PROBLEMS ON PROPERTIES OF DETERMINANTS(4 OR 3 marks)Method to evaluate the Determinants.(Created and DTP by KH VASUDEVA)

1. First Verify whether there is any common factor any row or column, try to get common factor by making transformations like R1+R2+R3 or C1+C2+C3 and R2-R1, R3-R1or C2-C1, C3-C1 etc,

2. Try to get one of the factors of RHS of Determinants by suitable transformations like R2-R1, R3-R1or C2-C1, C3-C1 etc.

3. Try to reduce all elements of any row or column to zero and also reduce the row or column containing elements as 1 to zeroes.

1.Proving Problems involving operations like R2-R1, R3-R1

a) 1 a a2 b) 1 a a3

1 b b2 =(a-b)(b-c)(c-a) 1 b b3 =(a-b)(b-c)(c-a)(a+b+c) 1 c c2 1 c c3

d) 1 b+c b2+c2 e) 1 a a2-bc 1 c+a c2+a2 =(a-b)(b-c)(c-a) 1 b b2-ca =0 1 a+b a2+b2 1 c c2-ab

2. Problems involving operations like C2-C1, C3-C1

a) 1 1 1 b) 1 1 1 bc ca ab =(a-b)(b-c)(c-a) a b c =(a-b)(b-c)(c-a)(a+b+c) b+c c+a a+b a3 b3 c3

c) 1 1 1 d) 1 1 1 a b c =(a-b)(b-c)(c-a) a2 b2 c2 =(a-b)(b-c)(c-a)(ab+bc+ca) a2 b2 c2 a3 b3 c3

3. Other types involving R 1-R2, R2-R3

a) a2+bc a 1 b) a2+2a 2a+1 1

b2+ca b 1 =-2(a-b)(b-c)(c-a) 2a+1 a+2 1 =(a-1)3 c2+ab c 1 3 3 1

4. Operations involving R 1+R2+R3 or C1+C2+C3

a)a+b+2c a b b) x p q c b+c+2a b =2(a+b+c)3 p x q =(x-p)(x-q)(x+p+q) c a c+a+2b p q x

c) x+a b c d) 1+a b c a x+b c =x2(x+a+b+c) a 1+b c = (1+a+b+c) a b x+c a b 1+c

e) a+b+c -c -b

- c a+b+c -a =2(a+b)(c+a)(a+b) (Hint: R1+R2+R3, &C2+C1,C3+C1) -b -a a+b+c

f) a-b-c 2a 2a

2b b-c-a 2b =(a+b+c)3 (Hint: R1+R2+R3,& C2-C1,C3-C1)

2c 2c c-a-b

1

g) a-2b-2c 3b 3c 3a b-2c-2a 3c =4(a+b+c)3

(Hint: C1+C2+C3R 2-R1,R3-R1) 3a 3b c-2a-2b

If w is an imaginary cube roots of unity then evaluateh) 1 w w2

w w2 1 (Hint: C1+C2+C3 and expand through first column) w2 1 w

5. Other problems a) 0 ab2 ac2 -a2 ab ac a2b 0 bc2 =2a3b3c3 ab -b2 bc =4a2b2c2

a2c b2c 0 ac -bc -c2

Note: Evaluate by taking common factors a,b,and c outside the determinant.

c) a2 bc ac+c2

a2+ab b2 ac =4a2b2c2(Note: Take commonfactors ab,c outside the determinant

ab b2+bc c2 and apply R1

+R2

-R3

and C1-C

3 OR C

1-(C

2+C

3). Common factor should be

taken at every time)

d) b+c a a b c+a b =4abc(Hint: apply R1-(R2+R3) and evaluate) c c c+b e) b+c c b c c+a a =4abc (Hint: apply C1-(C2+C3) and evaluate) b a a+b

f) bc a a2

ca b b2 =(a-b)(b-c)(c-a)(ab+bc+ca)( hint:apply R2-R1, R3-R1, take out b-a ab c c2 from R2 and c-a from R3, then apply R3-R2 , take out (c-b) from

R3 , and apply C3-C1)

6. P rove that a) b+c c+a a+b a b c q+r r+p p+q = 2 p q r y+z z+x x+y x y z

b) a+b b+c c+a a b c b+c c+a a+b = 2 b c a c+a a+b b+c c a b

c) b+c c+a a+b a b c c+a a+b b+c = 2 b c a a+b b+c c+a c a b

Hint: Take C1+C2+C3 and take commonfactor 2 outside the determinant 2) Take C2-C1 and C3-C1 3) take again C1+C2+C3 4)interchange C1 and C2

5) Interchange C1and C3 OR Apply C1+C2+C3 ,take 2 outside and apply C1-C2,C3-C1,C2-C3 ; C1↔C2 ,C2↔C3

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Miscellaneous problems.

7. a) a b+c a2

b-a a-b b2-a2 =(b-a)(c-a)(b-c)(a+b+c) (Hint: Take commonfactor (b-a) in R2,

c-a a-c c2-a2 (c-a) in R3, and apply C1+C2)

b) bc a a2

ca b b2 =(a-b)(b-c)(c-a)(ab+bc+ca) (Hint: Applyi) R2-R1&R3-R1 ii) R3-R2 ab c c2 iii) C3-C1 (take common factor at each step))

c) a-x a-y a-z b-x b-y b-z =0 (Hint: R2-R1,R3-R2) c-x c-y c-z

d) (b+c)2 a2 b2

b2 (c+a)2 b2 =2abc(a+b+c)3 (Hint: C1-C3&C2-C3, take out (a+b+c) c2 a2 (a+b)2 from C1&C2 and evaluate directly)

e) -bc b2+bc c2+bc a2+ac -ac c2+ac =(ab+bc+ca)3 (Hint: Multiply and divide R1by a,R2 a2+ab b2+ab -ab by b, R3 by c, take out a,b,c from C1, C2,C3 then apply R1+R2+R3,then C2-C1, C3-C1) f) b2 +c2 ab ac ab c2 +a2 bc (Ans: 4a2b2c2 ) Hint: Multiply C1 by a C2 by b C3 by c, ca cb a2 +b2

divide by abc, take out common factor abc and apply C1-C2 –C3 )

g) If a, b, c are in AP Show that x+1 x+2 x+a x+2 x+3 x+b = 0 (Hint : Apply R2-R1, R3-R1) x+3 x+4 x+c

h) Without expanding Prove that

1 a bc 1 a a2

1 b ca = 1 b b2 (Hint: Multiply R1by a,R2 by b ,R3 by c, 1 c ab 1 c c2 1/abc as common factor , take C2↔C3, C2↔C3)

i) 1+a 1 1 1 1+b 1 = abc(1+1/a+1/b+1/c) (Hint : Expand as usual get abc+bc+ca+ab, 1 1 1+c take abc as common factor)

j) 1+x 1 1 1 1+y 1 =0 where x#0,y#0 and Z#0, then show that 1+∑(1/x)=0 1 1 1+z (Hint: same as in above problem ,By condition abc#0)

k) If a,b,c are all different and a a2 1+a3

3

b b2 1+b3 =0 Prove that 1+abc=0

c c2 1+c3

(Hint: Expand using property" Sum of two terms in any row or colum can be expressed as the sum of two determinants, apply R1-R2, R2-R3after taking 1+abc as common factor.since a,b,c are different, (a-b)(b-c)(c-a)#0)

l) Prove that x x2 y+z

y y2 z+x =(x-y)(y-z)(z-x)(x+y+z) (Hint: Get expressions of RHS

z z2 x+y by R1-R2&R2-R3, then apply c1-c2, and then c1+c3

m) Using properties of Determinants Prove that

x +y x x

5x + 4y 4x 2x =x3 (Hint: Express first column as sum of two determinants, take

10x +8y 8x 3x common factor in each determinant and apply C2-C1, C3-C2)

n) If a,b,c (are all +ve ) are the pth qth rth terms respectively of a GP then Prove that loga p 1 logb q 1 =0 (Hint: a = Arp-1, b=Arq-1, c= Arr-1 A is first term r is C.R. Taking logc r 1 logerithmes write log a = logA + (p-1) logr, similarly for logb,

logc , then express as sum of two determinants, take common factor and apply C1-C2-C3)

8. Solving Determinants.

1. Find x without expanding

A) a) 1 2 3 b) 2 3 1 c) 1 -1 2

2 x 3 =0 4 x -2 =0 3 4 5 =0

3 4 3 1 3 3 1 x 2

B) a) x+1 x+2 3 b) x+2 3 4

3 x+2 x+1 =0 2 x+3 4 =0

x+1 2 x+3 2 3 x+4

c) x+2 x+6 x-1 d) 2x+7 x+4 x+3

x+6 x-1 x+2 =0 Prove that 2x+7=0 x+4 2x+6 x+2 =0

x-1 x+2 x+6 x+3 x+2 2x+5

e) 1 -2 x+3 (Answers: A) a) x=3 b) x=6 c) x=-1

1 x-2 3 =0 B) a) x=0 or 3 b) x=0 or -9 d) x=-2 ,-3,-4 e) x=0 or -2 x+1 -2 3

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