12bx+9c=0or ac=b2, i.e. a, b, c are in G.P.

Q13.

Answer :

(a) 0
Δ= 1 wn w2nw2n 1 wn wn w2n 1=1+ wn+ w2n wn w2nw2n+ 1+ wn 1 wn wn+ w2n+1 w2n 1 Appplying C1→C1+ C2+C3Now,1+w+w2=0 ∵ w is a complex cube root of unity⇒1+ wn+ w2n=0 ∵ n is not a multiple of 3⇒Δ=0 wn w2n0 1 wn 0 w2n 1 = 0

Q14.

Answer :

Ar = 1 r 2r2 n n2n nn+12 2n+1⇒∑r=1nAr = ∑r=1n1 ∑r=1n r ∑r=1n 2r∑r=1n2 n n2n nn+12 2n+1As ∑r=1n1 = 1 + 1 + 1 … + 1 (n times) = n ∑r=1n r = 1 + 2 + 3 + …+ n= nn + 12 Let S=∑r=1n 2r=2 + 22 + 23= … + 2n 2S = 22 + 23=…+ 2n + 2n+1⇒2S – S = S =∑r=1n 2r= 2n+1 – 2⇒∑r=1nAr= n nn+12 2n+1-22n n n2 n nn+12 2n+1 Applying R1→R1-R2∑r=1nAr= n -n nn+12 – nn+12 2n+1-2- 2n+12n n n2 n nn+12 2n+1=0 0 -22n n n2 n nn+12 2n+1=-2 × 2n n n nn+12 =-2n3 + n2 – n2=-2n3

Q15.

Answer :

(c) negativeDiscriminant D of ax2 + 2bx + c = 2b2 – 4ac < 0 Given⇒4b2 – 4ac < 0 ⇒b2 – ac < 0, where a>0 …(1)Δ= a b ax + b b c bx + cax + b bx + c 0= ax bx ax2 + bx b c bx + cax + b bx + c 0 Applying R1→x R1=1x ax + b bx + c ax2 + bx + bx + c b c bx + cax + b bx + c 0 Applying R1→R1+R2=1x 0 0 ax2 + 2bx + c b c bx + cax + b bx + c 0 Applying R1→R1-R3=1xax2 + 2bx + c bcax + b bx + c Expanding along R1=1xax2 + 2bx + cb2x + bc – acx – bc=1xax2 + 2bx + c x b2 – ac = ax2 + 2bx + cb2 – ac < 0 From eq. (1)⇒Δ < 0

 

Page 6.82 (Multiple Choice Questions)

Q16.

Answer :

(b) 0

52 53 5453 54 55 54 55 56 =52 × 53 × 54 1 5 521 5 521 5 52 Taking out common factors from R1, R2, R3=52 × 53 × 54 × 5 1 1 521 1 521 1 52 =52 × 53× 54× 0=0

Q17.

Answer :

(b) 10

log3 512 log4 3 log3 8 log4 9 × log2 3 log8 3 log34 log3 4=log3 29 log22 3 log3 23 log22 33 × log2 3 log23 3 log322 log3 22=9 log3 2 12log23 3 log3 2 12×2 log2 3 × log2 3 13 log2 3 2 log32 2 log3 2 ∵ log bm an = nm logb a=9 log3 2 × log2 3 – 3 log3 2 × 12log23 × log2 3 × 2 log3 2 – 13 log2 3 × 2 log32 ∵ logmn × lognm =1=9 – 32 × 2 – 23=152 × 43 = 10

Q18.

Answer :

(a) 0
x + 2 x + 3 x + 2ax + 3 x + 4 x + 2bx + 4 x + 5 x + 2c= 0 0 2a + c – 2bx + 3 x + 4 x + 2bx + 4 x + 5 x + 2c Applying R1→R1+R3-R2, R1→R1 – R2 = 0 0 0x+3 x+4 x+2bx+4 x+5 x+2c ∵ a, b, c are in A.P.= 0

Q19.

Answer :

(a) 0

A + B + C = π ⇒A + C = π – B, A + B = π – C and B + C = π – AThus the determinant becomes sin π sin π-B cos C-sin B 0 tan Acos π-C tan π-A 0= 0 sin B cos C-sin B 0 tan A-cos C -tan A 0 sin π=0, sin π-B=B, cos π-C=-cos C, tan π-A=-tan AIt is a skew symmetric matrix of the odd order 3. Thus, by property of determinants, we get∆ = 0⇒ 0 sin B cos C-sin B 0 tan A-cos C -tan A 0 = 0

Q20.

Answer :

(b) 2
Let ∆=cosec x sec x sec xsec x cosec x sec xsec x sec x cosec x=cosec x3 1 sec x cosec x sec x cosec x sec x cosec x 1 sec x cosec x sec x cosec x sec x cosec x 1=cosec x3 1 tan x tan x tan x 1 tan xtan x tan x 1=cosec x3 1 – tan x tan x – 1 0 0 1 – tanx tan x – 1tan x tan x 1 Applying R1 →R1-R2, R2 →R2-R3=cosec x3 1 – tan x2 1 -1 0 0 1 -1tan x tan x 1 Taking out 1 – tan x common from R1 and R2=cosec x31 – tan x211-1tan x 1 + tan x-1 01-1 Expanding along C1=cosec x3 1 – tan x2 1 + tan x + tan x=cosec x3 1 – tan x2 1 + 2 tan x ∆= 0cosec x3 1 – tan x2 1 + 2 tan x = 0⇒1 – tan x = 0, cosec x3= 0 and 1 + 2 tan x = 0ortan x = 1, cosec x = 0 and tan x = -1 2⇒- π4 ≤ x ≤ π4 tan x = 1, tan x = -12 are 2 real roots as cosec x = 0 has no solutionThus, there are 2 solutions.

Q21.

Answer :

(d) Det A∈ 2, 4

1 sin θ 1- sin θ 1 sin θ -1 – sin θ 1= 1 sin θ 2- sin θ 1 0 -1 – sin θ 0 Applying C3→C3 + C1= 2 × – sin θ 1 -1 – sin θ Expanding along C3=2 sin2 θ + 1Given: 0≤θ≤2π ⇒-1≤sin θ≤1⇒0≤sin2 θ≤1A=2sin2 θ + 1A =2 × 1 = 2 θ = 0 =2 × 2 = 4 θ = 2π ⇒Det A∈2, 4

Comments

  1. Vishal Awathe says:
    August 5, 2016 at 2:12 pm

    Chapter 7 and chapter 8 solutions are not there.

    Reply

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