Exercise 2.4 [Pages 59 - 60] Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 2 Matrices

Exercise 2.4 [Pages 59 - 60]

Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 2 Matrices Exercise 2.4 [Pages 59 - 60]

Exercise 2.4 | Q 1.1 | Page 59

QUESTION

Find AT,  if A = [13-45]

SOLUTION

A = [13-45]

∴ AT = [1-435].


Exercise 2.4 | Q 1.2 | Page 59

QUESTION

Find AT, if A = [2-61-405]

SOLUTION

A = [2-61-405]

∴ AT = [2-4-6015].


Exercise 2.4 | Q 2 | Page 59

QUESTION

If [aij]3×3, where aij = 2(i – j), find A and AT. State whether A and AT both are symmetric or skew-symmetric matrices?

SOLUTION

A = [aij]3×3=[a11a12a13a21a22a23a31a32a33]

Given, aij = 2 (i – j)
∴ a11 = 2(1 – 1) = 0,  a12 = 2(1 – 2) = – 2
a13 = 2(1 – 3) = – 4,  a21 = 2(2 – 1) = 2,
a22 = 2(2 – 2) = 0,     a23 = 2(2 – 3) = – 2,
a31 = 2(3 – 1) = 4,     a32 = 2(3 – 2) = 2,
a33 = 2(3 – 3) = 0

∴ A = [0-2-420-2420]

∴ AT = [024-20-2-4-20]

= -[0-2-420-2420]=-A

∴ AT = – A and A = – AT 
∴ A and AT both are skew-symmetric matrices.


Exercise 2.4 | Q 3 | Page 59

QUESTION

If A = [5-34-3-21], prove that (AT)T = A.

SOLUTION

A = [5-34-3-21]

∴ AT = [54-2-3-31]

∴ (AT)T = [5-34-3-21]
= A.


Exercise 2.4 | Q 4 | Page 59

QUESTION

If A = [12-52-34-549], prove that AT = A.

SOLUTION

A = [12-52-34-549]

∴ AT = [12-52-34-549]
=A.


Exercise 2.4 | Q 5.1 | Page 59

QUESTION

If A = [2-35-4-61],B=[214-1-33],C=[12-14-23], then show that (A + B)T = AT + BT.

SOLUTION

A = [2-35-4-61]+[214-1-33]

= [2+2-3+15+4-4-1-6-31+3]

= [4-29-5-94]

∴ (A + B)T = [49-9-2-54]        ...(i)

Now, AT = [25-6-3-41]and BT=[24-31-13]

∴ AT + BT = [25-6-3-41]+[24-31-13]

= [2+25+4-6-3-3+1-4-11+3]

= [49-9-2-54]         ...(ii)
From (i) and (ii, we get
(A + B)T = AT + BT.


Exercise 2.4 | Q 5.2 | Page 59

QUESTION

If A = [2-35-4-61],B=[214-1-33],C=[12-14-23], then show that (A – C)T = AT – CT.

SOLUTION

A – C = [2-35-4-61]-[12-14-23]

= [2-1-3-25+1-4-4-6+21-3]

= [1-56-8-4-2]

∴ (A – C)T = [16-4-5-8-2]              ...(i)

Now, AT = [25-6-3-41] and

CT = [1-1-2243]

∴ AT –  CT = [25-6-3-41]-[1-1-2243]

= [2-15+1-6+2-3-2-4-41-3]

= [16-4-5-8-2]          ...(ii)

From (i) and (ii), we get
(A – C)T = AT – CT.


Exercise 2.4 | Q 6 | Page 59

QUESTION

If A = [54-23] and B = [-134-1], then find CT, such that 3A – 2B + C = I, where I is e unit matrix of order 2.

SOLUTION

3A – 2B + C = I
∴ C = I + 2B – 3A

∴ C = [1001]+2[-134-1]-3[54-23]

= [1001]+[-268-2]-[1512-69]

= [1-2-150+6-120+8+61-2-9]

∴ C = [-16-614-10]

∴ C = [-1614-6-10].


Exercise 2.4 | Q 7.1 | Page 59

QUESTION

If A = [73004-2],B=[0-2321-4], then find AT + 4BT.

SOLUTION

A = [73004-2] and B=[0-2321-4]

∴ AT = [70340-2] and BT=[02-213-4]

AT + 4BT = [70340-2]+4[02-213-4]

= [70340-2]+[08-8412-16]

= [7+00+83-84+40+12-2-16]

= [78-5812-18].


Exercise 2.4 | Q 7.2 | Page 59

QUESTION

If A = [73004-2],B=[0-2321-4], then find 5AT – 5BT.

SOLUTION

A = [73004-2]andB=[0-2321-4]

∴ AT = [70340-2]andBT=[02-213-4]

5AT – 5B = 5(AT – BT)

= 5([70340-2]-[02-213-4])

= 5[7-00-23+24-10-3-2+4]

= 5[7-253-32]

= [35-102515-1510].


Exercise 2.4 | Q 8 | Page 59

QUESTION

If A = [101312],B=[21-435-2]and C=[023-1-10], verify that (A + 2B + 3C)T = AT + 2BT+ CT.

SOLUTION

A + 2B + 3C

= [101312]+2[21-435-2]+3[023-1-10]

= [101312]+[42-861-4]+[069-3-30]

= [1+4+00+2+61-8+93+6-31+10-32-4+0]

∴ A + 2B + 3C = [58268-2]

∴ [A + 2B + 3C]T = [56882-2]   ...(i)

Now, AT = [130112],BT=[2315-4-2]

and CT = [0-12-130]

∴ AT + 2BT + 3CT 

= [130112]+2[2315-4-2]+3[0-12-130]

= [130112]+[46210-8-4]+[0-36-390]

= [1+4+03+6+30+2+61+10-31-8+92-4+0]

∴ AT + 2BT + 3CT = [56882-2]     ...(iii)

From (i) and (ii), we get
[A + 2B + 3C]T = AT + 2BT + 3CT.


Exercise 2.4 | Q 9 | Page 59

QUESTION

If A = [-121-32-3] and B = [21-32-13], prove that (A + BT)T = AT + B.

SOLUTION

A = [-121-32-3]and B=[21-32-13]

∴ AT = [-1-3221-3]and BT=[2-3-1123]

∴ A + BT = [-121-32-3]+[2-31123]

= [-1+22-31-1-3+12+2-3+3]

= [1-10-240]

∴ (A + BT)T = [1-2-1400]       ...(i)

Now, AT + B = [-1-3221-3]+[21-32-13]

= [-1+2-3+12-32+21-1-3+3]

= [1-2-1400]                 ...(ii)
From (i) and (ii), we get
(A + BT)T = AT + B.


Exercise 2.4 | Q 10.1 | Page 59

QUESTION

Prove that A + AT is a symmetric and A – AT is a skew symmetric matrix, where A = [124321-2-32]

SOLUTION

A = [124321-2-32]

∴ AT = [13-222-3412]

∴ A + AT = [124321-2-32]+[13-222-3412]

= [1+12+34-23+22+21-3-2+4-3+12+2]

∴ A + AT = [25254-22-24]

∴ (A + AT)T = [25254-22-24]

∴ (A + AT)T = A + AT i.e., A + AT = (A + AT)T
∴ A + AT is a symmetric matrix.

A – AT = [124321-2-32]-[13-222-3412]

= [1-12-34+23-22-21+3-2-4-3-12-2]

∴ A – AT = [0-16104-6-40]

∴ (A – AT)T [01-6-10-4640]

= -[0-16104-6-40]

∴ (A – AT)T = –  (A – AT)
i.e., A – AT = –  (A – AT)T
∴ A – AT  is a skew symmetric matrix.


Exercise 2.4 | Q 10.2 | Page 59

QUESTION

Prove that A + AT is a symmetric and A – AT is a skew symmetric matrix, where A = [52-43-724-5-3]

SOLUTION

A = [52-43-724-5-3]

∴ AT = [5342-7-5-42-3]

∴ A + AT = [52-43-724-5-3]+[5342-7-5-42-3]

= [5+52+3-4+43+2-7-72-54-4-5+2-3-3]

∴ A + AT = [10505-14-30-3-6]

∴  (A + AT)T = A + AT i.e., A + AT = (A + AT)T 
∴ A + AT = is a symmetric matrix.

A – AT = [52-43-724-5-3]-[5342-7-5-42-3]

= [5-52-3-4-43-2-7+72+54+4-5-2-3+3]

= [0-1-81078-70]

∴ (A – AT)T = [018-10-7-870]

= [0-1-81078-70]

∴ (A – AT)T = – (A – AT)
i.e., A – AT = – (A – AT)T
∴ A – AT  is a skew symmetric matrix.


Exercise 2.4 | Q 11.1 | Page 59

QUESTION

Express each of the following matrix as the sum of a symmetric and a skew symmetric matrix [4-23-5].

SOLUTION

A square matrix A can be expressed as the sum of a symmetric and a skew-symmetric matrix as

A = 12(A+AT)+12(A-AT)

Let A = [4-23-5]

∴ AT = [43-2-5]

∴ A + AT = [4-23-5]+[43-2-5]

= [4+4-2+33-2-5-5]

= [811-10]

Also, A – AT = [4-23-5]-[43-2-5]

= [4-4-2-33+2-5+5]

= [0-550]

Let P = 12(A+AT)

= 12[811-10]

= [41212-5]
and
Q = 12(A-AT)

= 12[0-550]

= [0-52520]

∴ P is a symmetric matrix         ...[∵ aij = aij]

and Q is a skew-symmetric matrix.   ...[∵ aij = – aij]
∴ A = P + Q

∴ A = [41212-5]+[0-52520].


Exercise 2.4 | Q 11.2 | Page 59

QUESTION

Express each of the following matrix as the sum of a symmetric and a skew symmetric matrix [33-1-2-21-4-52].

SOLUTION

A square matrix A can be expressed as the sum of a symmetric and a skew-symmetric matrix as

A = 12(A+AT)+12(A-AT)

Let A = [33-1-2-21-4-52]

∴ AT = [3-2-43-2-5-112]

∴ A + AT = [33-1-2-21-4-52]+[3-2-43-2-5-112]

= [3+33-2-1-4-2+3-2-21-5-4-1-5+12+2]

= [61-51-4-4-5-44]

Also, A – AT = [33-1-2-21-4-52]-[3-2-43-2-5-112]

= [3-33+2-1+4-2-3-2+21+5-4+1-5-12-2]

= [053-506-3-60]

Let P = 12(A+AT)

= 12[61-51-4-4-5-44]

and Q = 12(A-AT)

= 12[053-506-3-60]

∴ P is a symmetric matrix          ...[∵ aij = aij]

and Q is a skew symmetric matrix.  ...[∵ aij = –  aij]
∴ A = P + Q

∴ A = 12[61-51-4-4-5-44]+12[053-506-3-60].


Exercise 2.4 | Q 12.1 | Page 60

QUESTION

If A = [2-13-241]and B=[03-42-11], verify that (AB)T = BTAT.

SOLUTION

A = [2-13-241]and B=[03-42-11]

∴ AT = [234-1-21]and BT=[023-1-41]

AB = [2-13-241][03-42-11]

= [0-26+1-8-10-49+2-12-20+212-1-16+1]

= [-27-9-411-14211-15]

∴ (AB)T = [-2-4271111-9-14-15]     ...(i)

BTAT = [023-1-41][234-1-21]

= [0-20-40+26+19+212-1-8-1-12-2-16+1]

= [-2-4271111-9-14-15]          ...(ii)
From (i) and (ii), we get
(AB)T = BTAT.


Exercise 2.4 | Q 12.2 | Page 60

QUESTION

If A = [2-13-241]and B=[03-42-11], verify that (BA)T = ATBT.

SOLUTION

BA = [03-42-11][2-13-241]

= [0+9-160-6-44-3+4-2+2+1]

∴ BA = [-7-1051]

∴ (BA)T = [-75-101]          ...(i)

ATBT = [234-1-21][023-1-41]

= [0+9-164-3+40-6-4-2+2+1]

= [-75-101]                ...(ii)
From (i) and (ii)
(BA)T = ATBT.

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