Transcript Document

1.5 Elementary Matrices and a Method for Finding
A1
An elementary row operation (sometimes called just a row operation)
on a matrix A is any one of the following three types of operations:
• Interchange of two rows of A
• Replacement of a row r of A by c r for some number c ≠ 0
• Replacement of a row r1 of A by the sum r1 + c r2 of that row and a
multiple of another row r2 of A
An n×n elementary matrix is a matrix produced by applying exactly
one elementary row operation to In
• Eij is the elementary matrix obtained by interchanging the i-th and jth rows of In
• Ei(c) is the elementary matrix obtained by multiplying the i-th row of
In by c ≠ 0
• Eij(c) is the elementary matrix obtained by adding c times the j-th
row to the i-th row of In, where i ≠ j
Examples:
1
 1 4 0   1 0 0  
1 0 
 
 0
,
0
1
0
,
0
1
0
,

 



0 2 0 0 1 0 0 1 0

 
 0

0 0 0

1 0 0
0 0 1

0 1 0
Theorem (Row Operations by Matrix Multiplication)
Suppose that E is an m×m elementary matrix produced by applying a
particular elementary row operation to Im, and that A is an m×n matrix.
Then EA is the matrix that results from applying that same elementary row
operation to A
Remark:
The above theorem is primarily of theoretical interest. Computationally, it is
preferable to perform row operations directly rather than multiplying on the
left by an elementary matrix.
Theorem
Every elementary matrix is invertible, and the inverse is also an elementary matrix.
Theorem (Equivalent Statements)
If A is an n×n matrix, then the following statements are equivalent, that is, all
true or all false
• A is invertible
• Ax = 0 has only the trivial solution
• The reduced row-echelon form of A is In
• A is expressible as a product of elementary matrices
To find the inverse of an invertible matrix A, we must find a sequence of
elementary row operations that reduces A to the identity and then perform this
same sequence of operations on In to obtain A1
 1 2 3
Example: Using Row Operations to Find the inverse of A   2 5 3 


 1 0 8


Solution:
• To accomplish this we shall adjoin the identity matrix to the right side of A,
thereby producing a matrix of the form [A | I ]
• We shall apply row operations to this matrix until the left side is reduced to
I; these operations will convert the right side to A1 , so that the final matrix
1
will have the form [I | A ]
1 2 3 1 0 0 
2 5 3 0 1 0


1 0 8 0 0 1 
Thus
Row operations
rref
1 0 0 40 16 9 
0 1 0 13 5 3


0 0 1 5 2 1
 40 16 9 


A1   13 5 3 
 5 2 1 


If and n X n matrix A is not invertible, then it cannot be reduced to In by elementary
row operations, i.e, the computation can be stopped.
Example:
1 6 4


A   2 4 1
 1 2 5 


 1 6 4 1 0 0
 2 4 1 0 1 0 


 1 2 5 0 0 1 
Row operations
rref
1 0 11/ 4 0 1/ 4 1/ 2 
0 1 9 / 8 0 1/ 8 1/ 4 


0 0
0
1 1
1 
Since we have obtained a row of zeros on the left side, A is not invertible.
1.6 Further Results on Systems of Equations and Invertibility
Theorem 1.6.1
Every system of linear equations has either no solutions, exactly one
solution, or in finitely many solutions.
Theorem 1.6.2
If A is an invertible n×n matrix, then for each n×1 matrix b, the system of
equations Ax = b has exactly one solution, namely, x = A1 b.
Remark: this method is less efficient, computationally, than Gaussian elimination,
But it is important in the analysis of equations involving matrices.
Example: Solve the system by using A1
x1  2 x2  3x3  5
2 x1  5 x2  3x3  3
 8 x3  17
x1
Let
 1 2 3


A   2 5 3 ,
 1 0 8


 x1 
5
x   x2  , b   3 
 x3 
17 
We showed in previous section that
Then
 40 16 9   5   1 


x  A1b   13 5 3   3    1
 5 2 1  17   2 

   
Thus x1 = 1, x2 = -1, x3 =2
 40 16 9 


A1   13 5 3 
 5 2 1 


Linear Systems with a Common Coefficient Matrix
To solve a sequence of linear systems, Ax = b1, Ax = b2, …, Ax = bk, with
common coefficient matrix A
1
1
• If A is invertible, then the solutions x1 = A b1, x2 = A b2 , …, xk =
Ab1 k
• A more efficient method is to form the matrix [ A | b1 | b2| … | bk ], then
reduce it to reduced row-echelon form we can solve all k systems at
once by Gauss-Jordan elimination (Here A may not be invertible)
Example: Solve the system
(a) x1  2 x2  3x3  4
2 x1  5 x2  3x3  5
x1
Solution:
 8 x3  9
(a) x1  2 x2  3 x3  1
2 x1  5 x2  3x3  6
x1
 8 x3  6
1 2 3 4 1 
2 5 3 5 6 


1 0 8 9 6 
Reducing this to reduced row-echelon form yields (details on board)
1 0 0 1 2 
0 1 0 0 1 


0 0 1 1 1
Thus, the solution of system (a) is x1=1, x2=0, x3=1 and the solution
Of the system (b) is x1=2, x2=1, x3= -1
Theorem 1.6.3
Let A be a square matrix
1
(a) If B is a square matrix satisfying BA = I, then B = A
(b) If B is a square matrix satisfying AB = I, then B = A1
Theorem 1.6.5
Let A and B be square matrices of the same size. If AB is invertible, then
A and B must also be invertible
Theorem 1.6.4 (Equivalent Statements)
If A is an n×n matrix, then the following statements are equivalent
• A is invertible
• Ax = 0 has only the trivial solution
• The reduced row-echelon form of A is In
• A is expressible as a product of elementary matrices
• Ax = b is consistent for every n×1 matrix b
• Ax = b has exactly one solution for every n×1 matrix b
A Fundamental Problem: Let A be a fixed mXn matrix. Find all mX1 matrices b such
Such that the system of equations Ax=b is consistent.
If A is an invertible matrix, then for every mXn matrix b, the linear system Ax=b has
1
The unique solution x= A b.
If A is not square, or if A is a square but not invertible, then theorem 1.6.2 does not
Apply. In these cases the matrix b must satisfy certain conditions in order for Ax=b
To be consistent.
Determine Consistency by Elimination
Example: What conditions must b1, b2, and b3 satisfy in order for the system of
equations
x1  x2  2 x3  b1
x1 
 x3  b2
2 x1  x2  3x3  b3
To be consistent?
1 1 2 b1 
Solution: The augumented matrix is 

1
0
1
b
2


 2 1 3 b3 
Which can be reduced to row-echelon form as follows:
1 1 2 b1 
1 0 1 b  R2-R1
2

 2 1 3 b3 
-R2
b1 
1 1 2
0 1 1 b  b  R3-2R1
2
1

 2 1 3
b3 
b1  R3+R2
1 1 2
0 1 1 b  b 
1
2 

0 1 1 b3  2b1 
b1 
1 1 2
0 1 1 b  b 
2
1 

0 1 1 b3  2b1 
b1
1 1 2

0 1 1

b

b
1
2


0 0 0 b3  b1  b2 
From the third row in the matrix that the system has a solution if and only if
b1, b2, and b3 satisfy the condition
B3 - b1 - b2 = 0 or
b3 = b1 + b2
Example: What conditions must b1, b2, and b3 satisfy in order for the system of
equations
x1  2 x2  3x3  b1
2 x1  5 x2  3 x3  b2
x1
 8 x3  b3
To be consistent?
Solution: The augumented matrix is
1 2 3 b1 
2 5 3 b 
2

1 0 8 b3 
0 0 40b1  16b2  9b3 
0 1 0 13b  5b  3b 
1
2
3 

0 0 1
5b1  2b2  b3 
Which can be reduced to row-echelon form :1
In this case, there are no restrictions on b1, b2 and b3. That is, the system
Has the unique solution
x1=-40b1+16b2+9b3,
x2=13b1-5b2-3b3,
x3=5b1-2b2-b3