Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

(I) Select the correct option from the given alternatives.

Question 1.
Given A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 2
\end{array}\right]\), I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) if A – λI is a singular matrix, then ___________
(a) λ = 0
(b) λ2 – 3λ – 4 = 0
(c) λ2 + 3λ – 4 = 0
(d) λ2 – 3λ – 6 = 0
Answer:
(b) λ2 – 3λ – 4 = 0
Hint:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) I Q1

Question 2.
Consider the matrices A = \(\left[\begin{array}{ccc}
4 & 6 & -1 \\
3 & 0 & 2 \\
1 & -2 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 4 \\
0 & 1 \\
-1 & 2
\end{array}\right]\), C = \(\left[\begin{array}{l}
3 \\
1 \\
2
\end{array}\right]\). Out of the given matrix products, ___________
(i) (AB)TC
(ii) CTC(AB)T
(iii) CTAB
(iv) ATABBTC
(a) Exactly one is defined
(b) Exactly two are defined
(c) Exactly three are defined
(d) all four are defined
Answer:
(c) Exactly three are defined
Hint:
A is of order 3 × 3, B is of order 3 × 2 and C is of order 3 × 1.
(AB)TC is of order 2 × 1.
CTC and (AB)T are of different orders.
CTC (AB)T is not defined.
CTAB is of order 1 × 2.
ATABBTC is of order 3 × 1.

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Question 3.
If A and B are square matrices of equal order, then which one is correct among the following?
(a) A + B = B + A
(b) A + B = A – B
(c) A – B = B – A
(d) AB = BA
Answer:
(a) A + B = B + A
Hint:
Matrix addition is commutative.
∴ A + B = B + A

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
a & 2 & b
\end{array}\right]\) is a matrix satisfying the equation AAT = 9I, where I is the identity matrix of order 3, then the ordered pair (a, b) is equal to ___________
(a) (2, -1)
(b) (-2, 1)
(c) (2, 1)
(d) (-2, -1)
Answer:
(d) (-2, -1)

Question 5.
If A = \(\left[\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right]\) and |A3| = 125, then α = ___________
(a) ±3
(b) ±2
(c) ±5
(d) 0
Answer:
(a) ±3
Hint:
|A3| = 125
|A|3 = 53 …….[∵ |An| = |A|n, n ∈ N]
∴ |A| = 5
\(\left|\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right|=5\)
α2 – 4 = 5
α2 = 9
∴ α = ± 3

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Question 6.
If \(\left[\begin{array}{ll}
5 & 7 \\
x & 1 \\
2 & 6
\end{array}\right]-\left[\begin{array}{cc}
1 & 2 \\
-3 & 5 \\
2 & y
\end{array}\right]=\left[\begin{array}{cc}
4 & 5 \\
4 & -4 \\
0 & 4
\end{array}\right]\), then ___________
(a) x = 1, y = -2
(b) x = -1, y = 2
(c) x = 1, y = 2
(d) x = -1, y = -2
Answer:
(c) x = 1, y = 2

Question 7.
If A + B = \(\left[\begin{array}{ll}
7 & 4 \\
8 & 9
\end{array}\right]\) and A – B = \(\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]\), then the value of A is ___________
(a) \(\left[\begin{array}{ll}
3 & 1 \\
4 & 3
\end{array}\right]\)
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
6 & 2 \\
8 & 6
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
7 & 6 \\
8 & 12
\end{array}\right]\)
Answer:
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)

Question 8.
If \(\left[\begin{array}{cc}
x & 3 x-y \\
z x+z & 3 y-w
\end{array}\right]=\left[\begin{array}{ll}
3 & 2 \\
4 & 7
\end{array}\right]\), then ___________
(a) x = 3, y = 7, z = 1, w = 14
(b) x = 3, y = -5, z = -1, w = -4
(c) x = 3, y = 6, z = 2, w = 7
(d) x = -3, y = -7, z = -1, w = -14
Answer:
(a) x = 3, y = 7, z = 1, w = 14

Question 9.
For suitable matrices A, B, the false statement is ___________
(a) (AB)T = ATBT
(B) (AT)T = A
(C) (A – B)T = AT – BT
(D) (A + B)T = AT + BT
Answer:
(a) (AB)T = ATBT
Hint:
(AB)T = BTAT

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Question 10.
If A = \(\left[\begin{array}{cc}
-2 & 1 \\
0 & 3
\end{array}\right]\) and f(x) = 2x2 – 3x, then f(A) = ___________
(a) \(\left[\begin{array}{cc}
14 & 1 \\
0 & -9
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
-14 & 1 \\
0 & 9
\end{array}\right]\)
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
-14 & -1 \\
0 & -9
\end{array}\right]\)
Answer:
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) I Q10

(II) Answer the following questions.

Question 1.
If A = diag[2, -3, -5], B = diag[4, -6, -3] and C = diag [-3, 4, 1], then find
i. B + C – A
ii. 2A + B – 5C.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q1
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q1.1
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q1.2

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Question 2.
If f(α) = A = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\), find
i. f(-α)
ii. f(-α) + f(α)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q2

Question 3.
Find matrices A and B, where
(i) 2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) and A + 3B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\)
(ii) 3A – B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 0 & 5
\end{array}\right]\) and A + 5B = \(\left[\begin{array}{ccc}
0 & 0 & 1 \\
-1 & 0 & 0
\end{array}\right]\)
Solution:
i. Given equations are
2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) …….(i)
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q3
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q3.1
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q3.2

Question 4.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
3 & -2 \\
-1 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & 4 & 1 \\
2 & -1 & -3
\end{array}\right]\), verify
i. (A + 2BT)T = AT + 2B
ii. (3A – 5BT)T = 3AT – 5B.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q4
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q4.1
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q4.2

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Question 5.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) and A + AT = I, where I is a unit matrix of order 2 × 2, then find the value of α.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q5

Question 6.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & 2 \\
-1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 3 & 2 \\
4 & -1 & -3
\end{array}\right]\), show that AB is singular.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q6
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q6.1

Question 7.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 4 & 6 \\
1 & 2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
-3 & 2 & -1 \\
-2 & 1 & 0
\end{array}\right]\), show that AB and BA are both singular matrices.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q7

Question 8.
If A = \(\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right]\), show that BA = 6I.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q8

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Question 9.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
0 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -2
\end{array}\right]\), verify that |AB| = |A|.|B|.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q9

Question 10.
If Aα = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that Aα . Aβ = Aα+β
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q10

Question 11.
If A = \(\left[\begin{array}{cc}
1 & \omega \\
\omega^{2} & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
\omega^{2} & 1 \\
1 & \omega
\end{array}\right]\), where ω is a complex cube root of unity, then show that AB + BA + A – 2B is a null matrix.
Solution:
ω is the complex cube root of unity.
ω3 = 1
ω3 – 1 = 0
(ω – 1) (ω2 + ω + 1) = 0
ω = 1 or ω2 + ω + 1 = 0
But, ω is a complex number.
1 + ω + ω2 = 0 …….(i)
AB + BA + A – 2B
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q11
which is a null matrix.

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Question 12.
If A = \(\left[\begin{array}{lrr}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & -3
\end{array}\right]\), show that A2 = A.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q12

Question 13.
If A = \(\left[\begin{array}{ccc}
4 & -1 & -4 \\
3 & 0 & -4 \\
3 & -1 & -3
\end{array}\right]\), show that A2 = I.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q13

Question 14.
If A = \(\left[\begin{array}{cc}
3 & -5 \\
-4 & 2
\end{array}\right]\), show that A2 – 5A – 14I = 0.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q14

Question 15.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
-1 & 2
\end{array}\right]\), show that A – 4A + 3I = 0.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q15

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Question 16.
If A = \(\left[\begin{array}{cc}
-3 & 2 \\
2 & -4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & x \\
y & 0
\end{array}\right]\) and (A + B)(A – B) = A2 – B2, find x and y.
Solution:
(A + B)(A – B) = A2 – B2
A2 – AB + BA – B2 = A2 – B2
-AB + BA = 0
AB = BA
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q16
By equality of matrices, we get
2 – 4x = -3x
∴ x = 2 and 2y = 2x
y = x
∴ y = 2
∴ x = 2, y = 2

Question 17.
If A = \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\), show that (A + B)(A – B) ≠ A2 – B2.
Solution:
We have to prove that
(A + B) . (A – B) ≠ A2 – B2
i.e., to prove that A(A – B) + B(A – B) ≠ A2 – B2
i.e., to prove that A2 – AB + BA – B2 ≠ A2 – B2
i.e., to prove that AB ≠ BA.
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q17
∴ AB ≠ BA

Question 18.
If A = \(\left[\begin{array}{ll}
2 & -1 \\
3 & -2
\end{array}\right]\), find A3.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q18
∴ A2 = I
Multiplying throughout by A, we get
A3 = A . I
∴ A3 = A

Question 19.
Find x, y if,
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q19
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q19.1
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q19.2

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Question 20.
Find x, y, z if
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q20
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q20.1
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q20.2

Question 21.
If A = \(\left[\begin{array}{ccc}
2 & 1 & -3 \\
0 & 2 & 6
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & 0 & -2 \\
3 & -1 & 4
\end{array}\right]\), find ABT and ATB.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q21
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q21.1

Question 22.
If A = \(\left[\begin{array}{cc}
2 & -4 \\
3 & -2 \\
0 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 1 & 0
\end{array}\right]\), show that (AB)T = BTAT.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q22

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Question 23.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), prove that An = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\), for all n ∈ N.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q23
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q23.1

Question 24.
If A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), prove that An = \(\left[\begin{array}{cc}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\), for all n ∈ N.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q24
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q24.1

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Question 25.
Two farmers Shantaram and Kantaram cultivate three crops rice, wheat, and groundnut. The sale (in Rupees) of these crops by both the farmers for the month of April and may 2008 is given below,
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q25
Find
(i) the total sale in rupees for two months of each farmer for each crop.
(ii) the increase in sales from April to May for every crop of each farmer.
Solution:
(i) Total sale for Shantaram:
For rice = 15000 + 18000 = ₹ 33000.
For wheat = 13000 + 15000 = ₹ 28000.
For groundnut = 12000 + 12000 = ₹ 24000.
Total sale for Kantaram:
For rice = 18000 + 21000 = ₹ 39000
For wheat = 15000 + 16500 = ₹ 31500
For groundnut = 8000 + 16000 = ₹ 24000

Alternate method:
Matrix form
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q25.1
∴ The total sale of April and May of Shantaram in ₹ is ₹ 33000 (rice), ₹ 28000 (wheat), ₹ 24000 (groundnut), and that of Kantaram in ₹ is ₹ 39000(rice), ₹ 31500(wheat), and ₹ 24000 (groundnut).

(ii) Increase in sale from April to May for Shantaram:
For rice = 18000 – 15000 = ₹ 3000
For wheat = 15000 – 13000 = ₹ 2000
For groundnut = 12000 – 12000 = ₹ 0
Increase in sale from April to May for Kantaram:
For rice = 21000 – 18000 = ₹ 3000
For wheat = 16500 – 15000 = ₹ 1500
For groundnut = 16000 – 8000 = ₹ 8000

Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

Alternate method:
Matrix form
Maharashtra Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) II Q25.2
∴ The increase in sales for Shantaram from April to May in each crop is ₹ 3000 (rice), ₹ 2000(wheat), 0 (groundnut), and that for Kantaram is ₹ 3000 (rice), ₹ 1500 (wheat), and ₹ 8000 (groundnut).