Matrix multiplication, inverse
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Transcript of Matrix multiplication, inverse
Announcements
Ï Quiz 2 on Wednesday Jan 27 on sections 1.4, 1.5, 1.7 and 1.8
Ï If you have any grading issues with quiz 1, please discuss withme asap.
Ï Test 1 will be on Feb 1, Monday in class. More details later.
Last Class..
Two matrices are equal if
Ï they have the same size
Ï 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 Aand B .
A+B is de�ned only if A and B are of the same size.
Last Class..
Two matrices are equal if
Ï they have the same size
Ï 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 Aand B .
A+B is de�ned only if A and B are of the same size.
Last Class..
Two matrices are equal if
Ï they have the same size
Ï 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 Aand B .
A+B is de�ned only if A and B are of the same size.
Last Class..
Two matrices are equal if
Ï they have the same size
Ï 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 Aand B .
A+B is de�ned only if A and B are of the same size.
Matrix Multiplication
De�nitionSuppose A is an m×n matrix and let B be another matrix of sizen×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
]
Matrix Multiplication
For multiplication of 2 matrices to be possible,
Ï The number of columns in the �rst matrix (multiplicant) mustbe same as the number of rows in the second matrix(multiplier).
Ï If both matrices are square matrices, you can always multiplythem either way (that is AB or BA)
Ï For arbitrarily sized matrices, it is possible that you can �ndAB but not �nd BA. Or you could �nd BA but not AB .Sometimes, both AB and BA will not exist.
Ï Thus in general AB 6=BA
Matrix Multiplication
For multiplication of 2 matrices to be possible,
Ï The number of columns in the �rst matrix (multiplicant) mustbe same as the number of rows in the second matrix(multiplier).
Ï If both matrices are square matrices, you can always multiplythem either way (that is AB or BA)
Ï For arbitrarily sized matrices, it is possible that you can �ndAB but not �nd BA. Or you could �nd BA but not AB .Sometimes, both AB and BA will not exist.
Ï Thus in general AB 6=BA
Matrix Multiplication
For multiplication of 2 matrices to be possible,
Ï The number of columns in the �rst matrix (multiplicant) mustbe same as the number of rows in the second matrix(multiplier).
Ï If both matrices are square matrices, you can always multiplythem either way (that is AB or BA)
Ï For arbitrarily sized matrices, it is possible that you can �ndAB but not �nd BA. Or you could �nd BA but not AB .Sometimes, both AB and BA will not exist.
Ï Thus in general AB 6=BA
Matrix Multiplication
For multiplication of 2 matrices to be possible,
Ï The number of columns in the �rst matrix (multiplicant) mustbe same as the number of rows in the second matrix(multiplier).
Ï If both matrices are square matrices, you can always multiplythem either way (that is AB or BA)
Ï For arbitrarily sized matrices, it is possible that you can �ndAB but not �nd BA. Or you could �nd BA but not AB .Sometimes, both AB and BA will not exist.
Ï Thus in general AB 6=BA
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
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
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
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
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
Questions
Suppose A is a 2×5 matrix and B is a 5×3 matrix, what are thesizes of AB and BA?
A︷ ︸︸ ︷[ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗
]B︷ ︸︸ ︷
∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗
=
AB︷ ︸︸ ︷[ ∗ ∗ ∗∗ ∗ ∗
]
2×5 5×3
Match
Size of AB
2×3
The product BA is not de�ned here.
Questions
Suppose A is a 2×5 matrix and B is a 5×3 matrix, what are thesizes of AB and BA?
A︷ ︸︸ ︷[ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗
]B︷ ︸︸ ︷
∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗
=
AB︷ ︸︸ ︷[ ∗ ∗ ∗∗ ∗ ∗
]
2×5 5×3
Match
Size of AB
2×3
The product BA is not de�ned here.
Questions
Suppose A is a 2×5 matrix and B is a 5×3 matrix, what are thesizes of AB and BA?
A︷ ︸︸ ︷[ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗
]B︷ ︸︸ ︷
∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗
=
AB︷ ︸︸ ︷[ ∗ ∗ ∗∗ ∗ ∗
]
2×5 5×3
Match
Size of AB
2×3
The product BA is not de�ned here.
Questions
Suppose A is a 2×5 matrix and B is a 5×3 matrix, what are thesizes of AB and BA?
A︷ ︸︸ ︷[ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗
]B︷ ︸︸ ︷
∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ ∗
=
AB︷ ︸︸ ︷[ ∗ ∗ ∗∗ ∗ ∗
]
2×5 5×3
Match
Size of AB
2×3
The product BA is not de�ned here.
The Row-Column rule
The following is the more e�cient and easy way for computing AB .
If the product AB is de�ned, 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 thesum of the products of the entries from the �rst row of A and thesecond column of B .
Example
Find AB where A=[2 3 −10 1 2
]and B =
0 2−2 34 2
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
−10 ∗
∗ ∗
2×0
3×(−2
)
(−1)×4
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
−10 ∗
∗ ∗
2×0
3×(−2
)
(−1)×4
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
−10 ∗
∗ ∗
2×0
3×(−2
)
(−1)×4
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
−10 ∗
∗ ∗
2×0
3×(−2
)
(−1)×4
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
10 11
∗ ∗
2×2
3×3
(−1)×2
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
10 11
∗ ∗
2×2
3×3
(−1)×2
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
10 11
∗ ∗
2×2
3×3
(−1)×2
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
10 11
∗ ∗
2×2
3×3
(−1)×2
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
10 11
6 ∗
0×0
1× (
−2)
2×4
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
10 11
6 ∗
0×0
1× (
−2)
2×4
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
10 11
6 ∗
0×0
1× (
−2)
2×4
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
10 11
6 ∗
0×0
1× (
−2)
2×4
+
+
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
−10 11
6 7
0×2
1×3
2×2
++
C =A×B : 2 rows 2 columns
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
−10 11
6 7
0×2
1×3
2×2
++
C =A×B : 2 rows 2 columns
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
−10 11
6 7
0×2
1×3
2×2
++
C =A×B : 2 rows 2 columns
2 3 −1
0 1 2
A : 2 rows 3 columns
0 2
−2 3
4 2
B : 3 rows 2 columns
−10 11
6 7
0×2
1×3
2×2
++
C =A×B : 2 rows 2 columns
Identity Matrix
Ï A diagonal matrix with 1's on the diagonals. (Thisautomatically means that it is a square matrix).
Ï Use the notation In for the identity matrix of order n×n.
Ï A matrix A of suitable order multiplied with the identity matrixgives back A.
Example [1 00 1
]︸ ︷︷ ︸
I2 1 0 00 1 00 0 1
︸ ︷︷ ︸
I3
Problem 2, sec 2.1 Parts 3 and 4
If B =[7 −5 11 −4 −3
], C =
[1 2−2 1
]and E =
[ −53
]. Find CB
and EB .
Obviously the product EB cannot be found (E has 1 column and B
has 2 rows).
CB =[
1 2−2 1
][7 −5 11 −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
]
Problem 2, sec 2.1 Parts 3 and 4
If B =[7 −5 11 −4 −3
], C =
[1 2−2 1
]and E =
[ −53
]. Find CB
and EB .
Obviously the product EB cannot be found (E has 1 column and B
has 2 rows).
CB =[
1 2−2 1
][7 −5 11 −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
]
Problem 2, sec 2.1 Parts 3 and 4
If B =[7 −5 11 −4 −3
], C =
[1 2−2 1
]and E =
[ −53
]. Find CB
and EB .
Obviously the product EB cannot be found (E has 1 column and B
has 2 rows).
CB =[
1 2−2 1
][7 −5 11 −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
]
Problem 2, sec 2.1 Parts 3 and 4
If B =[7 −5 11 −4 −3
], C =
[1 2−2 1
]and E =
[ −53
]. Find CB
and EB .
Obviously the product EB cannot be found (E has 1 column and B
has 2 rows).
CB =[
1 2−2 1
][7 −5 11 −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
]
Problem 4, sec 2.1
If A= 9 −1 3
−8 7 −6−4 1 8
, �nd A−5I3 and (5I3)A.
Since I3 = 1 0 0
0 1 00 0 1
, 5I3 = 5 0 0
0 5 00 0 5
,A−5I3 =
9 −1 3−8 7 −6−4 1 8
− 5 0 0
0 5 00 0 5
= 4 −1 3
−8 2 −6−4 1 3
Problem 4, sec 2.1
If A= 9 −1 3
−8 7 −6−4 1 8
, �nd A−5I3 and (5I3)A.
Since I3 = 1 0 0
0 1 00 0 1
, 5I3 = 5 0 0
0 5 00 0 5
,A−5I3 =
9 −1 3−8 7 −6−4 1 8
− 5 0 0
0 5 00 0 5
= 4 −1 3
−8 2 −6−4 1 3
Problem 4, sec 2.1
(5I3)A= 5 0 0
0 5 00 0 5
9 −1 3−8 7 −6−4 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)+0.80.9+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
Problem 10, sec 2.1
A=[
2 −3−4 6
], B =
[8 45 5
]and C =
[5 −23 1
]. Verify that
AB =AC but B 6=C .
Solution: This example shows that the usual cancellation rule doesnot apply to matrices in general. (Usually, if ab = ac , you cancancel a on both sides (if a 6= 0 ) and this will give b = c .)
AB =[
2 −3−4 6
][8 45 5
]=
[1 −7−2 14
]AC =
[2 −3−4 6
][5 −23 1
]=
[1 −7−2 14
]Thus AB =AC but clearly, B 6=C .
Problem 10, sec 2.1
A=[
2 −3−4 6
], B =
[8 45 5
]and C =
[5 −23 1
]. Verify that
AB =AC but B 6=C .
Solution: This example shows that the usual cancellation rule doesnot apply to matrices in general. (Usually, if ab = ac , you cancancel a on both sides (if a 6= 0 ) and this will give b = c .)
AB =[
2 −3−4 6
][8 45 5
]=
[1 −7−2 14
]AC =
[2 −3−4 6
][5 −23 1
]=
[1 −7−2 14
]Thus AB =AC but clearly, B 6=C .
Transpose of a Matrix
Ï If A is an m×n matrix, then the transpose of A is denoted byAT .
Ï AT is an n×m matrix whose columns are the rows of A. (Orswap the rows and columns of A).
Ï If A=AT , then A is called a symmetric matrix.
Ï (AT )T =A.
Example
If A= 1 2
3 45 6
, then AT =[1 3 52 4 6
]
Transpose of a Matrix
Ï If A is an m×n matrix, then the transpose of A is denoted byAT .
Ï AT is an n×m matrix whose columns are the rows of A. (Orswap the rows and columns of A).
Ï If A=AT , then A is called a symmetric matrix.
Ï (AT )T =A.
Example
If A= 1 2
3 45 6
, then AT =[1 3 52 4 6
]
Transpose of a Matrix
Ï If A is an m×n matrix, then the transpose of A is denoted byAT .
Ï AT is an n×m matrix whose columns are the rows of A. (Orswap the rows and columns of A).
Ï If A=AT , then A is called a symmetric matrix.
Ï (AT )T =A.
Example
If A= 1 2
3 45 6
, then AT =[1 3 52 4 6
]
Transpose of a Matrix
Ï If A is an m×n matrix, then the transpose of A is denoted byAT .
Ï AT is an n×m matrix whose columns are the rows of A. (Orswap the rows and columns of A).
Ï If A=AT , then A is called a symmetric matrix.
Ï (AT )T =A.
Example
If A= 1 2
3 45 6
, then AT =[1 3 52 4 6
]
Transpose of a Matrix
Ï If A is an m×n matrix, then the transpose of A is denoted byAT .
Ï AT is an n×m matrix whose columns are the rows of A. (Orswap the rows and columns of A).
Ï If A=AT , then A is called a symmetric matrix.
Ï (AT )T =A.
Example
If A= 1 2
3 45 6
, then AT =[1 3 52 4 6
]
Powers of a Matrix
If A is an n×n matrix, then Ak is A multiplied k times
Example
If A=[1 23 4
], then A2 =
[1 23 4
][1 23 4
]=
[7 1015 22
]Also, A3 =A2A=
[7 1015 22
][1 23 4
]=
[37 5481 118
]
Please do this and convince yourself that this is true.
Powers of a Matrix
If A is an n×n matrix, then Ak is A multiplied k times
Example
If A=[1 23 4
], then A2 =
[1 23 4
][1 23 4
]=
[7 1015 22
]
Also, A3 =A2A=[
7 1015 22
][1 23 4
]=
[37 5481 118
]
Please do this and convince yourself that this is true.
Powers of a Matrix
If A is an n×n matrix, then Ak is A multiplied k times
Example
If A=[1 23 4
], then A2 =
[1 23 4
][1 23 4
]=
[7 1015 22
]Also, A3 =A2A=
[7 1015 22
][1 23 4
]=
[37 5481 118
]
Please do this and convince yourself that this is true.
Powers of a Matrix
If A is an n×n matrix, then Ak is A multiplied k times
Example
If A=[1 23 4
], then A2 =
[1 23 4
][1 23 4
]=
[7 1015 22
]Also, A3 =A2A=
[7 1015 22
][1 23 4
]=
[37 5481 118
]
Please do this and convince yourself that this is true.
Section 2.2 Inverse of a Matrix
Recall that for numbers the inverse (or the multilpicative inverse) isits reciprocal.
Ï The inverse of 8 is 1/8 or 8−1.
Ï Satis�es 8−1.8= 1 and 8.8−1 = 1
Ï Can generalize the concept of inverse to a matrix
Ï The matrix involved must be a square matrix
Ï No slanted line notation for matrix inverses.
Section 2.2 Inverse of a Matrix
Recall that for numbers the inverse (or the multilpicative inverse) isits reciprocal.
Ï The inverse of 8 is 1/8 or 8−1.Ï Satis�es 8−1.8= 1 and 8.8−1 = 1
Ï Can generalize the concept of inverse to a matrix
Ï The matrix involved must be a square matrix
Ï No slanted line notation for matrix inverses.
Section 2.2 Inverse of a Matrix
Recall that for numbers the inverse (or the multilpicative inverse) isits reciprocal.
Ï The inverse of 8 is 1/8 or 8−1.Ï Satis�es 8−1.8= 1 and 8.8−1 = 1
Ï Can generalize the concept of inverse to a matrix
Ï The matrix involved must be a square matrix
Ï No slanted line notation for matrix inverses.
Section 2.2 Inverse of a Matrix
Recall that for numbers the inverse (or the multilpicative inverse) isits reciprocal.
Ï The inverse of 8 is 1/8 or 8−1.Ï Satis�es 8−1.8= 1 and 8.8−1 = 1
Ï Can generalize the concept of inverse to a matrix
Ï The matrix involved must be a square matrix
Ï No slanted line notation for matrix inverses.
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 sizen×n.
Ï Such a matrix C is called the inverse of A.
Ï The inverse of a matrix is unique.
Ï We denote the inverse of A by A−1
Ï Thus if A is an n×n matrix, AA−1 =A−1A= In
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 sizen×n.
Ï Such a matrix C is called the inverse of A.
Ï The inverse of a matrix is unique.
Ï We denote the inverse of A by A−1
Ï Thus if A is an n×n matrix, AA−1 =A−1A= In
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 sizen×n.
Ï Such a matrix C is called the inverse of A.
Ï The inverse of a matrix is unique.
Ï We denote the inverse of A by A−1
Ï Thus if A is an n×n matrix, AA−1 =A−1A= In
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 sizen×n.
Ï Such a matrix C is called the inverse of A.
Ï The inverse of a matrix is unique.
Ï We denote the inverse of A by A−1
Ï Thus if A is an n×n matrix, AA−1 =A−1A= In
Section 2.2 Inverse of a Matrix
Not every n×n matrix has an inverse
Ï If A−1 exists, we say that A is invertible.
Ï A matrix that is not invertible is also called a singular matrix.
Ï Invertible matrices are also called non-singular matrices.
De�nition
Given a matrix A=[
a b
c d
], the quantity ad −bc is called the
determinant of A.
Section 2.2 Inverse of a Matrix
Not every n×n matrix has an inverse
Ï If A−1 exists, we say that A is invertible.
Ï A matrix that is not invertible is also called a singular matrix.
Ï Invertible matrices are also called non-singular matrices.
De�nition
Given a matrix A=[
a b
c d
], the quantity ad −bc is called the
determinant of A.
Section 2.2 Inverse of a Matrix
Not every n×n matrix has an inverse
Ï If A−1 exists, we say that A is invertible.
Ï A matrix that is not invertible is also called a singular matrix.
Ï Invertible matrices are also called non-singular matrices.
De�nition
Given a matrix A=[
a b
c d
], the quantity ad −bc is called the
determinant of A.
Finding Inverse of a 2×2 Matrix
Let A=[
a b
c d
]. If ad −bc 6= 0 then A is invertible and
A−1 = 1
ad −bc
[d −b−c a
].
So if the determinant of A (or det A) is equal to 0, A−1 does notexist.
Finding Inverse of a 2×2 Matrix
Let A=[
a b
c d
]. If ad −bc 6= 0 then A is invertible and
A−1 = 1
ad −bc
[d −b−c a
].
So if the determinant of A (or det A) is equal to 0, A−1 does notexist.
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 6= 0, swap the main diagonal elements of A.
3. Then change the sign of both o� diagonal elements (don'tswap 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
AA−1 =[1 00 1
]6. This method will not work for 3×3 or bigger matrices.
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 6= 0, swap the main diagonal elements of A.
3. Then change the sign of both o� diagonal elements (don'tswap 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
AA−1 =[1 00 1
]6. This method will not work for 3×3 or bigger matrices.
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 6= 0, swap the main diagonal elements of A.
3. Then change the sign of both o� diagonal elements (don'tswap 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
AA−1 =[1 00 1
]6. This method will not work for 3×3 or bigger matrices.
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 6= 0, swap the main diagonal elements of A.
3. Then change the sign of both o� diagonal elements (don'tswap 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
AA−1 =[1 00 1
]6. This method will not work for 3×3 or bigger matrices.
Finding Inverse of a 2×2 Matrix
Find the inverse of
A=[1 23 4
]Solution: Here detA= (1)(4)− (2)(3)=−2 6= 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
A−1 =[ −2 13/2 −1/2
]
Finding Inverse of a 2×2 Matrix
Find the inverse of
A=[1 23 4
]Solution: Here detA= (1)(4)− (2)(3)=−2 6= 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
A−1 =[ −2 13/2 −1/2
]
Finding Inverse of a 2×2 Matrix
Find the inverse of
A=[1 23 4
]Solution: Here detA= (1)(4)− (2)(3)=−2 6= 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
A−1 =[ −2 13/2 −1/2
]
What's the use?
Remember solving the matrix equation Ax= b for suitable A and x?
TheoremIf A is an n×n invertible matrix, then for each vector b in Rn, the
equation Ax= b has a unique solution x=A−1b
So, �nd the inverse and multiply with the vector b to get the vectorx.
What's the use?
Remember solving the matrix equation Ax= b for suitable A and x?
TheoremIf A is an n×n invertible matrix, then for each vector b in Rn, the
equation Ax= b has a unique solution x=A−1b
So, �nd the inverse and multiply with the vector b to get the vectorx.
Example
Use the inverse of the previous example to solve{x1 + 2x2 = 23x1 + 4x2 = 4
Solution: Based on the previous theorem,
x=[
x1x2
]=
[ −2 13/2 −1/2
]︸ ︷︷ ︸
A−1
[24
]︸ ︷︷ ︸
b
=[01
]
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−1A−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?)
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−1A−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?)
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−1A−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?)
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−1A−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?)
Example
De�nitionElementary Matrix: A matrix obtained by doing one row operationon an identity matrix.
Example
For I3 = 1 0 0
0 1 00 0 1
the following are elementary matrices.
E1 = 0 0 1
0 1 01 0 0
︸ ︷︷ ︸
R1←→R3
, E2 = 4 0 0
0 1 00 0 1
︸ ︷︷ ︸
4R1
, E3 = 1 0 0
0 1 00 1 1
︸ ︷︷ ︸
R2+R3
Example
De�nitionElementary Matrix: A matrix obtained by doing one row operationon an identity matrix.
Example
For I3 = 1 0 0
0 1 00 0 1
the following are elementary matrices.
E1 = 0 0 1
0 1 01 0 0
︸ ︷︷ ︸
R1←→R3
,
E2 = 4 0 0
0 1 00 0 1
︸ ︷︷ ︸
4R1
, E3 = 1 0 0
0 1 00 1 1
︸ ︷︷ ︸
R2+R3
Example
De�nitionElementary Matrix: A matrix obtained by doing one row operationon an identity matrix.
Example
For I3 = 1 0 0
0 1 00 0 1
the following are elementary matrices.
E1 = 0 0 1
0 1 01 0 0
︸ ︷︷ ︸
R1←→R3
, E2 = 4 0 0
0 1 00 0 1
︸ ︷︷ ︸
4R1
,
E3 = 1 0 0
0 1 00 1 1
︸ ︷︷ ︸
R2+R3
Example
De�nitionElementary Matrix: A matrix obtained by doing one row operationon an identity matrix.
Example
For I3 = 1 0 0
0 1 00 0 1
the following are elementary matrices.
E1 = 0 0 1
0 1 01 0 0
︸ ︷︷ ︸
R1←→R3
, E2 = 4 0 0
0 1 00 0 1
︸ ︷︷ ︸
4R1
, E3 = 1 0 0
0 1 00 1 1
︸ ︷︷ ︸
R2+R3