Unit #1 Matrices and determinants (Exercise 1.3, 1.4, 1.5 & 1.7)
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Transcript of Unit #1 Matrices and determinants (Exercise 1.3, 1.4, 1.5 & 1.7)
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7/30/2019 Unit #1 Matrices and determinants (Exercise 1.3, 1.4, 1.5 & 1.7)
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Mudassar Nazar Notes Page 1
Question# 5
For Matrices A = , B = and C = , verify the
following rules:
(i) A + C = C + ASolution
L.H.S = A + C
= +
=
=
R.H.S = C + A
= +
=
=
Hence,
A + C = C + A
(ii) A +B = B + A
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Mudassar Nazar Notes Page 2
Solution
L.H.S = +
=
=
L.H.S = +
=
=
Hence,
A+B = B+A
(iii) B + C = C + BSolution
L.H.S = B + C
= +
=
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Mudassar Nazar Notes Page 4
=
R.H.S = 2A + B
= 2 +
= +
=
=
Hence,
A + ( B + A) = 2A + B
(v) (C - B) +A = C +(A - B)
Solution
L.H.S = +
= +
= +
=
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Mudassar Nazar Notes Page 5
=
R.H.S = C + (A B )
= +
= +
= +
=
=
Hence,
(CB) +A = C + (A B )
(vi) 2A +B = A + ( A +B)
Solution
L.H.S = 2A + B
= 2 +
= +
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Mudassar Nazar Notes Page 6
=
=
R.H.S = A + (A + B)
= +
= +
= +
=
=
Hence,
2A + B = A + ( A+ B)
(vii) (C B ) A = (C A ) B
Solution
L.H.S. = (C
B )
A = ( C
A )
B
= -
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Mudassar Nazar Notes Page 7
= -
= -
=
=
R.H.S = ( C A ) - B
= -
= -
= -
=
=
Hence,
( C B ) A = ( C A ) B
(viii) ( A + B ) + C = A + ( B + C)
Solution
L.H.S = ( A + B ) + C
= +
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Mudassar Nazar Notes Page 8
= +
= +
=
=
R.H.S = A + ( B + C )
= +
= +
= +
=
=
Hence,
( A + B ) + C = A + ( B + C)
(ix) A + ( B
C ) = ( A
C ) + B
Solution
L.H.S = A + ( B C )
= +
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Mudassar Nazar Notes Page 9
= +
= +
=
=
R.H.S = ( A C ) + B
= +
= +
= +
=
=
Hence,
A + ( B C ) = ( A C ) + B
(x) 2A + 2B = 2 ( A + B )
Solution
L.H.S = 2A + 2B
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Mudassar Nazar Notes Page 10
= 2 + 2
= +
=
=
R.H.S = 2 ( A + B )
= 2
= 2
= 2
=
Hence,
2A + 2B = 2 ( A + B )
Question# 8
I f A = , B = , then verify that
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Mudassar Nazar Notes Page 11
(i) (A + B )t = At + BtSolution
L.H.S = (A + B )t
=t
=t
=t
=
R.H.S = At+ B
t
=t+
t
= +
=
=
Hence,
(A + B )t= A
t+ B
t
(ii) (A - B )t = At - BtSolution
L.H.S = (A - B )t
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Mudassar Nazar Notes Page 12
=t
=t
=t
=
R.H.S = At- B
t
=t-
t
= -
=
=
Hence,
(A - B )t
= At
- Bt
(iii) A + Atis symmetric
Solution
A =
At
=
A + At
= +
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Mudassar Nazar Notes Page 13
=
=
(A + At)
t=
(A + At)
t= (A + A
t)
Hence,
A + Atis symmetric
(iv) A - Atis skew-symmetric
Solution
A =
At
=
A At
= -
=
=
(A At)
t=
= -
(A At)
t= - ( AA
t)
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Mudassar Nazar Notes Page 14
Hence,
A - Atis skew-symmetric
(v) B +Btis symmetric
Solution
B =
Bt
=
B + Bt
= +
=
=
(B + Bt)
t=
(B + Bt)
t= (B+ B
t)
Hence,
B + Btis symmetric
(vi) B - Bt is skew-symmetricSolution
B =
Bt
=
B Bt
= -
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Mudassar Nazar Notes Page 15
=
=
(B Bt)
t=
= -
(BBt)
t= - ( B B
t)
Hence,
B - Btis skew-symmetric
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Mudassar Nazar Notes Page 16
Question # 5
Let A = , B = and C = verify whether
(i) AB BASolution
L.H.S = AB
=
=
=
R.H.S = BA
=
=
=
Hence,
AB BA
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Mudassar Nazar Notes Page 17
(ii) A(BC) = (AB)CSolution
L.H.S =
=
=
=
=
R.H.S = (AB)C
=
=
=
=
=
Hence,
A(BC) = (AB)C
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Mudassar Nazar Notes Page 18
(iii) A(B+C) =AB + ACSolution
L.H.S = A(B+C)
=
=
=
=
=
R.H.S = AB + AC
= +
=
=
=
=
Hence,
A( B+C) = AB + AC
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Mudassar Nazar Notes Page 19
(iv) A(B-C) = (A-B)C
Solution
L.H.S = A(B-C)
=
=
=
=
=
L.H.S = AB AC
= -
= -
= -
=
=
Hence,
A(B-C) = AB AC
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Mudassar Nazar Notes Page 20
Exercise 1.5 Question # (6)
If A = , B = , D = , then verify that
(i) (AB)-1 = B-1 A-1 (ii) (DA)-1 = A-1 D-1(i) Solution
AB =
=
=
=
= 0 (-48)
= 0 + 48
= 48
Adj (AB) =
L.H.S = (AB)-1
= . Adj (AB)
=
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Mudassar Nazar Notes Page 21
=
=
=
= 4- (-2)
= 4 + 2
= 6
Adj B =
B-1
= Adj B
B-1
=
=
Adj A =
A-1
= Adj A
R.H.S = B-1 A-1
=
=
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Mudassar Nazar Notes Page 22
=
=
=
=
Hence
(AB)-1
= B-1
A-1
(ii) (DA)-1 = A-1 D-1Solution
DA =
=
=
=
= 44 (-20)
= 44 + 20
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Mudassar Nazar Notes Page 23
= 64
Adj (DA) =
L.H.S = (DA)-1
= . Adj (DA)
=
=
=
=
= 8- 0
= 8
Adj A =
A-1
= Adj A
A-1 =
=
= 6-(-2)
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Mudassar Nazar Notes Page 24
= 6 + 2
= 8
Adj D =
D-1
= Adj D
=
R.H.S = A-1 D-1
=
=
=
=
=
=
Hence
(DA)-1
= A-1
D-1
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Mudassar Nazar Notes Page 25
Review Exercise 1: Question # 7
If A = , B = , then verify that
(i) (AB)t = Bt At (ii) (AB)-1 = B-1 A-1(i) Solution
L.H.S = (AB)t
=t
=t
=t
=
R.H.S = Bt
At
=t t
=
=
=
Hence
(AB)t
= Bt
At
(ii) (AB)-1 = B-1 A-1
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Mudassar Nazar Notes Page 26
Solution
AB =
=
AB =
=
= 0-10
= -10
= -10
Adj (AB) =
L.H.S = (AB)-1
= Adj (AB)
=
=
= -10 (-12)
= -10 + 12
= 2
Adj B =
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= Adj B
=
=
= -3 2
= -5
= -5
Adj A =
= Adj A
=
R.H.S = B-1
A-1
=
=
=
=
Hence(AB)
-1= B
-1A
-1
