Questio
Chapter 1: Relations and functions.
n
Question numbers 1 to 5 are of each 1 mark and question numbers 6 to 10 are of 4
numbers
marks each.
1.
Let A = {1, 2, 3}, B = {4, 5, 6} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to
B. State whether f is one-one or not.
Answer
Since f(1) = 4, f(2) = 5, f(3) = 6 that is different elements of doain have different
f-images in the range.
Therefore f is one-one.
2.
1
If 𝑓: 𝐑 → 𝐑 be defined by 𝑓(𝑥) = (3 − 𝑥 3 )3 , then find 𝑓𝑜𝑓(𝑥).
Answer
𝑓𝑜𝑓(𝑥) = 𝑓 {(3 −
3.
1
𝑥 3 )3 }
= [3 − {(3 −
1
1 3 3
𝑥 3 )3 } ]
1
= [3 − 3 + 𝑥 3 ]3 = 𝑥
If binary operation * is defined as 𝑎 ∗ 𝑏 = 𝑎. 𝑏 + 𝑎 + 1, then evaluate (3 ∗ 4) ∗ (5 ∗
6).
Answer
4.
(3 ∗ 4) ∗ (5 ∗ 6) = (3 × 4 + 3 + 1) × (5 × 6 + 5 + 1) = 16 × 36 = 576
If * is a binary operation on positive rational numbers and defined by 𝑎 ∗ 𝑏 =
𝑎𝑏
4
Answer
, ∀𝑎, 𝑏 ∈ 𝑄 +, find the identity element.
Let e be the binary element such that𝑥 ∗ 𝑒 = 𝑥 = 𝑒 ∗ 𝑥 ∀ 𝑥 ∈ 𝑄 +
Thus
𝑥𝑒
4
=𝑥=
𝑒𝑥
4
𝑜𝑟 𝑒 = 4 ∈ 𝑄 +
5.
Find the domain of definition of the function f(x) = log|x|.
Answer
F(x) = log|x|, for real domain, 𝑥 ≠ 0, the domain of , 𝑓(𝑥)𝑖𝑠 (−∞, 0) ∪ (0, ∞)
6.
If 𝑓(𝑥) = 𝑥 3 − 𝑥 and 𝑔(𝑥) = 𝑆𝑖𝑛 2𝑥, then find the value of (𝑓𝑜𝑔) ( )
Answer
(𝑓𝑜𝑔)(𝑥) = 𝑓(𝑔(𝑥))
𝜋
12
= 𝑓(𝑆𝑖𝑛 2𝑥)
= 𝑆𝑖𝑛3 2𝑥 − 𝑆𝑖𝑛 2𝑥
𝜋
𝜋
𝜋
12
6
6
Therefore (𝑓𝑜𝑔) ( ) = 𝑆𝑖𝑛3 − 𝑆𝑖𝑛
1 3 1
3
= ( ) − = −
2
2
8
7.
Show that the function 𝑓: 𝐑 → 𝐑 defined by 𝑓(𝑥) =
2𝑥−1
3
, 𝑥 ∈ 𝑹 is one-one and onto
function. Also find inverse of the function f.
Answer
The function 𝑓: 𝐑 → 𝐑 defined by 𝑓(𝑥) =
I.
2𝑥−1
3
,𝑥 ∈ 𝑹
F is one-one: Let 𝑥1 , 𝑥2 ∈ 𝑅 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑓(𝑥1 ) = 𝑓( 𝑥2 )
⇒
2𝑥1 − 1 2𝑥2 − 1
=
3
3
⇒ 𝑥1 = 𝑥2
(Page 1 of 12)
Therefore f is One-One.
II.
F is Onto: Let 𝑦 ∈ 𝑹 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑓(𝑥) = 𝑦
2𝑥 − 1
3
3𝑦 + 1
⟹𝑥=
∈𝑹
2
3𝑦 + 1
(
)−1
3𝑦 + 1
2
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑓 (
)=2
= 𝑦.
2
3
⟹𝑦=
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒𝑓 𝑖𝑠 𝑂𝑛𝑡𝑜.
𝑇ℎ𝑢𝑠 𝑓 𝑖𝑠 𝑜𝑛𝑒 − 𝑜𝑛𝑒 𝑎𝑛𝑑 𝑂𝑛𝑡𝑜.
𝑆𝑖𝑛𝑐𝑒 𝑓(𝑥) = 𝑦 ⟹ 𝑥 = 𝑓 −1 (𝑦) =
⟹ 𝑓 −1 (𝑥) =
8.
3𝑥 + 1
2
Examine which of the following is a binary operation (i) 𝑎 ∗ 𝑏 =
𝑏=
Answer
3𝑦 + 1
2
𝑎+𝑏
2
𝑎+𝑏
2
, 𝑎, 𝑏 ∈ 𝑁, (𝑖𝑖)𝑎 ∗
∈ 𝑄. For binary operation check the commutative and associative property.
𝑎+𝑏
, 𝑎, 𝑏 ∈ 𝑁 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑏𝑖𝑛𝑎𝑟𝑦 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛
2
1+2 3
𝑏𝑒𝑐𝑎𝑢𝑠𝑒1,2 ∈ 𝐍 𝑎𝑛𝑑 1 ∗ 2 =
= ∉𝐍
2
2
𝑎∗𝑏 =
And for all 𝑎, 𝑏 ∈ 𝑄 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑎 ∗ 𝑏 =
𝑎+𝑏
2
∈𝑄
Therefore it is binary operation on Rational numbers.
Commutativity: 𝑎 ∗ 𝑏 =
⟹ 𝑎∗𝑏 =
𝑎+𝑏
2
, 𝑎, 𝑏 ∈ 𝑄
𝑏+𝑎
, 𝑎, 𝑏 ∈ 𝑄 ⟹ 𝑎 ∗ 𝑏 = 𝑏 ∗ 𝑎, 𝑖𝑛 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠.
2
Associativity: 𝑎 ∗ (𝑏 ∗ 𝑐) = 𝑎 ∗ (
𝑏+𝑐
2
) , 𝑓𝑜𝑟 𝑎, 𝑏, 𝑐 ∈ 𝑄
𝑏+𝑐
2 = 2𝑎 + 𝑏 + 𝑐 𝑓𝑜𝑟 𝑎, 𝑏, 𝑐 ∈ 𝑄
𝑎 ∗ (𝑏 ∗ 𝑐) =
2
4
𝑏+𝑐
𝑎 ∗ (𝑏 ∗ 𝑐) = 𝑎 ∗ (
) , 𝑓𝑜𝑟 𝑎, 𝑏, 𝑐 ∈ 𝑄
2
𝑎+𝑏
+𝑐
𝑎+𝑏
2
(𝑎 ∗ 𝑏) ∗ 𝑐 =
∗𝑐 =
𝑓𝑜𝑟 𝑎, 𝑏, 𝑐 ∈ 𝑄
2
2
𝑎+
(𝑎 ∗ 𝑏) ∗ 𝑐 =
𝑎 + 𝑏 + 2𝑐
𝑓𝑜𝑟 𝑎, 𝑏, 𝑐 ∈ 𝑄
4
Therefore, 𝑎 ∗ (𝑏 + 𝑐) ≠ (𝑎 ∗ 𝑏) ∗ 𝑐
Or binary operation is not Associative.
9.
Let X be non-empty set. P(X) be its power set. Let ‘*’ be an operation defined on
(Page 2 of 12)
elements of P(x) by A ∗ B = A ∩ B ∀ A, B ∈ P(X), then (i) Prove that * is a binary
operation in P(X). (ii) Is * commutative? (iii) Is * Associative? (iv) Find the identity
element in P(X) w.r.t *.
Ans
wer
Given: ∗: 𝑃(𝑋) × 𝑃(𝑋) → 𝑃(𝑋)𝑑𝑒𝑓𝑖𝑛𝑒𝑑𝑎𝑠 𝐴 ∗ 𝐵 = 𝐴 ∩ 𝐵 ∀ 𝐴, 𝐵 ∈ 𝑃(𝑋)
(i)
Let A and B are any two non-empty subsets of X, then we know that A ∩
B=X
Now for any two non empty elements of P(X), we have A ∩ B ∈ P(X) ⟹
A ∗ B ∈ P(X)for A, B ∈ P(X)
Therefore, ‘*’ is binary operation in P(X).
(ii)
For any two non-empty subsets A and B of X, then we know that
A∩B=B∩A ⟹ A∗B=B∗A
Therefore ‘*’ is commutative.
(iii)
For any three non-empty subsets A, B, C∈ 𝑃(𝑋)
We have (𝐴 ∩ 𝐵) ∩ 𝐶 = 𝐴 ∩ (𝐵 ∩ 𝐶) ⟹ (𝐴 ∗ 𝐵) ∗ 𝐶 = 𝐴 ∗ (𝐵 ∗ 𝐶)
Therefore * is associative.
(iv)
Let I be the identity element in P(X).
A*I = I*A = A ∀𝐴 ∈ 𝑃(𝑋)
⟹𝐴∩𝐼 =𝐼∩𝐴 =𝐴
Therefore 𝑋 ∩ 𝐼 = 𝐼 ∩ 𝑋 = 𝑋 ∈ 𝑃(𝑋)
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑋 ⊂ 𝐼 𝑎𝑛𝑑 𝐼 ⊂ 𝑋
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐼 = 𝑋.
Thus I is the identity element.
10.
Given a non-empty set X, let ∗ : 𝑃(𝑋) × 𝑃(𝑋) → 𝑃(𝑋) be define as 𝐴 ∗ 𝐵 = (𝐴 − 𝐵) ∪
(𝐵 − 𝐴), ∀ 𝐴, 𝐵 ∈ 𝑃(𝑋). Show that the empty set ∅ is the identity fr the operation *
and all the elements A of P(X) are invertible with 𝐴−1 = 𝐴.
Answer
Given let ∗ : 𝑃(𝑋) × 𝑃(𝑋) → 𝑃(𝑋)be define as 𝐴 ∗ 𝐵 = (𝐴 − 𝐵) ∪ (𝐵 − 𝐴), ∀ 𝐴, 𝐵 ∈ 𝑃(𝑋)
I Part: Let∅ be an identity element.
∴ 𝐴 ∗ ∅ = ∅ ∗ 𝐴 = 𝐴 ∀ 𝐴 ∈ 𝑃(𝑋)
𝑆𝑜 𝐴 ∗ ∅ = (𝐴 − ∅) ∪ (∅ − 𝐴)
= (𝐴 ∩ ∅′ ) ∪ (∅ ∩ 𝐴′ )
= (𝐴 ∪ 𝐔) ∪ ∅
= A∪∅ =A
⟹ ∅ is the identity element.
II Part: Let 𝐴 ∈ 𝑃(𝑋)
∴ 𝐴 ∗ 𝐴 = (𝐴 − 𝐴) ∪ (𝐴 − 𝐴)
(Page 3 of 12)
= ∅∪∅= ∅
𝑆𝑜 𝐴 ∗ 𝐴 = ∅ ∀ 𝐴 ∈ 𝑃(𝑋)
⟹ 𝐸𝑣𝑒𝑟𝑦 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑃(𝑋)𝑖𝑠 𝑖𝑛𝑣𝑒𝑟𝑡𝑖𝑏𝑙𝑒 𝑎𝑛𝑑 𝐴−1 = 𝐴
Chapter 2: Inverse Trigonometric Functions
Question numbers 11 to 15 are of each 1 mark and question numbers 16 to 20 are of
4 marks each.
11.
What is the principle value of cos −1 (𝐶𝑜𝑠
Answer
cos −1 (𝐶𝑜𝑠
= cos −1 (𝐶𝑜𝑠
= cos −1 (𝐶𝑜𝑠
2𝜋
3
) + sin−1 (𝑆𝑖𝑛
2𝜋
3
)?
2𝜋
2𝜋
) + sin−1 (𝑆𝑖𝑛 )
3
3
2𝜋
𝜋
) + sin−1 (𝑆𝑖𝑛 (𝜋 − ))
3
3
2𝜋
𝜋
2𝜋 𝜋
) + sin−1 (𝑆𝑖𝑛 ) =
+ = 𝜋
3
3
3
3
12.
Find the value of 𝐶𝑜𝑠(sin−1 𝑥)
Answer
Let 𝑦 = 𝐶𝑜𝑠(sin−1 𝑥) 𝑎𝑛𝑑 sin−1 𝑥 = 𝜃
∴ 𝑆𝑖𝑛𝜃 = 𝑥
∴ 𝑦 = 𝐶𝑜𝑠𝜃 = √1 − 𝑆𝑖𝑛2 𝜃 = √1 − 𝑥 2
13.
If 𝑥 +
1
𝑥
= 2, then find the principle value of sin−1 𝑥 .
Answer
1
= 2 ⟹ 𝑥 2 + 1 = 2𝑥
𝑥
𝑥+
⟹ (𝑥 − 1)2 = 0 ⟹ 𝑥 = 1
𝜋
⟹ sin−1 𝑥 = sin−1 1 = .
2
14.
Express 𝑆𝑖𝑛(cot −1 𝑥) 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑥.
Answer
Let 𝑦 = 𝑆𝑖𝑛(cot −1 𝑥) 𝑎𝑛𝑑 cot −1 𝑥 = 𝜃
∴ 𝑥 = 𝐶𝑜𝑡𝜃
∴ 𝑦 = 𝑆𝑖𝑛𝜃 =
15.
Answer
1
1
1
=
=
𝐶𝑜𝑠𝑒𝑐𝜃
√1 + 𝐶𝑜𝑡 2 𝜃
√1 + 𝑥 2
For what value of x, sin−1 𝑥 − cos −1 𝑥 =
Given that sin−1 𝑥 − cos −1 𝑥 =
𝜋
6
𝜋
6
but sin−1 𝑥 + cos −1 𝑥 =
Add both relations
∴ 2sin−1 𝑥 =
4𝜋
6
=
2𝜋
3
∴ sin−1 𝑥 =
𝑜𝑟 𝑥 = 𝑆𝑖𝑛
16.
1
5√2
5
7
Find the value of 2 tan−1 ( ) + sec −1 (
𝜋
3
𝜋
√3
=
3
2
1
) + 2 tan−1 ( )
(Page 4 of 12)
8
𝜋
2
Answer
1
5√2
5
7
Sol: 2 tan−1 ( ) + sec −1 (
1
) + 2 tan−1 ( )
8
1
1
5√2
= 2 [tan−1 ( ) + tan−1 ( )] + sec −1 (
)
5
8
7
= 2 tan−1
1 1
+
5 8
1 1
1− ×
5 8
+ tan−1 √(
5√2
7
2
) −1
1
1
= 2 tan−1 + tan−1
3
7
1
2×
3 + tan−1 1
= tan−1
1
7
1−
9
3
1
= tan−1 + tan−1
4
7
3 1
+
𝜋
−1
= tan ( 4 7 ) = tan−1 1 =
3 1
4
1− ×
4 7
17.
Answer
Solve for x: cos −1 (
Solution: cos −1 (
𝑥 2 −1
𝑥 2 +1
𝑥 2 −1
𝑥 2 +1
) + tan−1 (
) + tan−1 (
2𝑥
𝑥 2 −1
2𝑥
𝑥 2 −1
)=
)=
2𝜋
3
2𝜋
3
2
∴ cos −1 [− (
1−𝑥
2𝑥
2𝜋
)] + tan−1 [− (
)] =
1 + 𝑥2
1 − 𝑥2
3
∴ π − cos −1 (
1 − 𝑥2
2𝑥
2𝜋
) − tan−1 (
)=
2
2
1+𝑥
1−𝑥
3
∴ 𝜋 − 2 tan−1 𝑥 − 2 tan−1 𝑥 =
2𝜋
3
𝜋
12
𝜋
∴ 𝑥 = 𝑡𝑎𝑛 = tan 15
12
∴ tan−1 𝑥 =
= tan(60 − 45) =
18.
Prove that tan
−1
𝑡𝑎𝑛60−𝑡𝑎𝑛45
1+𝑡𝑎𝑛60.𝑡𝑎𝑛45
−1
1 + tan
=
√3− 1
√3+1
2 + tan−1 3 = 𝜋
Answer
𝐿𝐻𝑆 = tan−1 1 + tan−1 2 + tan−1 3
π
∴ LHS = + tan−1 2 + tan−1 3
4
𝜋
2
+
3
𝑥+𝑦
∴ 𝐿𝐻𝑆 = + 𝜋 + tan−1 (
) 𝑆𝑖𝑛𝑐𝑒 tan−1 𝑥 + tan−1 𝑦 = 𝜋 + tan−1
, 𝑖𝑓 𝑥𝑦 > 1
4
1−2×3
1 − 𝑥𝑦
𝜋
𝜋
∴ 𝐿𝐻𝑆 = + 𝜋 + tan−1 (−1) = + 𝜋 − tan−1 (1)
4
4
𝜋
𝜋
∴ 𝐿𝐻𝑆 = + 𝜋 − = 𝜋 = 𝑅𝐻𝑆
4
4
19.
If 𝑥, 𝑦, 𝑧 ∈ [−1, 1]𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 sin−1 𝑥 + sin−1 𝑦 + sin−1 𝑧 =
𝑥 2012 + 𝑦 2013 + 𝑧 2014 −
3𝜋
2
, 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓
9
𝑥 2012
(Page 5 of 12)
+
𝑦 2013
+ 𝑧 2014
Answer
𝜋
𝜋
2
2
We know that − ≤ sin−1 𝑥 ≤
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ [−1, 1]
∴ 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 & 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 sin−1 𝑥 𝑎𝑟𝑒
𝜋
𝜋
& − 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦.
2
2
𝜋
𝜋
𝜋
, sin−1 𝑦 = , sin−1 𝑧 = 𝑜𝑟 𝑥 = 1 = 𝑦 = 𝑧
2
2
2
9
9
𝑥 2012 + 𝑦 2013 + 𝑧 2014 − 2012
=1+1+1−
=3−3=0
𝑥
+ 𝑦 2013 + 𝑧 2014
1+1+1
∴ sin−1 𝑥 =
20.
𝑥
𝑦
𝑥2
𝑏
𝑎2
If cos −1 + cos −1 = 𝛼, 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡
𝑎
Answer
+
Since cos −1
∴ cos −1 [
𝑦2
𝑏2
−
2𝑥𝑦
𝑎𝑏
𝐶𝑜𝑠𝛼 = 𝑆𝑖𝑛2 𝛼
𝑥
𝑦
+ cos −1 = 𝛼
𝑎
𝑏
𝑥𝑦
𝑥2
𝑦2
− √1 − 2 . √1 − 2 ] = 𝛼
𝑎𝑏
𝑎
𝑏
∴ [
𝑥𝑦
𝑥2
𝑦2
− √1 − 2 . √1 − 2 ] = 𝐶𝑜𝑠𝛼
𝑎𝑏
𝑎
𝑏
∴
𝑥𝑦
𝑥2
𝑦2
− 𝐶𝑜𝑠𝛼 = √1 − 2 . √1 − 2
𝑎𝑏
𝑎
𝑏
Square both the sides
𝑥 2𝑦2
2𝑥𝑦
𝑥2
𝑦2
2
+
𝐶𝑜𝑠
𝛼
−
𝐶𝑜𝑠𝛼
=
(1
−
)
(1
−
)
𝑎2 𝑏 2
𝑎𝑏
𝑎2
𝑏2
∴
𝑥 2𝑦2
2𝑥𝑦
𝑥 2 𝑦2 𝑥 2𝑦2
2
+
𝐶𝑜𝑠
𝛼
−
𝐶𝑜𝑠𝛼
=
1
−
−
+
𝑎2 𝑏 2
𝑎𝑏
𝑎2 𝑏 2 𝑎2 𝑏 2
∴
𝑥 2 𝑦 2 2𝑥𝑦
+
−
𝐶𝑜𝑠𝛼 = 1 − 𝐶𝑜𝑠 2 𝛼
𝑎2 𝑏 2
𝑎𝑏
∴
𝑥 2 𝑦 2 2𝑥𝑦
+
−
𝐶𝑜𝑠𝛼 = 𝑆𝑖𝑛2 𝛼
𝑎2 𝑏 2
𝑎𝑏
Chapter 3: Matrices
Question numbers 21 to 25 are of each 1 mark and question numbers 26 to 30 are
of 4 marks each and question numbers 31 to 35 are of 6 marks each.
21.
𝑎+𝑏
If [
5
Answer
Since [
6
2
]= [
5
𝑎𝑏
𝑎+𝑏
5
2
] , 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑎 𝑎𝑛𝑑 𝑏.
8
6
2
]= [
5
𝑎𝑏
2
]
8
∴ 𝑎 + 𝑏 = 6, 𝑎𝑏 = 8 𝑜𝑛 𝑠𝑜𝑙𝑣𝑖𝑛𝑔, 𝑎 = 2, 4.
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑤ℎ𝑒𝑛 𝑎 = 2, 𝑏 = 4 𝑎𝑛𝑑 𝑤ℎ𝑒𝑛 𝑎 = 4 , 𝑏 = 2.
22.
Answer
Construct a 2 x 2 matrix whose element 𝑎𝑖𝑗 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 𝑎𝑖𝑗 =
9
[2
25
2
8
]
18
(Page 6 of 12)
(2𝑖+𝑗)2
2
23.
−1
If 𝐴 = [ 0
0
0
−1
0
0
0 ] 𝑓𝑖𝑛𝑑 𝐴2
−1
Answer
−1 0
𝐴2 = [ 0 −1
0
0
2 3
] 𝑓𝑖𝑛𝑑 𝐴𝐵
3 1
24.
If 𝐴 = [
Answer
−3 −4 1
𝐴𝐵 = [
]
8 13 9
𝐶𝑜𝑠𝑥 −𝑆𝑖𝑛𝑥
If 𝐴 = [
] , 𝑓𝑖𝑛𝑑 𝐴. 𝐴′
𝑆𝑖𝑛𝑥 𝐶𝑜𝑠𝑥
25.
Answer
1
2
−2
1
] 𝑎𝑛𝑑 𝐵 = [
3
2
𝐶𝑜𝑠𝑥
Since 𝐴 = [
𝑆𝑖𝑛𝑥
−𝑆𝑖𝑛𝑥
𝐶𝑜𝑠𝑥
] 𝑡ℎ𝑒𝑛 𝐴′ = [
𝐶𝑜𝑠𝑥
−𝑆𝑖𝑛𝑥
∴ 𝐴𝐴′ = [
26.
If 𝐴 = [
Answer
3
Since 𝐴 = [
−4
3
−4
𝐶𝑜𝑠𝑥
𝑆𝑖𝑛𝑥
0
0 ] = −𝐴
−1
𝑆𝑖𝑛𝑥
]
𝐶𝑜𝑠𝑥
−𝑆𝑖𝑛𝑥
𝐶𝑜𝑠𝑥
].[
𝐶𝑜𝑠𝑥
−𝑆𝑖𝑛𝑥
𝑆𝑖𝑛𝑥
1
]=[
𝐶𝑜𝑠𝑥
0
0
]
1
−5
] , 𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝐴2 − 5𝐴 − 14𝐼 = 0. 𝐻𝑒𝑛𝑐𝑒 𝑓𝑖𝑛𝑑 𝐴−1 .
2
−5
3
] 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐴2 = [
2
−4
−5
3
].[
2
−4
−5
]
2
29 −25
]
−20 24
−25
3
] − 5[
24
−4
∴ 𝐴2 = [
29
∴ 𝐴2 − 5𝐴 − 14𝐼 = [
−20
0
= [
0
1
−5
] − 14 [
0
2
0
]
1
0
] = 0 Hence proved.
0
Pre-multiplying 𝐴2 − 5𝐴 − 14𝐼 = 0 𝑏𝑦 𝐴−1 , 𝑤𝑒 𝑔𝑒𝑡
𝐴−1 𝐴2 − 5𝐴−1 𝐴 − 14𝐴−1 𝐼 = 𝐴−1 0
𝑜𝑟 𝐴 − 5𝐼 − 14𝐴−1 = 0
𝑜𝑟 𝐴−1 =
1
(𝐴 − 5𝐼)
14
1
1
3 −5
{[
] − 5[
0
14 −4 2
1
−2 −5
𝑜𝑟 𝐴−1 =
[
]
14 −4 −3
𝑜𝑟 𝐴−1 =
27.
Represent the given matrix as the sum of symmetric and skew symmetric matrix
3
𝐴 = [−2
−4
Answer
0
]}
1
3 −1
−2 1 ]
−5 2
3
3 −1
3
Since 𝐴 = [−2 −2 1 ] 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐴′ = [3
−4 −5 2
1
Let 𝑃 =
1
2
(𝐴 + 𝐴′ ) =
3
{[−2
2
−4
1
3
−2
−5
−1
3
1 ] + [3
2
1
−2
−2
1
−2
−2
1
−4
−5]}
2
1
2
−
3
∴𝑃=
1
2
5
[− 2
−4
−5]
2
−2
−2
(Page 7 of 12)
5
2
−2
2 ]
1
2
3
∴ 𝑃′ =
1
2
5
[− 2
−2
′
5
2
−
=
−2
−2
1
2
3
2 ]
1
2
5
[− 2
−
5
2
−2 = 𝑃
−2
−2
2 ]
∴ 𝑃 𝑖𝑠 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑎𝑡𝑟𝑖𝑥.
Let 𝑄 =
1
2
(𝐴 − 𝐴′ ) =
3
{[−2
2
−4
3
−2
−5
1
−1
3
1 ] − [3
2
1
−2
−2
1
0
5
2
3
2
0
3
−3
2]
5
2
3
[− 2
5 3
2 2
∴𝑄= −
0
5
2
3
[− 2
∴ 𝑄′ = −
,
0
5
2
3
[2
3
2
3 =
0
−3
2]
5
2
0
5
2
3
[− 2
−4
−5]}
2
∴ 𝑄′ = − −
−
5
2
−
0
3
2
−3
3
2 ]
3 = −𝑄
0
−3
2]
∴ 𝑄 𝑖𝑠 𝑠𝑘𝑒𝑤 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑎𝑡𝑟𝑖𝑥.
3
𝑁𝑜𝑤 𝑃 + 𝑄 =
3
= [−2
−4
28.
If 𝐴 = [
3
1
3
−2
−5
1
2
5
[− 2
1
2
−
5
2
5
2
3
2
0
3
−3
2]
0
5
2
3
[− 2
−2 + −
−2
−2
2 ]
−1
1 ]=𝐴
2
Hence the result.
−4
1 + 2𝑛
] 𝑡ℎ𝑒𝑛 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝐴𝑛 = [
−1
𝑛
−4𝑛
], where n is any positive
1 − 2𝑛
integer.
Answer
We shall prove it by using principle of Mathematical Induction.
For n = 1, 𝐴1 = [
1+2
1
−4
3
]=[
1−2
1
−4
], which is true.
−1
Therefore given statement is true for n = 1.
Let us suppose that given statement is true for n = k.
1 + 2𝑘
∴ 𝐴𝑘 = [
𝑘
−4𝑘
] 𝑖𝑠 𝑡𝑟𝑢𝑒.
1 − 2𝑘
Let us take n = k + 1
(Page 8 of 12)
∴ 𝐴𝑘+1 = 𝐴𝑘 . 𝐴 = [
𝐴𝑘+1 = [
∴ 𝐴𝑘+1 = [
3 + 6𝑘 − 4𝑘
3𝑘 + 1 − 2𝑘
1 + 2(𝑘 + 1)
𝑘+1
−4𝑘 3
][
1 − 2𝑘 1
1 + 2𝑘
𝑘
−4
]
−1
−4 − 8𝑘 + 4𝑘
3 + 2𝑘
]= [
−4𝑘 − 1 + 2𝑘
1+𝑘
−4 − 4𝑘
]
−1 − 2𝑘
−4(𝑘 + 1)
] which is true for n = k +1
1 − 2(𝑘 + 1)
Therefore by PMI, it is true for every positive integer k.
29.
Find the values of x, y and z if the given matrix A satisfy the equation A.A’ =I
0
where 𝐴 = [𝑥
𝑥
Answer
2𝑦
𝑦
−𝑦
𝑧
−𝑧]
𝑧
0
𝑆𝑖𝑛𝑐𝑒 𝐴 = [𝑥
𝑥
2𝑦
𝑦
−𝑦
𝑧
0
−𝑧] 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐴′ = [2𝑦
𝑧
𝑧
𝑥
𝑦
−𝑧
𝑥
−𝑦]
𝑧
Given that AA’=I
0
⟹ [𝑥
𝑥
2𝑦
𝑦
−𝑦
𝑥2 + 𝑥2
⟹ [ 𝑥𝑦 − 𝑥𝑦
−𝑧𝑥 + 𝑧𝑥
𝑧
0
−𝑧] [2𝑦
𝑧
𝑧
𝑥
𝑦
−𝑧
𝑥𝑦 − 𝑥𝑦
4𝑦 2 + 𝑦 2 + 𝑦 2
2𝑦𝑧 − 𝑦𝑧 − 𝑦𝑧
2𝑥 2
⟹[ 0
0
0
6𝑦 2
0
1
𝑥
−𝑦] = [0
𝑧
0
0
1
0
0
0]
1
−𝑥𝑧 + 𝑥𝑧
1
2𝑦𝑧 − 𝑦𝑧 − 𝑦𝑧] = [0
0
𝑧2 + 𝑧2 + 𝑧2
0
1
0 ] = [0
0
3𝑧 2
0
1
0
0
0]
1
⟹ 2𝑥 2 = 1, 6𝑦 2 = 1 𝑎𝑛𝑑 3𝑧 2 = 1
⟹𝑥=±
1
√2
,𝑦 = ±
1
√6
,𝑧 = ±
1
√3
30.
Prove that inverse of every square matrix, if it exists, is unique.
Answer
Let ‘A’ be any invertible square matrix of order n.
Let B and C are two inverse of A.
Therefore; AB = BA = I - - - - - (1)
And AC = CA = I - - - - - - - (2)
From (1), since AB = I
Pre-multiply it by C, we get,
C(AB) = CI
Or
(CA)B = C
[Since C(AB) = (CA)B and CI = C]
Or
IB = C
[from (2), CA = I]
Or
B = C
Hence inverse of a matrix is unique.
(Page 9 of 12)
0
1
0
0
0]
1
31.
1
Using elementary operations find the inverse of 𝐴 = [2
3
Answer
Using row operations; therefore, A = IA
1
[2
3
1
0
−1
−1
3]=
2
1
[0
0
0
1
0
1
0
−1
0
0] 𝐴
1
𝑢𝑠𝑖𝑛𝑔 𝑅3 → 𝑅3 + 𝑅1
1
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 [2
4
1
0
0
−1
1
3 ] = [0
1
1
0
1
0
0
0] 𝐴
1
0
1
−2
1
0] 𝐴
1
𝐴𝑝𝑝𝑙𝑦 𝑅1 → 𝑅1 + 𝑅3 𝑎𝑛𝑑 𝑅3 → 𝑅3 − 2𝑅2
5
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 [2
0
𝐴𝑝𝑝𝑙𝑦 𝑅3 𝑎𝑠
1
0
0
0
2
3 ] = [0
1
−5
𝑅3
, 𝑤𝑒 𝑔𝑒𝑡
5
5
[2
0
1
0
0
2
0
0
3 ] = [1
−1
5
0
1
2
−
5
1
0
1] 𝐴
5
5
[2
0
1
0
0
2
3
0
0 ]= 5
1
−1
[5
0
1
−
5
2
−
5
1
3
5 𝐴
1
5]
𝐴𝑝𝑝𝑙𝑦 𝑅2 → 𝑅2 + 3𝑅3
𝐴𝑝𝑝𝑙𝑦 𝑅1 → 𝑅1 − 2𝑅2
1
[2
0
𝐴𝑝𝑝𝑙𝑦 𝑅2 →
4
5
0
3
0 ]=
5
−1
1
[5
2
5
1
−
5
2
−
5
1
0
0
4
5
0
3
0] =
10
1
1
[− 5
2
5
1
−
10
2
5
1
0
0
1
2
0
3
0] =
10
1
1
[− 5
1
2
1
−
10
2
5
1
0
0
1
5
3
𝐴
5
1
5 ]
−
𝑅2
𝑎𝑛𝑑 𝑅1 → (−1)𝑅1
2
1
[1
0
1
5
3
𝐴
10
1
− ]
5
−
𝐴𝑝𝑝𝑙𝑦𝑅1 → 𝑅1 − 𝑅2
0
[1
0
𝐴𝑝𝑝𝑙𝑦 𝑅1 ↔ 𝑅2
(Page 10 of 12)
1
2
3
𝐴
10
1
− ]
5
−
−1
3]
2
1
[0
0
3
10
0
1
0] =
2
1
1
[− 5
0
1
0
1
10
1
2
2
5
3
10
1
− 𝐴
2
1
− ]
5
−
3
Therefore 𝐴−1 =
2
[−
1
3
10
1
10
1
−
10
1
1
2
2
5
5
−
32.
1
Using elementary operations find the inverse of 𝐴 = [−3
2
Answer
Let us apply row operations, that is A = IA
1
Therefore [−3
2
3
0
5
−2
1
−5] = [0
0
0
𝐴𝑝𝑝𝑙𝑦 𝑅2 → 𝑅2 + 3𝑅1 ,
0
1
0
2
1
− ]
5
0
0] 𝐴
1
𝑤𝑒 𝑔𝑒𝑡
1
[0
2
3
9
5
−2
1
−11] = [3
0
0
0
1
0
0
0] 𝐴
1
𝐴𝑝𝑝𝑙𝑦 𝑅3 → 𝑅3 − 2𝑅1 , 𝑤𝑒 𝑔𝑒𝑡
1
[0
0
3
9
−1
−2
1
−11] = [ 3
4
−2
0
1
0
0
0] 𝐴
1
1
[0
0
3
1
−1
−2
1
21 ] = [−13
4
−2
0
1
0
0
8] 𝐴
1
1
[0
0
0
1
−1
10
−5
21] = [−13
4
−2
0
1
0
3
8] 𝐴
1
𝐴𝑝𝑝𝑙𝑦 𝑅2 → 𝑅2 + 8𝑅3 , 𝑤𝑒 𝑔𝑒𝑡
𝐴𝑝𝑝𝑙𝑦 𝑅1 → 𝑅1 + 3𝑅3 , 𝑤𝑒 𝑔𝑒𝑡
𝐴𝑝𝑝𝑙𝑦 𝑅3 → 𝑅3 + 𝑅2 , 𝑤𝑒 𝑔𝑒𝑡
1
[0
0
𝐴𝑝𝑝𝑙𝑦 𝑅3 →
0
1
0
10
−5
21] = [−13
25
−15
0
1
1
3
8] 𝐴
9
0
1
1
25
3
8
9 ]𝐴
25
𝑅3
, 𝑤𝑒 𝑔𝑒𝑡
25
1
[0
0
0
1
0
−5
10
−13
21] = [ 3
−
1
5
0
1
0
1
0
21] = −13
3
1
[−5
𝐴𝑝𝑝𝑙𝑦 𝑅1 → 𝑅1 − 10𝑅3 , 𝑤𝑒 𝑔𝑒𝑡
1
[0
0
𝐴𝑝𝑝𝑙𝑦 𝑅2 → 𝑅2 − 21𝑅3 , 𝑤𝑒 𝑔𝑒𝑡
(Page 11 of 12)
2
5
1
1
25
−
3
5
8 𝐴
9
25 ]
−
3
0
5
−2
−5]
0
2
5
5
15
1
25
1
1
[0
0
0
1
0
0
2
0] = −
5
1
3
[− 5
1
Therefore 𝐴
−1
𝐴−1
3
7
4
= −
7
[ 1
−2
−1
𝐴
= 11
[4
Answer
2
5
−
5
11
3
15
1
25
9
5
25
25
3
14
3
14
1
−
2
1
2
−1
1
−
2
−4
3
−12
(Page 12 of 12)
]
5
4
6
3
1]
2
1
2
1
2
1
− ]
2
−
−1
0
−2
4
2]
7
1
−6
−2]
3
Using elementary operations find the inverse of 𝐴 = [2
0
3
𝐴−1 = [−2
8
3
5
2
Using elementary operations find the inverse of 𝐴 = [4
3
Answer
35.
−
2
Using elementary operations find the inverse of 𝐴 = [3
1
Answer
34.
−
2
= −5
[−
33.
3
5
11
𝐴
25
9
25 ]
−
3
−2]
9
0
3
4
−1
0]
1