Chapter 2 - Matrices EXERCISE 2.2 [PAGES 46 - 47] Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board

Chapter 2 - Matrices EXERCISE 2.2 [PAGES 46 - 47]

Exercise 2.2 | Q 1.1 | Page 46

QUESTION

If A = [2-35-4-61],B=[-122203]and C=[43-14-21], Show that A + B = B + A

SOLUTION

A + B = [2-35-4-61]+[-122203]

[2-1-3+25+2-4+2-6+01+3]

∴ A + B = [1-17-2-64]       ....(i)

B + A = [-122203]+[2-35-4-61]

[-1+22-32+52-40-63+1]

∴ B + A = [1-17-2-64]       ....(ii)

From (i) and (ii), we get
A + B = B + A.

Exercise 2.2 | Q 1.2 | Page 46

QUESTION

If A = [2-35-4-61],B=[-122203]and C=[43-14-21], Show that (A + B) + C = A + (B + C)

SOLUTION

(A + B) + C = {2-35-4-61]+[-122203]}+[43-14-21]

[2-1-3+25+2-4+2-6+01+3]+[43-14-21]

[1-17-2-64]+[43-14-21]

[1+4-1+37-1-2+4-6-24+1]

∴ (A+ B) + C = [5262-85]      ....(i)

A + (B + C) = [2-35-4-61]+{[-12220]+[43-14-21]}

[2-35-4-61]+[-1+42+32-12+40-23+1]

[2-35-4-61][3516-24]

[2+3-3+55+1-4+6-6-21+4]

[5262-85]         ....(ii)

From (i) and (ii), we get
(A + B) + C = A + (B + C).

Exercise 2.2 | Q 2 | Page 46

QUESTION

If A = [1-253],B=[1-34-7] , then find the matrix A − 2B + 6I, where I is the unit matrix of order 2.

SOLUTION

A – 2B + 6I = [1-253] 2[1-34-7]+6[1001]

[1-253]-[2-68-14]+[6006]

[1-2+6-2+6+05-8+0 +14+6]

[54-323].

Exercise 2.2 | Q 3 | Page 46

QUESTION

If A = [12-3-37-80-61],B=[9-12-42540-3] then find the matrix C such that A + B + C is a zero matrix.

SOLUTION

A + B + C is a zero martix.
∴ A + B + C = 0
∴ C = – (A + B)

-{[123-37-80-61]+[9-12-42540-3]}

-[1+92-1-3+2-3-47+2-8+50+4-6+01-3]

-[101-1-79-34-6-2]

[10-11-7-93-46-2].

Exercise 2.2 | Q 4 | Page 46 

QUESTION

If A = [1-23-5-60],B=[-1-24215]and C=[24-1-4-36], find the matrix X such that 3A – 4B + 5X = C.

SOLUTION

3A – 4B + 5X = C
∴ 5X =C + 4B – 3A

[24-1-4-36]+4[-1-24215]-3[1-23-5-60]

[24-1-4-36]+[-4-8168420]-[3-69-15-180]

[2-4- 34-8+6-1+16-9-4+8+5-3+4+ 186+20-0]

= 5X = [-526191926]

∴ X = 15[-526191926]

[-12565195195265].

Exercise 2.2 | Q 5 | Page 46

QUESTION

If A = [51-4320], find (AT)T.

SOLUTION

A = [51-4320]

∴ AT = [5312-40]

∴ (AT)T = [51-4320] = A.

 Exercise 2.2 | Q 6 | Page 46

QUESTION

If A = [731-2-41591], find (AT)T.

SOLUTION

A = [731-2-41591]

∴ AT = [7-253-49111]

∴ (AT)T = [731-2-41591] = A.

Exercise 2.2 | Q 7 | Page 47

QUESTION

Find a, b, c, if [135ab-5-7-4c0] is a symmetric matrix.

SOLUTION

Let A = [135ab-5-7-4c0]

∴ AT = [1b435-5ca-70]

Since A is a symmetric matrix,
A = AT

∴ [135ab-5-7-4c0]

[1b-435-5ca-70]

∴ By equality of matrices, we get

a = – 4, b = 35, c = – 7.

Exercise 2.2 | Q 8 | Page 47

QUESTION

Find x, y, z if [0-5ixy0z32-20] is a skew symmetric matrix.

SOLUTION

Let A = [0-5ixy0z32-20]

∴ AT = [0-5ixy0z32-20]

Since A is a skew-symmetric matrix,
A = AT

∴ [0-5ixy0z32-20]

[0-y-325i02-x-z0]

∴ By equality of matrices, we get

x = -32,y=5i,z=2

Exercise 2.2 | Q 9.1 | Page 47

QUESTION

For each of the following matrices, find its transpose and state whether it is symmetric, skew- symmetric or neither.

[12-52-34-549]

SOLUTION

Let A = [12-52-34-549]

∴ AT = [12-52-34-549]

∴ AT = A i.e., A = AT
∴ A is a symmetric matrix.

Exercise 2.2 | Q 9.2 | Page 47

QUESTION

For each of the following matrices, find its transpose and state whether it is symmetric, skew- symmetric or neither.

[251-546-1-63]

SOLUTION

Let A = [251-546-1-63]

∴ AT = [2-5-1-54-6163]

∴ AT = [-251-5-46-1-6-3]

∴ A ≠ AT and A ≠ – AT
∴ A is neither symmetric nor skew – symmetric matrix.

Exercise 2.2 | Q 9.3 | Page 47

QUESTION

For each of the following matrices, find its transpose and state whether it is symmetric, skew- symmetric or neither.

[01+2ii-2-1-2i0-72-i70]

SOLUTION

Let A = [01+2ii-2-1-2i0-72-i70]

∴ AT = [01-2i2-i-1+2i07i-2-70]

∴ AT = [01+2ii-2-1-2i0-72-i70]

∴ AT = – A i.e. A = – AT
∴ A is a skew-symmetric matrix.

Exercise 2.2 | Q 10 | Page 47

QUESTION

Construct the matrix A = [aij]3×3 where aij = i − j. State whether A is symmetric or skew-symmetric.

SOLUTION

A = [aij]3x3  

∴ A = [a11a12a13a21a22a23a31a32a33]

Given, aij = i – j
∴ a11 = 1 – 1 = 0, a12 = 1 – 2 = – 1, a13 = 1 – 3 = – 2
a21 = 2 – 1= 1, a22 = 2 – 2 = 0, a23 = 2 – 3 = – 1,
a31 = 3 – 1 = 2, a32 = 3 - 2 = 1, a33 = 3 – 3 = 0

∴ A = [0-1-210-1210]

∴ A[012-101-2-10]

-[0-1-210-1210]

∴ AT = – A i.e., A = – AT
∴ A is a skew-symmetric matrix.

Exercise 2.2 | Q 11 | Page 47

QUESTION

Solve the following equations for X and Y, if 3X − Y = [1-1-11]  and X – 3Y = [0-10-1].

SOLUTION

Given equations are

3X – Y = [1-1-11]          ...(i)

and X – 3Y = [0-10-1]     ...(ii)

By (i) x 3 – (ii), we get

8X = 3[1-1-11]-[0-10-1]

3[3-3-33]-[0-10-1]

[3-0-3+1-3-03+1]

∴ 8X = [3-2-34]

∴ X = 18[3-2-34]

∴ X = [38-28-3848]

[38-14-3812]

By (i) – (ii) x 3, we get

8Y = [1-1-11]-3[0-10-1]

[1-1-11]-[0-30-3]

[1-0-1+3-1-01+3]

∴ 8Y = [12-114]

∴ Y = 18[12-114]

[1828-1848]

[1814-1812].

Exercise 2.2 | Q 12 | Page 47

QUESTION

Find matrices A and B, if 2A – B = [6-60-421] and A – 2B = [328-21-7].

SOLUTION

Given equations are

2A – B = [6-60-421]           ...(i)

and A – 2B = [328-21-7]    ...(ii)

By (i) – (ii) x 2, we get

3B = [6-60-421]-2[328-21-7] 

[6-60-421]-[6416-42-14]

[6-6-6-4 0-16-4+42-21+14]

∴ 3B = [0-10-160015]

∴ B = 13[0-10-160015]

∴ B = [0-103-163005]

By (i) x 2 –  (ii), we get

3A = 2[6-60-421]-[328-21-7]

[12-120-842]-[3 28-21-7]

[12-3-12-20-8-8+24 -12+7]

∴ 3A = [9-14-8-639]

∴ A = 13[914-8-639]

∴ A = [3-143-83-213].

Exercise 2.2 | Q 13 | Page 47

QUESTION

Find x and y, if [2x+y-1134y4][-1 64303]=[3556187].

SOLUTION

[2x+y-1134y4][-1 64303]=[3556187]

∴ [2x+y-1-1+61+43+34y+04+3]=[3556187]

∴ [x+y-15564y7]=[3556187]

∴ By equality of matrices, we get
2x + y – 1 = 3 and 4y = 18

∴ 2x + y = 4 and y = 184=92

∴ 2+92 = 4

∴ 2x = 4-92

∴ 2x = -12

∴ x = -14 and y = 92.

Exercise 2.2 | Q 14 | Page 47

QUESTION

If [2a+b3a-bc+2d2c-d]=[234-1], find a, b, c and d.

SOLUTION

[2a+b3a-bc+2d2c-d]=[234-1]

∴ By  equality of matrices, we get
2a + b = 2        ....(i)
3a – b = 3        ....(ii)
c + 2d = 4       ....(iii)
2c –d = – 1      ....(iv)
Adding (i) and (ii), we get
5a = 5
∴ a = 1
Substituting a = 1 in (i), we get
2(1) + b = 2
∴ b = 0
By (iii) + (iv) x 2, we get
5c = 2

∴ c = 25
Substituting c = 25 i (iii), we get

25+2d = 4

∴ 2d = 4-25

∴ 2d = 185

∴ d = 95.

Exercise 2.2 | Q 15.1 | Page 47

QUESTION

There are two book shops own by Suresh and Ganesh. Their sales ( in Rupees) for books in three subject - Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B. July sales ( in Rupees) :
Physics Chemistry Mathematics
A = [560067508500665070558905][SureshGanesh]
August Sales (in Rupees :
B = [6650705589057000750010200][SureshGanesh]
Find the increase in sales in Rupees from July to August 2017.

SOLUTION


Increase  sales rrupees from july to August 2017.
For Suresh :
Increase in sales for Physics books
= 6650 – 5600 = ₹ 1050
Increase in sales for  Chemistry books
= 7055 – 6750 = ₹ 305
Increase in sales for Mathematics books
= 8905 – 8500 = ₹ 405
For Ganesh :
Increase in sales for Physics books
= 7000 – 6650 = ₹ 350
Increase in sales for  Chemistry books
= 7500 – 7055 = ₹ 455
Increase in sales for Mathematics books
= 10200 – 8905 = ₹ 1295.

Exercise 2.2 | Q 15.2 | Page 47

QUESTION

There are two book shops own by Suresh and Ganesh. Their sales ( in Rupees) for books in three subject - Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B. July sales ( in Rupees) :
Physics Chemistry Mathematics
A = [560067508500665070558905][SureshGanesh]
August Sales (in Rupees :
B = [6650705589057000750010200][SureshGanesh]
If both book shops get 10% profit in the month of August 2017, find the profit for each book seller in each subject in that month.

SOLUTION

Both book shops got 10% profit in the month of August 2017.
For Suresh :
Profit for Physics books = 6650×10100 = ₹ 665

Profit for Chemistry books = 7055×10100 = ₹ 705.50

Profit for Mathematics books = 8905×10100 = ₹ 890.50
For Ganesh :
Profit for Physics books = 7000×10100 = ₹ 700

Profit for Chemistry books = 7500×10100 = ₹ 750

Profit for Mathematics books = 10200×10100 = ₹ 1020

HSC Mathematics Full Solution

Concept: Algebra of Matrices

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board

Chapter 2 Matrices

Exercise 2.2 [PAGES 46 - 47

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