RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 12 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 12. Students can also read RBSE Class 12 Maths Important Questions for exam preparation. Students can also go through RBSE Class 12 Maths Notes to understand and remember the concepts easily.

RBSE Class 12 Maths Solutions Chapter 3 Matrices Ex 3.2

Question 1.
Let
A = \(\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right],\) B = \(\left[\begin{array}{rr} 1 & 3 \\ -2 & 5 \end{array}\right]\), C = \(\left[\begin{array}{rr} -2 & 5 \\ 3 & 4 \end{array}\right]\)
Find each of the following:
(i) A + B
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 1

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

(ii) A - B
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 2

(iii) 3A - C
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 3

(iv) AB
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 4

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

(v) BA
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 5

Question 2.
Compute the following:
(i) \(\left[\begin{array}{rr} a & b \\ -b & a \end{array}\right]\) + \(\left[\begin{array}{ll} a & b \\ b & a \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 6

(ii) \(\left[\begin{array}{ll} a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2} \end{array}\right]+\left[\begin{array}{rr} 2 a b & 2 b c \\ -2 a c & -2 a b \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 7

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

(iii) \(\left[\begin{array}{rrr} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{array}\right]+\left[\begin{array}{ccr} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 8

(iv) \(\left[\begin{array}{cc} \cos ^{2} x & \sin ^{2} x \\ \sin ^{2} x & \cos ^{2} x \end{array}\right]+\left[\begin{array}{ll} \sin ^{2} x & \cos ^{2} x \\ \cos ^{2} x & \sin ^{2} x \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 9

Question 3.
Compute the indicated products:
(i) \(\left[\begin{array}{rr} a & b \\ -b & a \end{array}\right]\left[\begin{array}{rr} a & -b \\ b & a \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 10

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

(ii) \(\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]\left[\begin{array}{lll} 2 & 3 & 4 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 11

(iii) \(\left[\begin{array}{rr} 1 & -2 \\ 2 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 12

(iv) \(\left[\begin{array}{lll} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{array}\right]\left[\begin{array}{rrr} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 13

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

(v) \(\left[\begin{array}{rr} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{array}\right]\left[\begin{array}{rrr} 1 & 0 & 1 \\ -1 & 2 & 1 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 14

(vi) \(\left[\begin{array}{rrr} 3 & -1 & 3 \\ -1 & 0 & 2 \end{array}\right]\left[\begin{array}{rr} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 15

Question 4.
If A = \(\left[\begin{array}{rrr} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right]\), B = \(\left[\begin{array}{rrr} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{array}\right]\) and C = \(\left[\begin{array}{rrr} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{array}\right]\), then compute (A + B) and (B - C). Also verify that A + (B - C) = (A + B) - C.
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 16

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Question 5.
If A = \(\left[\begin{array}{ccc} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{array}\right]\) and B = \(\left[\begin{array}{ccc} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{array}\right]\) then compute 3A - 5B.
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 17
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 18

Question 6.
Simplify:
cos θ \(\left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\) + sin θ \(\left[\begin{array}{rr} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 19

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Question 7.
(i) X + Y = \(\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right]\) and X - Y = \(\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 20

(ii) 2X + 3Y = \(\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]\) and 3X + 2Y = \(\left[\begin{array}{rr} 2 & -2 \\ -1 & 5 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 21
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 22

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Question 8.
Find X if Y = \(\left[\begin{array}{ll} 3 & 2 \\ 1 & 4 \end{array}\right]\) and 2X + Y = \(\left[\begin{array}{rr} 1 & 0 \\ -3 & 2 \end{array}\right]\).
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 23

Question 9.
Find x and y if
2\(\left[\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 24
Comparing corresponding elements, we get
2 + y = 5,
⇒ y = 5 - 2,
⇒ y = 3,
∴ y = 3,

2x + 2 = 8
⇒ 2x = 8 - 2
⇒ 2x = 6
∴ x = 3

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Question 10.
Solve the equation for x, y, z and t, if
2\(\left[\begin{array}{ll} x & z \\ y & t \end{array}\right]+3\left[\begin{array}{rr} 1 & -1 \\ 0 & 2 \end{array}\right]=3\left[\begin{array}{ll} 3 & 5 \\ 4 & 6 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 25
Comparing corresponding elements, we get
2x + 3 = 9, 2z - 3 = 15, 2y = 12, 2t + 6 = 18
⇒ 2x = 9 - 3, 2z = 15 + 3, y = \(\frac{12}{2}\), 2t = 18 - 6
⇒ 2x = 6, 2z = 18, y = 6, 2t = 12
∴ x = 3, z = 9, y = 6, t = 6

Question 11.
If x\(\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \)+ y\(\left[\begin{array}{r} -1 \\ 1 \end{array}\right]\) = \(\left[\begin{array}{c} 10 \\ 5 \end{array}\right]\), then find the values of x and y.
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 26
⇒ 2x - y = 10 ........ (i)
3x + y = 5 .......... (ii)
Adding equation (i) and (ii), we get
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 27
Putting value of x in euqation (i), we get
2 × 3 - y = 10 ⇒ 6 - y = 10
⇒ - y = 10 - 6 ⇒ y = 4
⇒ y = - 4
Thus, x = 3, y = - 4

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Question 12.
Given
3\(\left[\begin{array}{cc} x & y \\ z & w \end{array}\right]=\left[\begin{array}{cc} x & 6 \\ -1 & 2 w \end{array}\right]+\left[\begin{array}{cc} 4 & x+y \\ z+w & 3 \end{array}\right]\)
find the value of x, y, z and w.
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 28
Comparing corresponding elements, we get
3x = x + 4,
3z = - 1 + z + w,
⇒ 3x - x = 4
⇒ 2x = 4,
3z - z = w - 1,
2z = w - 1,

3y = 6 + x + y
3w = 2w + 3
3y - y = 6 + x
2y = 6 + x
3w - 2w = 3
w = 3

∴ x = 2,
2y = 6 + 2, ⇒ 2y = 8,
y = 4

And w = 3, then 2z = w - 1
⇒ 2z = 3 - 1
⇒ 2z = 2
∴ z = 1
Thus, x = 2, y = 4, z = 1, w = 3

Question 13.
If F(x) = \(\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]\), show that F(x) F(y) = F(x + y).
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 29

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Question 14.
Show that
(i) \(\left[\begin{array}{rr} 5 & -1 \\ 6 & 7 \end{array}\right]\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right] \neq\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]\left[\begin{array}{rr} 5 & -1 \\ 6 & 7 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 30

(ii)
\(\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]\)\(\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 31

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Question 15.
If A = \(\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\), then find the value of A2 - 5A + 6I
Answer:
Here, A2 = AA
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 32
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 33

Question 16.
If A = \(\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\), then prove that:
A3 - 6A2 + 7A + 2I = 0
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 34
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 35

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Question 17.
If A = \(\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]\) and I = \(\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\) and A2 = KA - 2I, then find k.
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 36
⇒ 1 = 3k - 2, - 2 = - 2k
4 = 4k, - 4 = - 2k - 2
⇒ 3k = 2 + 1, 2k = 2
4k = 4, - 2k = - 4 + 2
⇒ 3k = 3, 2k = 2
4k = 4, - 2k = - 2
⇒ k = 1, k = 1

Question 18.
If \(\left[\begin{array}{cc} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{array}\right]\) and I is the identity matrix of order 2, show that I + A = (I - A)\(\left[\begin{array}{rr} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\)
Answer:
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 37
RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 38

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Question 19.
A trust fund has ₹ 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and the second bond pays 7% interest per year. Using matrix multiplication determine how to divide ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:
(a) ₹ 1,800
(b) ₹ 2,000
Answer:
Let x and (30000 - x) are two parts of ₹ 30,000.
Then matrix A = [x (30000 - x)]
Rate of interest are 5% and 7% or 0.05 and 0.07
Let matrix of rate R = \(\left[\begin{array}{l} 0.05 \\ 0.07 \end{array}\right]\)

(a) We have, total interest = ₹ 1800
= [x (30000 - x)] \(\left[\begin{array}{l} 0.05 \\ 0.07 \end{array}\right]\) = [1800]
⇒ [x × 0.05 + (30000 - x) × 0.07] - [1800]
⇒ [0.05x + (30000 × 0.07) - 0 07 × x] = [1800]
⇒ 0.05x + 2100 - 0.07× = 1800
⇒ 0.05x - 0.07x = 1800 - 2100
⇒ - 0.02x = - 300
⇒ x = \(\frac{-300}{-0.02}\) = \(\frac{30000}{2}\) = 15000
First bond = ₹ 15,000
Now 30000 - x = 30000 -15000
∴ Second bond = ₹ 15,000
Thus, each bond should be bought for ₹ 15,000 so that interest of ₹ 1800 is received.

(b) For interest ₹ 2,000, matrix equation will be of the following type
[x (3000 - x)] \(\left[\begin{array}{l} 0.05 \\ 0.07 \end{array}\right]\) = [2000]
⇒ [x × 0.05 + (30000 - x) × 0.07] = [2000]
⇒ [0.05x + (30000 × 0.07) - (x × 0.07]) = [2000]
⇒ 0.05x + 2100.00 - 0.07x = 2000
⇒ 0.05x - 0.07x = 2000 - 2100
⇒ - 0.02x = - 100
⇒ x = \(\frac{-100}{-0.02}\) = \(\frac{10000}{2}\) = 5000
∴ First bond = ₹ 5,000
Thus 30000 - x = 30000 - 5000 = 25000
∴ Second bond = ₹ 25,000
To get ₹ 2,000 as interest two bonds should be divided into ₹ 5,000 and ₹ 25,000.

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Question 20.
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the book shop will receive from selling all the books using matrix algebra.
Answer:
Number of books in school is as follows:

Subject

Number of books

Chemistry

10 dozen = 10 × 12 = 120

Physics

8 dozen = 8 × 12 = 96

Economics

10 dozen = 10 × 12 = 120

Let number of books are represented by A
∴ A = [120 96 120]
And selling prices are represented by R then
∴ R = \(\left[\begin{array}{l} 80 \\ 60 \\ 40 \end{array}\right]\)
Money obtained by selling the books
= AR = [120 96 120] \(\left[\begin{array}{l} 80 \\ 60 \\ 40 \end{array}\right]\)
= [(120 × 80) + (96 × 60) + (120 × 40)]
= [9600 + 5760 + 4800] = [20160]
Total money obtained = ₹ 20,160
Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k respectively. Choose the correct answer in exercises 21 and 22.

Question 21.
The restriction on n, k and p so that PY + WY will defined are:
(A) k = 3, p = n
(B) k is arbitary, p = 2
(C) p is arbitrary, k = 3
(D) k = 2, p = 3
Answer:
Order P matrix P = p × k
Order of martix Y = 3 × k
Order of matrix W = n × 3
Then for PY defined
Number of columns in P = number of rows in Y
k =3
and WY defined
Number of columns in W = number of rows in Y
(w = 3)
Now PY + WY defined
order of PY = p × k = p × 3
Then order of WY = n × k
Thus, PY + WY is defined if order of PY and WY are same
∴ p = n and k = 3
Thus (A) is correct.

RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2

Question 22.
If n = p, then order of matrix 7X - 5Z.
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n
Answer:
X Order of X = 2 × n
Z Order of Z = 2 × p
∵ p = n
Thus, order of 7X - 5Z = 2 × n,
Hence, choice (B) is correct.

Bhagya
Last Updated on Nov. 1, 2023, 9:23 a.m.
Published Oct. 31, 2023