Copyright © 2011 Pearson, Inc.
7.2Matrix Algebra
Slide 7.2 - 2 Copyright © 2011 Pearson, Inc.
What you’ll learn about
Matrices Matrix Addition and Subtraction Matrix Multiplication Identity and Inverse Matrices Determinant of a Square Matrix Applications
… and whyMatrix algebra provides a powerful technique to manipulate large data sets and solve the related problems that are modeled by the matrices.
Slide 7.2 - 3 Copyright © 2011 Pearson, Inc.
Matrix
Let m and n be positive integers. An m ×n matrix
(read "m by n matrix") is a rectangular array of
m rows and n columns of real numbers.
a11 a12 L a1n
a21 a22 L a2n
M M M
am1 am2 L amn
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
We also use the shorthand notation aij⎡⎣
⎤⎦ for this matrix.
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Matrix Vocabulary
Each element, or entry, aij, of the matrix uses
double subscript notation. The row subscript is
the first subscript i, and the column subscript is
j. The element aij is the ith row and the jth
column. In general, the order of an m n
matrix is m n.
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Example Determining the Order of a Matrix
What is the order of the following matrix?
1 4 5
3 5 6
⎡
⎣⎢
⎤
⎦⎥
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Example Determining the Order of a Matrix
The matrix has 2 rows and 3 columns
so it has order 2×3.
What is the order of the following matrix?
1 4 5
3 5 6
⎡
⎣⎢
⎤
⎦⎥
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Matrix Addition and Matrix Subtraction
Let A = aij⎡⎣
⎤⎦ and B= bij
⎡⎣
⎤⎦ be matrices of order m×n.
1. The sum A+ B is the m×n matrix
A+ B= aij +bij⎡⎣
⎤⎦.
2. The difference A−B is the m×n matrix
A−B= aij −bij⎡⎣
⎤⎦.
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Example Matrix Addition
1 2 3
4 5 6
⎡
⎣⎢
⎤
⎦⎥+
2 3 45 6 7
⎡
⎣⎢
⎤
⎦⎥
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Example Matrix Addition
A+ B=
2+1 2 +3 3+ 44 +5 5+6 6 +7
⎡
⎣⎢
⎤
⎦⎥=
3 5 79 11 13
⎡
⎣⎢
⎤
⎦⎥
1 2 3
4 5 6
⎡
⎣⎢
⎤
⎦⎥+
2 3 45 6 7
⎡
⎣⎢
⎤
⎦⎥
Slide 7.2 - 10 Copyright © 2011 Pearson, Inc.
Example Using Scalar Multiplication
3
1 2 3
4 5 6
⎡
⎣⎢
⎤
⎦⎥
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Example Using Scalar Multiplication
=
3⋅1 3⋅2 3⋅33⋅4 3⋅5 3⋅6
⎡
⎣⎢
⎤
⎦⎥=
3 6 912 15 18
⎡
⎣⎢
⎤
⎦⎥
3
1 2 3
4 5 6
⎡
⎣⎢
⎤
⎦⎥
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The Zero Matrix
The m×n matrix 0 =[0] consisting entirely of
zeros is the zero matrix.
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Additive Inverse
Let A = aij⎡⎣
⎤⎦ be any m×n matrix.
The m×n matrix B= aij⎡⎣
⎤⎦ consisting of the additive
inverses of the entries of A is the additive inverse of Abecause A+ B=0.
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Matrix Multiplication
Let A = aij⎡⎣
⎤⎦ be any m×r matrix and B= bij
⎡⎣
⎤⎦
be any r ×n matrix.
The product AB= cij⎡⎣
⎤⎦ is the m×n matrix where
cij =ai1b1 j+ai2b2 j + ...+ airbrj .
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Example Matrix Multiplication
Find the product AB if possible.
A =1 2 30 1 −1
⎡
⎣⎢
⎤
⎦⎥ and B=
1 02 10 −1
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
Slide 7.2 - 16 Copyright © 2011 Pearson, Inc.
Example Matrix Multiplication
A =1 2 30 1 −1
⎡
⎣⎢
⎤
⎦⎥ and B=
1 02 10 −1
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
The number of columns of A is 3 and the number of
rows of B is 3, so the product is defined.
The product AB = cij⎡⎣
⎤⎦ is a 2×2 matrix where
c11 = 1 2 3⎡⎣ ⎤⎦
120
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥=1⋅1+ 2⋅2 +3⋅0 =5,
Slide 7.2 - 17 Copyright © 2011 Pearson, Inc.
Example Matrix Multiplication
A =1 2 30 1 −1
⎡
⎣⎢
⎤
⎦⎥ and B=
1 02 10 −1
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
c12= 1 2 3⎡⎣ ⎤⎦
01−1
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥=1⋅0+ 2⋅1+3⋅−1=−1,
c21 = 0 1 −1⎡⎣ ⎤⎦
120
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥=0⋅1+1⋅2 +−1⋅0 =2,
Slide 7.2 - 18 Copyright © 2011 Pearson, Inc.
Example Matrix Multiplication
A =1 2 30 1 −1
⎡
⎣⎢
⎤
⎦⎥ and B=
1 02 10 −1
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
c22= 0 1 −1⎡⎣ ⎤⎦
01−1
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥=0⋅0+1⋅1+−1⋅−1=2.
Thus AB=5 −12 2
⎡
⎣⎢
⎤
⎦⎥.
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Identity Matrix
The n×n matrix In with 1's on the main diagonal and
0's elsewhere is the identity matrix of order n×n.
In =
1 0 0 L 00 1 0 L 00 0 1 L 0M M M 00 0 0 0 1
⎡
⎣
⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥
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Inverse of a Square Matrix
Let A = aij⎡⎣
⎤⎦ be an n×n matrix.
If there is a matrix B such that
AB=BA=In,
then B is the inverse of A. We write B=A−1.
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Inverse of a 2 × 2 Matrix
If ad −bc≠0, then
a bc d
⎡
⎣⎢
⎤
⎦⎥
−1
=1
ad−bcd −b−c a
⎡
⎣⎢
⎤
⎦⎥.
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Determinant of a Square Matrix
Let A = aij⎡⎣
⎤⎦ be a matrix of order n×n (n> 2).
The determinant of A, denoted by detA or |A|,is the sum of the entries in any row or any column
multiplied by their respective cofactors. For
example, expanding by the ith row gives
detA=|A|=ai1Ai1 + ai2Ai2 + ...+ ainAin.
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Inverses of n n Matrices
An n n matrix A has an inverse if and only if
det A ≠ 0.
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Example Finding Inverse Matrices
Determine whether the matrix has an inverse.
If so, find its inverse matrix.
A =5 18 3
⎡
⎣⎢
⎤
⎦⎥
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Example Finding Inverse Matrices
Since det A =ad−bc=5⋅3−1⋅8 =7 ≠0,we conclude that A has an inverse.
Use the formula A−1 =1
ad−bcd −b−c a
⎡
⎣⎢
⎤
⎦⎥=
17
3 −1−8 5
⎡
⎣⎢
⎤
⎦⎥
=
37
−17
−87
57
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
.
A =
5 18 3
⎡
⎣⎢
⎤
⎦⎥
Slide 7.2 - 26 Copyright © 2011 Pearson, Inc.
Example Finding Inverse Matrices
Check:
A−1A=
37
−17
−87
57
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
5 18 3
⎡
⎣⎢
⎤
⎦⎥=
37−87
37−37
−407
+407
−87+157
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
=1 00 1
⎡
⎣⎢
⎤
⎦⎥=I2
A =
5 18 3
⎡
⎣⎢
⎤
⎦⎥
Similarly, A−1A=I2 .
Slide 7.2 - 27 Copyright © 2011 Pearson, Inc.
Properties of Matrices
Let A, B, and C be matrices whose orders are such that the following sums, differences, and products are defined.1. Community propertyAddition: A + B = B + AMultiplication: Does not hold in general2. Associative propertyAddition: (A + B) + C = A + (B + C)Multiplication: (AB)C = A(BC)3. Identity propertyAddition: A + 0 = AMultiplication: A·In = In·A = A
Slide 7.2 - 28 Copyright © 2011 Pearson, Inc.
Properties of Matrices
Let A, B, and C be matrices whose orders are such that the following sums, differences, and products are defined.4. Inverse propertyAddition: A + (-A) = 0Multiplication: AA-1 = A-1A = In |A|≠05. Distributive propertyMultiplication over addition:A(B + C) = AB + AC (A + B)C = AC + BCMultiplication over subtraction:A(B – C) = AB – AC (A – B)C = AC – BC
Slide 7.2 - 29 Copyright © 2011 Pearson, Inc.
Quick Review
The points (a) (1,−3) and (b) (x, y) are reflectedacross the given line.Find the coordinates of the reflected points.1. The x-axis
2. The line y=x3. The line y=−xExpand the expression,4. sin(x+ y)5. cos(x+ y)
Slide 7.2 - 30 Copyright © 2011 Pearson, Inc.
Quick Review Solutions
The points (a) (1,−3) and (b) (x, y) are reflectedacross the given line.Find the coordinates of the reflected points.1. The x-axis (a) (1,3) (b) (x,−y)2. The line y=x (a) ( −3,1) (b) (y,x)3. The line y=−x (a) ( −3,−1) (b) (−y,−x)Expand the expression,4. sin(x+ y) sinxcosy+sinycosx5. cos(x+ y) cosxcosy−sinxsiny