The document discusses the following: 1. There will be a quiz on Jan 27 covering sections 1.4, 1.5, 1.7, and 1.8 and any issues with quiz 1 should be discussed asap. 2. Test 1 will be on Feb 1 in class with more details to come. 3. Matrix multiplication is defined only when the number of columns of the first matrix equals the number of rows of the second matrix.

1. Announcements
Quiz 2 on Wednesday Jan 27 on sections 1.4, 1.5, 1.7 and 1.8
If you have any grading issues with quiz 1, please discuss with
me asap.
Test 1 will be on Feb 1, Monday in class. More details later.

2. Last Class..
Two matrices are equal if
they have the same size
the corresponding entries are all equal

3. Last Class..
Two matrices are equal if
they have the same size
the corresponding entries are all equal
If A and B are m × n matrices, the sum A + B is also an m × n matrix

4. Last Class..
Two matrices are equal if
they have the same size
the corresponding entries are all equal
If A and B are m × n matrices, the sum A + B is also an m × n matrix
The columns of A + B is the sum of the corresponding columns of A
and B .

5. Last Class..
Two matrices are equal if
they have the same size
the corresponding entries are all equal
If A and B are m × n matrices, the sum A + B is also an m × n matrix
The columns of A + B is the sum of the corresponding columns of A
and B .
A + B is dened only if A and B are of the same size.

6. Matrix Multiplication
Denition
Suppose A is an m × n matrix and let B be another matrix of size
n × p with columns b1 , b2 , . . . , bp , the product AB is the m × p
matrix whose columns are Ab1 , Ab2 , . . . , Abp
In other words,
AB = A b1 b2 . . . bp = Ab1 Ab2 . . . Abp

7. Matrix Multiplication
For multiplication of 2 matrices to be possible,
The number of columns in the rst matrix (multiplicant) must
be same as the number of rows in the second matrix
(multiplier).

8. Matrix Multiplication
For multiplication of 2 matrices to be possible,
The number of columns in the rst matrix (multiplicant) must
be same as the number of rows in the second matrix
(multiplier).
If both matrices are square matrices, you can always multiply
them either way (that is AB or BA)

9. Matrix Multiplication
For multiplication of 2 matrices to be possible,
The number of columns in the rst matrix (multiplicant) must
be same as the number of rows in the second matrix
(multiplier).
If both matrices are square matrices, you can always multiply
them either way (that is AB or BA)
For arbitrarily sized matrices, it is possible that you can nd
AB but not nd BA. Or you could nd BA but not AB .
Sometimes, both AB and BA will not exist.

10. Matrix Multiplication
For multiplication of 2 matrices to be possible,
The number of columns in the rst matrix (multiplicant) must
be same as the number of rows in the second matrix
(multiplier).
If both matrices are square matrices, you can always multiply
them either way (that is AB or BA)
For arbitrarily sized matrices, it is possible that you can nd
AB but not nd BA. Or you could nd BA but not AB .
Sometimes, both AB and BA will not exist.
Thus in general AB = BA

11. Questions
1. Suppose A is a 6 × 4 matrix and B is a 3 × 4 matrix, does AB
exist?

12. Questions
1. Suppose A is a 6 × 4 matrix and B is a 3 × 4 matrix, does AB
exist? NO
2. What about BA?

13. Questions
1. Suppose A is a 6 × 4 matrix and B is a 3 × 4 matrix, does AB
exist? NO
2. What about BA?NO
3. Suppose A is a 6 × 4 matrix and B is a 4 × 3 matrix, does AB
exist?

14. Questions
1. Suppose A is a 6 × 4 matrix and B is a 3 × 4 matrix, does AB
exist? NO
2. What about BA?NO
3. Suppose A is a 6 × 4 matrix and B is a 4 × 3 matrix, does AB
exist? Yes
4. What about BA?

15. Questions
1. Suppose A is a 6 × 4 matrix and B is a 3 × 4 matrix, does AB
exist? NO
2. What about BA?NO
3. Suppose A is a 6 × 4 matrix and B is a 4 × 3 matrix, does AB
exist? Yes
4. What about BA? NO

16. Questions
Suppose A is a 2 × 5 matrix and B is a 5 × 3 matrix, what are the
sizes of AB and BA?
B
A AB
 
∗ ∗ ∗
 ∗ ∗ ∗ 
∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗
∗ ∗ ∗ =
 
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗

 
 ∗ ∗ ∗ 
∗ ∗ ∗
2×5 5×3
The product BA is not dened here.

17. Questions
Suppose A is a 2 × 5 matrix and B is a 5 × 3 matrix, what are the
sizes of AB and BA?
B
A AB
 
∗ ∗ ∗
 ∗ ∗ ∗ 
∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗
∗ ∗ ∗ =
 
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗

 
 ∗ ∗ ∗ 
∗ ∗ ∗
2×5 5×3
Match
The product BA is not dened here.

18. Questions
Suppose A is a 2 × 5 matrix and B is a 5 × 3 matrix, what are the
sizes of AB and BA?
B
A AB
 
∗ ∗ ∗
 ∗ ∗ ∗ 
∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗
∗ ∗ ∗ =
 
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗

 
 ∗ ∗ ∗ 
∗ ∗ ∗
2×5 5×3
Match
Size of AB
The product BA is not dened here.

19. Questions
Suppose A is a 2 × 5 matrix and B is a 5 × 3 matrix, what are the
sizes of AB and BA?
B
A AB
 
∗ ∗ ∗
 ∗ ∗ ∗ 
∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗
∗ ∗ ∗ =
 
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗

 
 ∗ ∗ ∗ 
∗ ∗ ∗
2×5 5×3 2×3
Match
Size of AB
The product BA is not dened here.

20. The Row-Column rule
The following is the more ecient and easy way for computing AB .
If the product AB is dened, the entry in row i and column j of AB
is the sum of the products of corresponding entries from
row i of A and column j of B
For example, The entry in the rst row, second column in AB is the
sum of the products of the entries from the rst row of A and the
second column of B .

21. Example
0 2
 
2 3 −1
Find AB where A = and B =  −2 3
0 1 2

4 2

22. B : 3 rows 2 columns
 

 0 2 

 
 
0 −2 3
 
2×
 
 
 
4 2
 
 
   

 2 3 −1 


 −10 ∗ 

   
   
0 1 2
 
   ∗ ∗ 
A : 2 rows 3 columns

23. B : 3 rows 2 columns
 

 0 2 

 
 
0 −2 3
 
2×
 
 
2)  
(− 4 2
 
 
3×
   

 2 3 −1 


 −10 ∗ 

   
   
0 1 2
 
   ∗ ∗ 
A : 2 rows 3 columns

24. B : 3 rows 2 columns
 

 0 2 

 
 
0 −2 3
 
2×
 
 
2)  
(− 4 2
 
4
 
3×
1 )×
(−
   

 2 3 −1 


 −10 ∗ 

   
   
0 1 2
 
   ∗ ∗ 
A : 2 rows 3 columns

25. B : 3 rows 2 columns
 

 0 2 

 
 
0 −2 3
 
2×
 
 
2)
+
 
(− 4 2
 
4
 
3×
+
1 )×
(−
   

 2 3 −1 


 −10 ∗ 

   
   
0 1 2
 
   ∗ ∗ 
A : 2 rows 3 columns

26. B : 3 rows 2 columns
 

 0 2 

 
2
 
−2 3
 
2×
 
 
 
4 2
 
 
   

 2 3 −1 


 10 11 

   
   

0 1 2   ∗ ∗ 
A : 2 rows 3 columns

27. B : 3 rows 2 columns
 

 0 2 

 
2
 
−2 3
 
2×
 
 
3
 
3× 4 2
 
 
   

 2 3 −1 


 10 11 

   
   

0 1 2   ∗ ∗ 
A : 2 rows 3 columns

28. B : 3 rows 2 columns
 

 0 2 

 
2
 
−2 3
 
2×
 
 
3
 
3× 4 2
 
2
 
(−
1) ×
   

 2 3 −1 


 10 11 

   
   

0 1 2   ∗ ∗ 
A : 2 rows 3 columns

29. B : 3 rows 2 columns
 

 0 2 

 
2
 
−2 3
 
2×
 
 
+ 3
 
3 ×+ 4 2
 
2
 
(−
1) ×
   

 2 3 −1 


 10 11 

   
   

0 1 2   ∗ ∗ 
A : 2 rows 3 columns

30. B : 3 rows 2 columns
 

 0 2 

 
 
−2 3
 
 
0  
 
0×
4 2
 
 
   

 2 3 −1 


 10 11 

   
 
6
 
0 1 2
 
   ∗ 
A : 2 rows 3 columns

31. B : 3 rows 2 columns
 

 0 2 

 
 
−2 3
 
 
0  
 
0×
4 2
 
2)  
(−
1×
   

 2 3 −1 


 10 11 

   
 
6
 
0 1 2
 
   ∗ 
A : 2 rows 3 columns

32. B : 3 rows 2 columns
 

 0 2 

 
 
−2 3
 
 
0  
 
0×
4 2
 
2)  
(−
1×
4
2×
   

 2 3 −1 


 10 11 

   
 
6
 
0 1 2
 
   ∗ 
A : 2 rows 3 columns

33. B : 3 rows 2 columns
 

 0 2 

 
 
−2 3
 
 
0  
 
0×
4 2
 
2)
+
 
(−
1×
+
4
2×
   

 2 3 −1 


 10 11 

   
 
6
 
0 1 2
 
   ∗ 
A : 2 rows 3 columns

34. B : 3 rows 2 columns
 

 0 2 

 
 
−2 3
 
 
2
0×  
 
4 2
 
 
   

 2 3 −1 


 −10 11 

   
   
0 1 2 6 7
   
 
A : 2 rows 3 columns

35. B : 3 rows 2 columns
 

 0 2 

 
 
−2 3
 
 
2
0×  
 
3 4 2
 
 
1×
   

 2 3 −1 


 −10 11 

   
   
0 1 2 6 7
   
 
A : 2 rows 3 columns

36. B : 3 rows 2 columns
 

 0 2 

 
 
−2 3
 
 
2
0×  
 
3 4 2
 
 
1×
2
2×
   

 2 3 −1 


 −10 11 

   
   
0 1 2 6 7
   
 
A : 2 rows 3 columns

37. B : 3 rows 2 columns
 

 0 2 

 
 
−2 3
 
 
2
0×  
 
3 4 2
+  
 
1×
2
+
2×
   

 2 3 −1 


 −10 11 

   
   
0 1 2 6 7
   
 
A : 2 rows 3 columns C = A × B : 2 rows 2 columns

38. Identity Matrix
A diagonal matrix with 1's on the diagonals. (This
automatically means that it is a square matrix).
Use the notation In for the identity matrix of order n × n.
A matrix A of suitable order multiplied with the identity matrix
gives back A.
Example
1 0
0 1
I2
1 0 0
 
 0 1 0 
0 0 1
I3

39. Problem 2, sec 2.1 Parts 3 and 4
7 −5 1 1 2 −5
If B = ,C= and E = . Find CB
1 −4 −3 −2 1 3
and EB .

40. Problem 2, sec 2.1 Parts 3 and 4
7 −5 1 1 2 −5
If B = ,C= and E = . Find CB
1 −4 −3 −2 1 3
and EB .
Obviously the product EB cannot be found (E has 1 column and B
has 2 rows).

41. Problem 2, sec 2.1 Parts 3 and 4
7 −5 1 1 2 −5
If B = ,C= and E = . Find CB
1 −4 −3 −2 1 3
and EB .
Obviously the product EB cannot be found (E has 1 column and B
has 2 rows).
1 2 7 −5 1
CB =
−2 1 1 −4 −3

42. Problem 2, sec 2.1 Parts 3 and 4
7 −5 1 1 2 −5
If B = ,C= and E = . Find CB
1 −4 −3 −2 1 3
and EB .
Obviously the product EB cannot be found (E has 1 column and B
has 2 rows).
1 2 7 −5 1
CB =
−2 1 1 −4 −3
1.7 + 2.1 1.(−5) + 2.(−4) 1.1 + 2.(−3)
=
(−2).7 + 1.1 (−2).(−5) + 1.(−4) (−2).1 + 1.(−3)
9 −13 −5
=
−13 6 −5

43. Problem 4, sec 2.1
9 −1 3
 
If A =  −8 7 −6 , nd A − 5I3 and (5I3 )A.
−4 1 8

44. Problem 4, sec 2.1
9 −1 3
 
If A =  −8 7 −6 , nd A − 5I3 and (5I3 )A.
−4 1 8
1 0 0 5 0 0
   
Since I3 =  0 1 0 , 5I3 =  0 5 0 ,
0 0 1 0 0 5
9 −1 3 5 0 0 4 −1 3
     
A − 5I3 =  −8 7 −6  −  0 5 0  =  −8 2 −6 
−4 1 8 0 0 5 −4 1 3

45. Problem 4, sec 2.1
5 0 0 9 −1 3
  
(5I3 )A =  0 5 0   −8 7 −6  =
0 0 5 −4 1 8
5.9 + 0.(−8) + 0.(−4) 5.(−1) + 0.7 + 0.1 5.3 + 0.(−6) + 0.8
 
 0.9 + 5.(−8) + 0.(−4) 0.(−1) + 5.7 + 0.1 0.3 + 5.(−6) + 0.8 
0.9 + 0.(−8) + 5.(−4) 0.(−1) + 0.7 + 5.1 0.3 + 0.(−6) + 5.8
45 −5 15
 
= −40 35 −30 
−20 5 40

46. Problem 10, sec 2.1
2 −3 8 4 5 −2
A= , B= and C = . Verify that
−4 6 5 5 3 1
AB = AC but B = C .

47. Problem 10, sec 2.1
2 −3 8 4 5 −2
A= , B= and C = . Verify that
−4 6 5 5 3 1
AB = AC but B = C .
Solution: This example shows that the usual cancellation rule does
not apply to matrices in general. (Usually, if ab = ac , you can
cancel a on both sides (if a = 0 ) and this will give b = c .)
2 −3 8 4 1 −7
AB = =
−4 6 5 5 −2 14
2 −3 5 −2 1 −7
AC = =
−4 6 3 1 −2 14
Thus AB = AC but clearly, B = C .

48. Transpose of a Matrix
If A is an m × n matrix, then the transpose of A is denoted by
AT .

49. Transpose of a Matrix
If A is an m × n matrix, then the transpose of A is denoted by
AT .
AT is an n × m matrix whose columns are the rows of A. (Or
swap the rows and columns of A).

50. Transpose of a Matrix
If A is an m × n matrix, then the transpose of A is denoted by
AT .
AT is an n × m matrix whose columns are the rows of A. (Or
swap the rows and columns of A).
If A = AT , then A is called a symmetric matrix.

51. Transpose of a Matrix
If A is an m × n matrix, then the transpose of A is denoted by
AT .
AT is an n × m matrix whose columns are the rows of A. (Or
swap the rows and columns of A).
If A = AT , then A is called a symmetric matrix.
(AT )T = A.

52. Transpose of a Matrix
If A is an m × n matrix, then the transpose of A is denoted by
AT .
AT is an n × m matrix whose columns are the rows of A. (Or
swap the rows and columns of A).
If A = AT , then A is called a symmetric matrix.
(AT )T = A.
Example
1 2
 
1 3 5
If A =  3 4 , then AT =
2 4 6
5 6

53. Powers of a Matrix
If A is an n × n matrix, then Ak is A multiplied k times

54. Powers of a Matrix
If A is an n × n matrix, then Ak is A multiplied k times
Example
1 2 1 2 1 2 7 10
If A = , then A2 = =
3 4 3 4 3 4 15 22

55. Powers of a Matrix
If A is an n × n matrix, then Ak is A multiplied k times
Example
1 2 1 2 1 2 7 10
If A = , then A2 = =
3 4 3 4 3 4 15 22
7 10 1 2 37 54
Also, A3 = A2 A = =
15 22 3 4 81 118

56. Powers of a Matrix
If A is an n × n matrix, then Ak is A multiplied k times
Example
1 2 1 2 1 2 7 10
If A = , then A2 = =
3 4 3 4 3 4 15 22
7 10 1 2 37 54
Also, A3 = A2 A = =
15 22 3 4 81 118
Please do this and convince yourself that this is true.

57. Section 2.2 Inverse of a Matrix
Recall that for numbers the inverse (or the multilpicative inverse) is
its reciprocal.
The inverse of 8 is 1/8 or 8−1 .

58. Section 2.2 Inverse of a Matrix
Recall that for numbers the inverse (or the multilpicative inverse) is
its reciprocal.
The inverse of 8 is 1/8 or 8−1 .
Satises 8−1 .8 = 1 and 8.8−1 = 1

59. Section 2.2 Inverse of a Matrix
Recall that for numbers the inverse (or the multilpicative inverse) is
its reciprocal.
The inverse of 8 is 1/8 or 8−1 .
Satises 8−1 .8 = 1 and 8.8−1 = 1
Can generalize the concept of inverse to a matrix

60. Section 2.2 Inverse of a Matrix
Recall that for numbers the inverse (or the multilpicative inverse) is
its reciprocal.
The inverse of 8 is 1/8 or 8−1 .
Satises 8−1 .8 = 1 and 8.8−1 = 1
Can generalize the concept of inverse to a matrix
The matrix involved must be a square matrix
No slanted line notation for matrix inverses.

61. Section 2.2 Inverse of a Matrix
Given an n × n matrix A, we want to nd another n × n matrix C
such that AC = In and CA = In where In is the identity matrix of size
n × n.
Such a matrix C is called the inverse of A.

62. Section 2.2 Inverse of a Matrix
Given an n × n matrix A, we want to nd another n × n matrix C
such that AC = In and CA = In where In is the identity matrix of size
n × n.
Such a matrix C is called the inverse of A.
The inverse of a matrix is unique.

63. Section 2.2 Inverse of a Matrix
Given an n × n matrix A, we want to nd another n × n matrix C
such that AC = In and CA = In where In is the identity matrix of size
n × n.
Such a matrix C is called the inverse of A.
The inverse of a matrix is unique.
We denote the inverse of A by A−1

64. Section 2.2 Inverse of a Matrix
Given an n × n matrix A, we want to nd another n × n matrix C
such that AC = In and CA = In where In is the identity matrix of size
n × n.
Such a matrix C is called the inverse of A.
The inverse of a matrix is unique.
We denote the inverse of A by A−1
Thus if A is an n × n matrix, AA−1 = A−1 A = In

65. Section 2.2 Inverse of a Matrix
Not every n × n matrix has an inverse
If A−1 exists, we say that A is invertible.

66. Section 2.2 Inverse of a Matrix
Not every n × n matrix has an inverse
If A−1 exists, we say that A is invertible.
A matrix that is not invertible is also called a singular matrix.
Invertible matrices are also called non-singular matrices.

67. Section 2.2 Inverse of a Matrix
Not every n × n matrix has an inverse
If A−1 exists, we say that A is invertible.
A matrix that is not invertible is also called a singular matrix.
Invertible matrices are also called non-singular matrices.
Denition
a b
Given a matrix A = , the quantity ad − bc is called the
c d
determinant of A.

68. Finding Inverse of a 2 ×2 Matrix
a b
Let A = . If ad − bc = 0 then A is invertible and
c d
1 d −b
A−1 =
ad − bc −c a
.

69. Finding Inverse of a 2 ×2 Matrix
a b
Let A = . If ad − bc = 0 then A is invertible and
c d
1 d −b
A−1 =
ad − bc −c a
.
So if the determinant of A (or det A) is equal to 0, A−1 does not
exist.

70. Finding Inverse of a 2 ×2 Matrix
Steps for a 2 × 2 matrix A
1. First check whether det A=0. If so, stop. A is not invertible.

71. Finding Inverse of a 2 ×2 Matrix
Steps for a 2 × 2 matrix A
1. First check whether det A=0. If so, stop. A is not invertible.
2. If det A = 0, swap the main diagonal elements of A.

72. Finding Inverse of a 2 ×2 Matrix
Steps for a 2 × 2 matrix A
1. First check whether det A=0. If so, stop. A is not invertible.
2. If det A = 0, swap the main diagonal elements of A.
3. Then change the sign of both o diagonal elements (don't
swap these)

73. Finding Inverse of a 2 ×2 Matrix
Steps for a 2 × 2 matrix A
1. First check whether det A=0. If so, stop. A is not invertible.
2. If det A = 0, swap the main diagonal elements of A.
3. Then change the sign of both o diagonal elements (don't
swap these)
4. Divide this matrix (after steps 2 and 3) by detA to give A−1 .
(This divides each element of the resultant matrix.)
5. If you want to check your answer, you can see whether
1 0
AA−1 =
0 1
6. This method will not work for 3 × 3 or bigger matrices.

74. Finding Inverse of a 2 ×2 Matrix
Find the inverse of
1 2
A=
3 4
Solution: Here detA = (1)(4) − (2)(3) = −2 = 0. So we can nd A−1 .

75. Finding Inverse of a 2 ×2 Matrix
Find the inverse of
1 2
A=
3 4
Solution: Here detA = (1)(4) − (2)(3) = −2 = 0. So we can nd A−1 .
Interchange the positions of 1 and 4. Change the signs of 2 and 3.
Then we get
4 −2
−3 1
.

76. Finding Inverse of a 2 ×2 Matrix
Find the inverse of
1 2
A=
3 4
Solution: Here detA = (1)(4) − (2)(3) = −2 = 0. So we can nd A−1 .
Interchange the positions of 1 and 4. Change the signs of 2 and 3.
Then we get
4 −2
−3 1
. Divide each element of the matrix by detA which is -2. This gives
−2 1
A−1 =
3/2 −1/2

77. What's the use?
Remember solving the matrix equation Ax = b for suitable A and x ?

78. What's the use?
Remember solving the matrix equation Ax = b for suitable A and x ?
Theorem
If A is an n × n invertible matrix, then for each vector b in Rn , the
equation Ax = b has a unique solution x = A−1 b
So, nd the inverse and multiply with the vector b to get the vector
x.

79. Example
Use the inverse of the previous example to solve
x1 + 2x2 = 2
3x1 + 4x2 = 4
Solution: Based on the previous theorem,
x1 −2 1 2 0
x= = =
x2 3/2 −1/2 4 1
A−1 b

80. Example
Theorem
1. If A is an invertible matrix, its inverse A−1 is also an invertible
matrix and (A−1 )−1 = A

81. Example
Theorem
1. If A is an invertible matrix, its inverse A−1 is also an invertible
matrix and (A−1 )−1 = A
2. If A is an invertible matrix, its transpose AT is also an
invertible matrix and (AT )−1 = (A−1 )T

82. Example
Theorem
1. If A is an invertible matrix, its inverse A−1 is also an invertible
matrix and (A−1 )−1 = A
2. If A is an invertible matrix, its transpose AT is also an
invertible matrix and (AT )−1 = (A−1 )T
3. If A and B are invertible matrices, their product AB is also
invertible

83. Example
Theorem
1. If A is an invertible matrix, its inverse A−1 is also an invertible
matrix and (A−1 )−1 = A
2. If A is an invertible matrix, its transpose AT is also an
invertible matrix and (AT )−1 = (A−1 )T
3. If A and B are invertible matrices, their product AB is also
invertible and (AB )−1 = B −1 A−1 . This is called the
shoes-socks principle. (Remember the order in which you
put your socks and shoes on and the order in which you
remove them?)

84. Example
Denition
Elementary Matrix: A matrix obtained by doing one row operation
on an identity matrix.
Example
1 0 0
 
For I3 =  0 1 0  the following are elementary matrices.
0 0 1

85. Example
Denition
Elementary Matrix: A matrix obtained by doing one row operation
on an identity matrix.
Example
1 0 0
 
For I3 =  0 1 0  the following are elementary matrices.
0 0 1
0 0 1
 
E1 =  0 1 0 ,
1 0 0
R 1←→R 3

86. Example
Denition
Elementary Matrix: A matrix obtained by doing one row operation
on an identity matrix.
Example
1 0 0
 
For I3 =  0 1 0  the following are elementary matrices.
0 0 1
0 0 1 4 0 0
   
E1 =  0 1 0 , E2 =  0 1 0 ,
1 0 0 0 0 1
R 1←→R 3 4R 1

87. Example
Denition
Elementary Matrix: A matrix obtained by doing one row operation
on an identity matrix.
Example
1 0 0
 
For I3 =  0 1 0  the following are elementary matrices.
0 0 1
0 0 1 4 0 0 1 0 0
     
E1 =  0 1 0 , E2 =  0 1 0 , E3 =  0 1 0 
1 0 0 0 0 1 0 1 1
R 1←→R 3 4R 1 R 2+R 3