Students can Download Maths Chapter 4 Determinants Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

karnataka 2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise

Question 1.
Prove that the determinant
\(\left| \begin{array}{rrr} { x } & { \sin \theta } & { \cos \theta } \\ { -\sin \theta } & { -x } & { 1 } \\ { 1 } & { 1 } & x \end{array} \right| is\quad independent\quad of\quad \theta \)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 1

KSEEB Solutions

Question 2.
With out expanding, prove that
\(\left|\begin{array}{lll}{\mathbf{a}} & {\mathbf{a}^{2}} & {\mathbf{b} \mathbf{c}} \\{\mathbf{b}} & {\mathbf{b}^{2}} & {\mathbf{c} \mathbf{a}} \\{\mathbf{c}} & {\mathbf{c}^{2}} & {\mathbf{a} \mathbf{b}}\end{array}\right|=\left|\begin{array}{ccc}{\mathbf{1}} & {\mathbf{a}^{2}} & {\mathbf{a}^{3}} \\{\mathbf{1}} & {\mathbf{b}^{2}} & {\mathbf{b}^{3}} \\{\mathbf{1}} & {\mathbf{c}^{2}} & {\mathbf{c}^{3}}\end{array}\right|\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 2
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 3

Question 3.
Evaluate
\(\left|\begin{array}{ccc}{\cos \alpha \cos \beta} & {\cos \alpha \sin \beta} & {-\sin \alpha} \\{\sin \alpha \cos \beta} & {\sin \alpha \sin \beta} & {\cos \alpha}\end{array}\right|\)
Answer:
expand along c3
-sin α (-sin α sin2β – sin α cos2β) + cosa (cosα cos2 β+cosα sin2β)
= sin2 α (sin2 β + cos2 β )+cos2 α (cos2 β + sin2 β)
= sin2α + cos2α= 1

Question 4.
It a, b, c are real, find the factors of the determination
\(\Delta=\left|\begin{array}{lll}{\mathbf{b}+\mathbf{c}} & {\mathbf{c}+\mathbf{a}} & {\mathbf{a}+\mathbf{b}} \\{\mathbf{c}+\mathbf{a}} & {\mathbf{a}+\mathbf{b}} & {\mathbf{b}+\mathbf{c}} \\{\mathbf{a}+\mathbf{b}} & {\mathbf{b}+\mathbf{c}} & {\mathbf{c}+\mathbf{a}}\end{array}\right|\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 4
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 5

Question 5.
Solve if a ≠ 0 and
\(\left|\begin{array}{ccc}{\mathbf{x}+\mathbf{a}} & {\mathbf{x}} & {\mathbf{x}} \\{\mathbf{x}} & {\mathbf{x}+\mathbf{a}} & {\mathbf{x}} \\{\mathbf{x}} & {\mathbf{x}} & {\mathbf{x}+\mathbf{a}}\end{array}\right|=\mathbf{0}\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 6

KSEEB Solutions

Question 6.
Prove that
\(\left|\begin{array}{ccc}{\mathbf{a}^{2}} & {\mathbf{b} \mathbf{c}} & {\mathbf{a c}+\mathbf{c}^{2}} \\{\mathbf{a}^{2}+\mathbf{a} \mathbf{b}} & {\mathbf{b}^{2}} & {\mathbf{a c}} \\{\mathbf{a b}} & {\mathbf{b}^{2}+\mathbf{b c}} & {\mathbf{c}^{2}}\end{array}\right|=4 \mathbf{a}^{2} \mathbf{b}^{2} \mathbf{c}^{2}\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 7
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 8

KSEEB Solutions

Question 7.
\(\begin{aligned}&\text { If } \mathbf{A}^{-1}=\left[\begin{array}{rrr}{3} & {-1} & {1} \\{-15} & {6} & {-5} \\{5} & {-2} & {2}\end{array}\right], \mathbf{B}=\left[\begin{array}{rrr}{1} & {2} & {-2} \\{-1} & {3} & {0} \\{0} & {-2} & {1}\end{array}\right] \text { find }\\&(\mathbf{A} \mathbf{B})^{-1}\end{aligned}\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 9

Question 8.
\(\text { Let } A=\left[\begin{array}{ccc}{1} & {-2} & {1} \\{-2} & {3} & {1} \\{1} & {1} & {5}\end{array}\right], \text { verify that }\)
(i) (adj A)1 = Adj (A1)
(ii) (A1)1 = A
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 10
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 11
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 12
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 13

Question 9.
Evaluate
\(\left[\begin{array}{ccc}{\mathbf{x}} & {\mathbf{y}} & {\mathbf{x}+\mathbf{y}} \\{\mathbf{y}} & {\mathbf{x}+\mathbf{y}} & {\mathbf{x}} \\{\mathbf{x}+\mathbf{y}} & {\mathbf{x}} & {\mathbf{y}}\end{array}\right]\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 14
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 15

KSEEB Solutions

Question 10.
\(\left[\begin{array}{ccc}{\mathbf{1}} & {\mathbf{x}} & {\mathbf{y}} \\{\mathbf{1}} & {\mathbf{x}+\mathbf{y}} & {\mathbf{y}} \\{\mathbf{1}} & {\mathbf{x}} & {\mathbf{x}+\mathbf{y}}\end{array}\right]\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 16

Prove the following using properties

Question 11.
\(\left[\begin{array}{lll}{\alpha} & {\alpha^{2}} & {\beta+\alpha} \\{\beta} & {\beta^{2}} & {\gamma+\alpha} \\{\gamma} & {\gamma^{2}} & {\alpha+\beta}\end{array}\right]=(\beta-\alpha)(\gamma-\alpha)(\alpha-\beta)\)
Answer:

2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 17

KSEEB Solutions

Question 12.
\(\begin{aligned}&\left[\begin{array}{lll}{\mathbf{x}} & {\mathbf{x}^{2}} & {\mathbf{1}+\mathbf{p} \mathbf{x}^{3}} \\{\mathbf{y}} & {\mathbf{y}^{2}} & {\mathbf{1}+\mathbf{p} \mathbf{y}^{3}} \\{\mathbf{z}} & {\mathbf{z}^{2}} & {\mathbf{1}+\mathbf{p}\mathbf{z}^{3}}\end{array}\right]=(\mathbf{x}-\mathbf{y})(\mathbf{y}-\mathbf{z})(\mathbf{z}-\mathbf{x})\\&(1+p x y z)\end{aligned}\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 18

Question 13.
\(\left[\begin{array}{rrr}{3 a} & {-a+b} & {-a+c} \\{-b+a} & {3 b} & {-b+c} \\{-c+a} & {-c+b} & {3 c}\end{array}\right]=3(a+b+c)(a b+b c+c a)\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 19
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 20

Question 14.
\(\left|\begin{array}{ccc}{1} & {1+p} & {1+p+q} \\{2} & {3+2 p} & {4+3 p+2 q} \\{3} & {6+3 p} & {10+6 p+3 q}\end{array}\right|=1\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 21

Question 15.
\(\left|\begin{array}{ccc}{\sin \alpha} & {\cos \alpha} & {\cos (\alpha+\delta)} \\{\sin \beta} & {\cos \beta} & {\cos (\beta+\delta)} \\{\sin \gamma} & {\cos \gamma} & {\cos (\gamma+\delta)}\end{array}\right|=0\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 22

KSEEB Solutions

Question 16.
Solve
\(\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4\)
\(\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1\)
\(\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 23
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 24

Question 17.
It a, b, c are in AP, then the determinant
\(\left|\begin{array}{ccc}{x+2} & {x+3} & {x+2 a} \\{x+3} & {x+4} & {x+2 b} \\{x+4} & {x+5} & {x+2 c}\end{array}\right|\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 25
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 26

KSEEB Solutions

Question 18.
If x, y, z are real, then the inverse of
\(\mathbf{A}=\left|\begin{array}{lll}{\mathbf{x}} & {\mathbf{0}} & {\mathbf{0}} \\{\mathbf{0}} & {\mathbf{y}} & {\mathbf{0}} \\{\mathbf{0}} & {\mathbf{0}} & {\mathbf{z}}\end{array}\right|\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 27

Question 19.
Let
\({ A }=\left[ \begin{array}{ccc} { { 1 } } & { \sin \theta } & { { 1 } } \\ { -\sin \theta } & { { 1 } } & { \sin \theta } \\ { -1 } & { -\sin \theta } & { { 1 } } \end{array} \right] { 0 }\leq \theta \leq 2\pi \)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 28

2nd PUC Maths Chapter 4 Determinants Miscellaneous Exercise Additional Questions and Answers

Try These Questions

Question 1.
\(\text { If } \mathbf{A}=\left[\begin{array}{lll}{1} & {2} & {2} \\{2} & {1} & {2} \\{2} & {2} & {1}\end{array}\right] \text { find } A^{-1}\)
and hence prove that A2 – 4A – 5 I = 0
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 29

Question 2.
use product
\(\left[\begin{array}{ccc}{1} & {-1} & {2} \\{0} & {2} & {-3} \\{3} & {-2} & {4}\end{array}\right]\left[\begin{array}{ccc}{-2} & {0} & {1} \\{9} & {2} & {-3} \\{6} & {1} & {-2}\end{array}\right]\)
solve x – y + 2z = 1
2y – 3z = 1
3x – 2y + 4z = 2
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 30
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 31

KSEEB Solutions

Question 3.
If the point (x, y), (a, 0), (0, b) are collinear, prove that
\(\frac{x}{a}+\frac{y}{b}=1\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 32

Question 4.
\(\text { If } \Delta=\left|\begin{array}{ccc}{\mathbf{1}} & {\mathbf{1}} & {\mathbf{1}} \\{\mathbf{1}} & {\mathbf{1}+\mathbf{x}} & {\mathbf{1}} \\{\mathbf{1}} & {\mathbf{1}} & {\mathbf{1}+\mathbf{y}}\end{array}\right|, \text { then } \Delta \text { is }\)
(a) Divisible by neither x nor y
(b) Divisible by y but not x
(c) Divisible by x but not y
(d) Divisible by both x and y [AIEEE 2007]
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 33

Question 5.
If a, b, c are positive and unequal, show that the value of the determinant
\(\left|\begin{array}{lll}{\mathbf{a}} & {\mathbf{b}} & {\mathbf{c}} \\{\mathbf{b}} & {\mathbf{c}} & {\mathbf{a}} \\{\mathbf{c}} & {\mathbf{a}} & {\mathbf{b}}\end{array}\right|\) is always negative [Kerala CET 09]
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 34
since a, b, c are positive a + b + c is positive and (a – b)2 is always positive,
Δ is always negative.

Question 6.
Using properties of matrices prove that 1+abc is a factor of [CET]
\(\left|\begin{array}{lll}{\mathbf{a}} & {\mathbf{a}^{2}} &{\mathbf{1}+\mathbf{a}^{3}} \\{\mathbf{b}} & {\mathbf{b}^{2}} & {\mathbf{1}+\mathbf{b}^{3}} \\{\mathbf{c}} &{\mathbf{c}^{2}} &{\mathbf{1}+\mathbf{c}^{3}}\end{array}\right|\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 35

Question 7.
For what values of y is the matrix  [CBSE 2011]
\(\mathbf{A}=\left|\begin{array}{rr}{\mathbf{y}^{2}+6} & {\mathbf{2} \mathbf{y}} \\{\mathbf{y}+3} & {\mathbf{2}} \end{array}\right|\)singular.
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 36

KSEEB Solutions

Question 8.
Find y
\(\left[\begin{array}{cc}{x} & {x-y} \\{2 x+y} & {7}\end{array}\right]=\left[\begin{array}{ll}{3} & {1} \\{8} & {7}\end{array}\right]\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 37

Question 9.
If A be a square matrix of order 3 x 3 then find |KA|
|KA| = kn |A| when n is the order [CBSE 2007]
Answer:
|KA| = k3 |A|

Question 10.
Evaluate  [PUC 2009]
\(\left|\begin{array}{cccc}{\sin 30} & {\cos } & {30} \\{-\sin 60} & {\cos } & {60}\end{array}\right|\)
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 38

Question 11.
From the matrix equation AB = AC, we can conclude that B = C provided
(A) A is singular
(B) A is non singular
(C) A is symmetric
(D) A is skew symmetric [IIT]
Answer:
AB = AC
A-1 (AB) = A-1 (AC) ⇒ (A-1 A) B = (A-1 A) C ⇒ IB = IC ⇒ B = C
AB = AC ⇒ B = C if A1 exist
ie,. A is non singular
correct Answer is (B)

Question 12.
A2 – A + I = 0, Find the inverse if A [IIT]
Answer:
A2 – A + I = 0
Pre – multiply by A-1
A-1 A2 – A-1 A + A-1  I = 0
A -1 + A1 = 0
A-1= I – A

KSEEB Solutions

Question 13.
It is a triangle
\(\mathbf{A B C},\left|\begin{array}{lll}{\mathbf{1}} & {\mathbf{a}} & {\mathbf{b}} \\{\mathbf{1}} & {\mathbf{c}} & {\mathbf{a}} \\{\mathbf{1}} & {\mathbf{b}} & {\mathbf{c}}\end{array}\right|=\mathbf{0}\)
then find the value of sin2A+ sin2B+ sin2C [Karnataka CET 2002]
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 39

Question 14.
Let A, B be two square matrices such that A + B = AB, then  [Karnataka CET]
(A) AB = BA
(B) AB = -BA
(C) AB + 2BA = 0
(D) None of these
Ans:
A + B = AB A + B – AB = 0
A + B – AB + I = 0+1
(A – I)(B – I) = 1
⇒ A -1 inverse of B -1
(B -1) (A -1) = I
BA – B – A + = I
BA – (B + A) ⇒ BA = B + A
= A + B  ∴ AB = BA
Answer (A)

Question 15.
\(\mathbf{A}(\alpha)=\left[\begin{array}{cc}{\cos \alpha} & {\sin \alpha} \\{-\sin \alpha} & {\cos \alpha}\end{array}\right]\),
then show that A(α)A(β) = A (α+β) [Kerala CET]
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 40

KSEEB Solutions

Question 16.
\(\mathbf{A}=\left[\begin{array}{ll}{\mathbf{1}} & {-\mathbf{1}} \\{\mathbf{2}} & {-\mathbf{1}}\end{array}\right], \mathbf{B}=\left[\begin{array}{ll}{\mathbf{a}} & {-\mathbf{1}} \\{\mathbf{b}} & {-\mathbf{1}}\end{array}\right] \text { and }(\mathbf{A}+\mathbf{B})^{2}\)
= A2 + B2 Find the values of a, b
Answer:
2nd PUC Maths Question Bank Chapter 4 Determinants Miscellaneous Exercise 41