Unit 1 Matrices and Determinants Examples
24
Mudassar Nazar Notes Page 1 Unit # 1 Matrices and Determinants Examples Example # 1 If A = and B = , then find A + B Solution A + B = + A + B = A + B = Example # 2 If A = and B = , then find A – B Solution A – B = - A – B = A – B = Example # 3 If A = and B = , then find A + B Solution
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Transcript of Unit 1 Matrices and Determinants Examples
Mudassar Nazar Notes Page 1
Unit # 1 Matrices and Determinants Examples
Example # 1
If A = and B = , then find A + B
Solution
A + B = +
A + B =
A + B =
Example # 2
If A = and B = , then find A – B
Solution
A – B = -
A – B =
A – B =
Example # 3
If A = and B = , then find A + B
Solution
Mudassar Nazar Notes Page 2
A + B = +
A + B =
A + B =
Example # 4
If A = and B = , then find A + B
Solution
A + B = +
A + B =
A + B =
Example # 5
If A = and B = , then find A – B
Solution
A – B = -
A – B =
Mudassar Nazar Notes Page 3
A – B =
Example # 6
If A = and B = , then verify Commutative Law Under Addition
Solution
A + B = +
A + B =
A + B =
And
B + A = +
B + A =
B + A =
Hence ,
A + B = B + A
Mudassar Nazar Notes Page 4
Example # 7
If A = , B = , C = , then verify Associative Law Under Addition
i.e., (A + B ) + C = A + ( B + C )
Solution
L.H.S = ( A + B ) + C
= +
= +
= +
=
=
R.H.S = A + ( B + C)
= +
= +
= +
Mudassar Nazar Notes Page 5
=
=
Hence,
(A + B) + C = A + ( B + C )
Example # 8
If A = and B = , then find AB
Solution
AB =
AB =
AB =
Example # 9
If A = and B = , then find AB
Solution
AB =
AB =
Mudassar Nazar Notes Page 6
AB =
Example # 10
If A = , B = and C = then verify Associative Law Under Multiplication
i.e., (AB) C = A (BC)
Solution
L.H.S = (AB) C
=
=
=
=
=
R.H.S = A(BC)
=
=
=
=
=
Hence,
(AB)C = A(BC)
Mudassar Nazar Notes Page 7
Example # 11
If A = , B = and C = then verify Distributive Law of Multiplication over
Addition.
i.e., (i) A(B+C) = AB + AC (Left Distributive Law)
(ii) (A+B)C = AC + BC (Right Distributive Law)
Solution (1) A(B+C) = AB +AC
L.H.S = A(B +C)
=
=
=
= —
=
R.H.S = AB + AC
= +
= +
= +
=
=
Mudassar Nazar Notes Page 8
Hence,
A(B+C) = AB + AC
(2) (A+B)C = AC + BC
L.H.S = (A +B)C
=
=
=
=
=
R.H.S = AC + BC
= +
= +
= +
=
=
Hence,
(A + B)C = AC + BC
Mudassar Nazar Notes Page 9
Example # 12
If A = , B = and C = then verify Distributive Law of Multiplication over
Subtraction..
i.e., (i) A(B – C ) = AB - AC (Left Distributive Law)
(ii) (A -B)C = AC - BC (Right Distributive Law)
Solution (1) A(B-C) = AB -AC
L.H.S = A(B -C)
=
= –
=
=
=
R.H.S = AB - AC
= -
= -
= -
=
=
Mudassar Nazar Notes Page 10
Hence,
A(B -C) = AB - AC
(2) (A-B)C = AC - BC
L.H.S = (A -B)C
=
=
=
=
=
R.H.S = AC - BC
= -
= -
= -
=
=
Hence,
(A - B)C = AC - BC
Mudassar Nazar Notes Page 11
Example # 13
If A = and B = , then verify whether Commutative Law of Multiplication holds or not?
i.e., AB = BA
Solution
L.H.S = AB
=
=
=
R.H.S = BA
=
=
=
Hence,
AB BA
Commutative Law under Multiplication does not hold in general
Mudassar Nazar Notes Page 12
Example # 14
If A = and B = , then verify whether Commutative Law of Multiplication holds or not?
i.e., AB = BA
Solution
L.H.S = AB
=
=
=
R.H.S = BA
=
=
=
Hence,
AB = BA
Commutative Law under Multiplication hold in particular case.
Mudassar Nazar Notes Page 13
Example # 15
If A = and B = , then prove that B is identity matrix of A
Or
Prove that AB = A = BA
Solution
AB =
=
=
BA =
=
=
Hence,
AB = A = BA
Example # 16
If A = and B = , then verify law of transpose of product
Or
Prove that (AB)t = BtAt
Solution L.H.S = (AB)t
Mudassar Nazar Notes Page 14
=
=
=
=
R.H.S = BtAt
=
=
=
=
Hence,
(AB)t = BtAt
Example # 17
If A = , then prove that AA-1 = I = A-1A
Solution
A =
=
= - 6 – (-1)
= -6 +1
= -5
Mudassar Nazar Notes Page 15
Adj A =
A-1 =
A-1 =
A-1 =
AA-1 =
AA-1 =
AA-1 = –
AA-1 =
AA-1 =
AA-1 = I
A-1A =
A-1A = –
A-1A = –
A-1A = I
Hence,
AA-1 = I = A-1A
Mudassar Nazar Notes Page 16
Example # 18
If A = and B = , then verify (AB)-1 = A-1B-1
Solution
AB =
AB =
AB =
=
= 3 – 0
= 3
= 3
Adj (AB)=
L.H.S = (AB)-1
=
=
=
=
= 0 – (-3)
= 3
Mudassar Nazar Notes Page 17
Adj B =
B-1 =
B-1 =
B-1 =
=
= 0 – (-1)
= 0+1
= 1
Adj A =
A-1 =
A-1 =
A-1 =
R.H.S = A-1B-1
=
=
=
=
Hence,
Mudassar Nazar Notes Page 18
(AB)-1 = A-1B-1
Example # 19
If A = and B = , then prove that A + B = 0 = B + A
Or
Prove that A and B are Additive Inverses of each other.
Solution
A + B = +
A + B =
A + B =
A + B = 0
B + A = +
B + A =
B + A =
B + A = 0
Hence,
A + B = 0 = B + A
Mudassar Nazar Notes Page 19
Example # 20
If B = , then find determinant of B
Solution
=
= 3 – (-2)
= 3+2
=5
Example # 21
If M = , then find determinant of M
i.e.,
Solution
=
= 6 – 6
= 0
Mudassar Nazar Notes Page 20
Example # 22
Solve the following system by using matrix inversion method.
4x – 2y = 8
3x + y = -4
Solution
4x – 2y = 8
3x + y = -4
=
Let
M =
=
= 4 – (-6)
= 4 + 6
= 10
Adj M =
M-1 =
M-1 =
M-1 =
= M-1
=
Mudassar Nazar Notes Page 21
=
=
=
Hence,
x = 0 and y = -4
Example # 23
Solve the following system of linear equation by Cramer’s Rule.
3x – 2y = 1
-2x + 3y = 2
Solution
3x – 2y = 1
-2x + 3y = 2
Let
A = , Ax = , Ay =
=
= 9 – 4
= 5
= 5
=
= 3 –(- 4)
= 3+4
Mudassar Nazar Notes Page 22
= 7
=
= 6 –(-2)
= 6+2
= 8
X = and y =
X = and y =
Example # 24
The length of a rectangle is 6 cm less than three times of its width. The perimeter of the rectangle is
140 cm. Find the dimensions of the rectangle.
Solution
Let the width of rectangle be x and length be y
According to first condition
Y = 3x – 6
6 = 3x – y
3x – y = 6
According to the second condition
Perimeter = 140 cm
2(x + y) = 140
X + y =
Mudassar Nazar Notes Page 23
x + y = 70
The matrix form is
=
Let
M =
=
= 3 – (-1)
= 3 +1
= 4
Adj M =
M-1 =
M-1 =
M-1 =
= M-1
=
=
=
=
Mudassar Nazar Notes Page 24
Hence,
x =19 and y = 51
Width = 19 cm
Length= 51 cm
