Determinants(Miscellaneous Ex.)

Miscellaneous Exercise

⇒ |A| = x(-x2-1)-sinθ(-xsinθ-cosθ)+cosθ(-sinθ+xcosθ)

⇒ |A| = -x3-x+xsin2θ+sinθcosθ-sinθcosθ+xcos2θ

⇒ |A| = -x3-x+xsin2θ+xcos2θ

⇒ |A| = -x3-x+x(sin2θ+cos2θ)

We know that sin2θ+cos2θ = 1

⇒ |A| = -x3-x+x

⇒ |A| = -x3

Hence determinant is independent of θ.

2. Without expanding the determinant, prove that

Solun:- Taking L.H.S:-

=R.H.S.   (Hence Proved……)

⇒ |A| = cosαcosβ(cosαcosβ-0)-cosαsinβ(-cosαsinβ-0)-sinα(-sinαsin2β-sinαcos2β)

⇒ |A| = cos2αcos2β+cos2αsin2β+sin2α

⇒ |A| = cos2α(cos2β+sin2β)+sin2α

We know that cos2β+sin2β = 1

⇒ |A| = cos2α+sin2α

⇒ |A| = 1

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

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

⇒ 2(a+b+c)[1(a-b)(b-a)-(b-c)(a-c)] = 0

⇒ 2(a+b+c)[-(a-b)(a-b)-(b-c)(a-c)] = 0

⇒ 2(a+b+c)[-(a2+b2-2ab)-(ab-bc-ac+c2)] = 0

⇒ 2(a+b+c)[-b2-a2+2ab-ab+bc+ac-c2] = 0

⇒ 2(a+b+c)[-b2-a2+ab+bc+ac-c2] =0

So a+b+c =0 

And -2b2-2a2+2ab+2bc+2ac-2c2 = 0

⇒ (a-b)2+(b-c)2+(c-a)2 = 0

Squares are non-negative

⇒ (a-b) = 0

⇒ (b-c) = 0

⇒ (c-a) = 0

So a=b=c

⇒ (3x+a)[1(0+a2)] = 0

⇒ x = -a/3 and a=0

Given a doesn’t equal to 0 then

⇒ x = -a/3

= abc[a(bc-ab-ac)-c(ac+bc-ab)+(a+c){(ab+ac+b2+bc)-b2}]

= abc[abc-a2b-a2c-ac2-bc2+abc+a2b+a2c+abc+abc+ac2+bc2]

= abc(4abc)

= 4a2b2c2

Cofactor = (-1)i+jMij

⇒ |B| = 1(3)-2(-1)-2(2) = 1

We know that (AB)-1 = B-1A-1

= (AB)-1

(i) [adj A]-1 = adj (A-1)       (ii) (A-1)-1 = A

Solun:- We know that transpose of cofactor matrix is adjoint matrix.

Cofactor = (-1)i+jMij

⇒ |A| = 1(14)+2(-11)+1(-5) = -13

We know that A-1 = 1/|A| (adj A)

⇒ (adj A)-1 = 1/|adj A| [adj(adj A)]

⇒ (adj A)-1 = 1/|adj A| (adj B) …………....(3)

Calculating adj of adj A hence calculate adjoint of B:-

According to equation 3:-

⇒ (adj A)-1 = 1/|adj A| (adj B)

Calculating adjoint of A-1 hence adjoint of C:-

So (adj A)-1 = adj (A-1)   (Hence Proved....)

⇒ |A-1| = -13

So (A-1)-1 = A  (Hence Proved……….)

⇒ |A| = (2x+2y)(xy-x2-y2)

⇒ |A| = - 2(x+y)(x2+y2-xy)

We know that (x+y) (x2+y2-xy) = x3+y3

⇒ |A| = - 2(x3+y3)

* Exercises 11 to 15 are not in the revised syllabus 2020-21

16. Solve the system of the following equations

Solun:- Let 1/x = u; 1/y = v;  1/z = w;

Equations are:- 2u+3v+10w = 4

4u-6v+5w = 1

6u+9v-20w = 2

Solving these equations using matrix method:-

Represent these equations in matrix form:-  AX = B

We know that X = A-1B

Cofactor = (-1)i+j Mij

⇒ |A| = 1200

⇒ x=2;  y=3;  z=5

17. If a,b,c are in A.P. , then the determinant

(A) 0     (B) 1     (C) x     (D) 2x

(Expand by C1)

⇒ |A| = (2)[(-1){(2c-2b)-(c-a)}]

⇒ |A| = (-2)[c-2b+a]

Given a, b, c are in A.p. 2b=a+c

⇒ c-2b+a = 0

Hence |A| = 0

18. If x,y,z are non-zero real numbers, then the inverse of matrix

Solun:- We know that inverse of a matrix exists if the matrix is nonsingular i.e. |A| is not equal to zero.

Hence inverse of matrix A exists.

Answer is (A)

⇒ Det(A) ∈ [2,4]  (Answer is ……...D)

Download PDF of Miscellaneous Exercise

See Also:-

Notes of Determinants

Exercise 4.1

Exercise 4.3

Exercise 4.4

Exercise 4.5

Exercise 4.6

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