Matrix Inverse, IMT
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Transcript of Matrix Inverse, IMT
Announcements
Ï Quiz 2 on Wednesday Jan 27 on sections 1.4, 1.5, 1.7 and 1.8
Ï Test 1 will be on Feb 1, Monday in class. More details later.
Last Class
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 not
exist.
Last Class
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 not
exist.
Last Class
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'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
AA−1 =[1 0
0 1
]6. This method will not work for 3×3 or bigger matrices.
Last Class
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'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
AA−1 =[1 0
0 1
]6. This method will not work for 3×3 or bigger matrices.
Last Class
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'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
AA−1 =[1 0
0 1
]6. This method will not work for 3×3 or bigger matrices.
Last Class
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'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
AA−1 =[1 0
0 1
]6. This method will not work for 3×3 or bigger matrices.
Example 7(a) section 2.2, One Case
Let A=[1 2
5 12
]. Find A−1 and use it to solve the equation
Ax= b3 where b3 =[2
6
]
Solution: Here detA= (1)(12)− (5)(2)= 2 6= 0. So we can �nd A−1.
Interchange the positions of 1 and 12. Change the signs of 2 and 5.
Then we get [12 −2−5 1
]. Divide each element of the matrix by detA which is 2. This gives
A−1 =[
6 −1−5/2 1/2
]
x=[
x1x2
]=
[6 −1
−5/2 1/2
]︸ ︷︷ ︸
A−1
[2
6
]︸ ︷︷ ︸
b
=[
6
−2]
Example 7(a) section 2.2, One Case
Let A=[1 2
5 12
]. Find A−1 and use it to solve the equation
Ax= b3 where b3 =[2
6
]
Solution: Here detA= (1)(12)− (5)(2)= 2 6= 0. So we can �nd A−1.Interchange the positions of 1 and 12. Change the signs of 2 and 5.
Then we get [12 −2−5 1
].
Divide each element of the matrix by detA which is 2. This gives
A−1 =[
6 −1−5/2 1/2
]
x=[
x1x2
]=
[6 −1
−5/2 1/2
]︸ ︷︷ ︸
A−1
[2
6
]︸ ︷︷ ︸
b
=[
6
−2]
Example 7(a) section 2.2, One Case
Let A=[1 2
5 12
]. Find A−1 and use it to solve the equation
Ax= b3 where b3 =[2
6
]
Solution: Here detA= (1)(12)− (5)(2)= 2 6= 0. So we can �nd A−1.Interchange the positions of 1 and 12. Change the signs of 2 and 5.
Then we get [12 −2−5 1
]. Divide each element of the matrix by detA which is 2. This gives
A−1 =[
6 −1−5/2 1/2
]
x=[
x1x2
]=
[6 −1
−5/2 1/2
]︸ ︷︷ ︸
A−1
[2
6
]︸ ︷︷ ︸
b
=[
6
−2]
Algorithm to �nd A−1
To �nd the inverse of a square matrix A of any size,
1. Write the augmented matrix with A and I (the identity matrix
of proper size) written side by side.
2. Do proper row reductions on both A and I till A is reduced to
I .
3. This changes I to a new matrix which is A−1
4. If you cannot reduce A to I (if you get a row full of zeros for
example), A−1 does not exist
Algorithm to �nd A−1
To �nd the inverse of a square matrix A of any size,
1. Write the augmented matrix with A and I (the identity matrix
of proper size) written side by side.
2. Do proper row reductions on both A and I till A is reduced to
I .
3. This changes I to a new matrix which is A−1
4. If you cannot reduce A to I (if you get a row full of zeros for
example), A−1 does not exist
Algorithm to �nd A−1
To �nd the inverse of a square matrix A of any size,
1. Write the augmented matrix with A and I (the identity matrix
of proper size) written side by side.
2. Do proper row reductions on both A and I till A is reduced to
I .
3. This changes I to a new matrix which is A−1
4. If you cannot reduce A to I (if you get a row full of zeros for
example), A−1 does not exist
Algorithm to �nd A−1
To �nd the inverse of a square matrix A of any size,
1. Write the augmented matrix with A and I (the identity matrix
of proper size) written side by side.
2. Do proper row reductions on both A and I till A is reduced to
I .
3. This changes I to a new matrix which is A−1
4. If you cannot reduce A to I (if you get a row full of zeros for
example), A−1 does not exist
Example 30, section 2.2
Let A=[5 10
4 7
]. Find A−1 using the algorithm.
Solution: Start with the augmented matrix 5 10 1 0
4 7 0 1
Divide R1 by 5 1 2 1/5 0
4 7 0 1
Example 30, section 2.2
Let A=[5 10
4 7
]. Find A−1 using the algorithm.
Solution: Start with the augmented matrix 5 10 1 0
4 7 0 1
Divide R1 by 5 1 2 1/5 0
4 7 0 1
Example 30, section 2.2
1 2 1/5 0
4 7 0 1
R2-4R1
1 2 1/5 0
0 −1 −4/5 1
R1+2R2
=⇒[1 0 −7/5 2
0 −1 −4/5 1
]Divide R2 by -1 =⇒
[1 0 −7/5 2
0 1 4/5 −1]= [
I A−1 ]
Example 30, section 2.2
1 2 1/5 0
4 7 0 1
R2-4R1
1 2 1/5 0
0 −1 −4/5 1
R1+2R2
=⇒[1 0 −7/5 2
0 −1 −4/5 1
]
Divide R2 by -1 =⇒[1 0 −7/5 2
0 1 4/5 −1]= [
I A−1 ]
Example 30, section 2.2
1 2 1/5 0
4 7 0 1
R2-4R1
1 2 1/5 0
0 −1 −4/5 1
R1+2R2
=⇒[1 0 −7/5 2
0 −1 −4/5 1
]Divide R2 by -1 =⇒
[1 0 −7/5 2
0 1 4/5 −1]= [
I A−1 ]
Example of a 3×3 Matrix
Let A= 1 0 −2
3 −1 −4−2 3 −4
. Find A−1 using the algorithm.
Solution: Start with the augmented matrix
1 0 −2 1 0 0
3 −1 −4 0 1 0
−2 3 −4 0 0 1
R2-3R1
1 0 −2 1 0 0
0 −1 2 −3 1 0
−2 3 −4 0 0 1
R3+2R1
Example of a 3×3 Matrix
Let A= 1 0 −2
3 −1 −4−2 3 −4
. Find A−1 using the algorithm.
Solution: Start with the augmented matrix1 0 −2 1 0 0
3 −1 −4 0 1 0
−2 3 −4 0 0 1
R2-3R1
1 0 −2 1 0 0
0 −1 2 −3 1 0
−2 3 −4 0 0 1
R3+2R1
Example of a 3×3 Matrix
1 0 −2 1 0 0
0 −1 2 −3 1 0
0 3 −8 2 0 1
R3+3R2
=⇒ 1 0 −2 1 0 0
0 −1 2 −3 1 0
0 0 −2 −7 3 1
Divide R3 by -2
=⇒ 1 0 −2 1 0 0
0 −1 2 −3 1 0
0 0 1 7/2 −3/2 −1/2
Example of a 3×3 Matrix
1 0 −2 1 0 0
0 −1 2 −3 1 0
0 3 −8 2 0 1
R3+3R2
=⇒ 1 0 −2 1 0 0
0 −1 2 −3 1 0
0 0 −2 −7 3 1
Divide R3 by -2
=⇒ 1 0 −2 1 0 0
0 −1 2 −3 1 0
0 0 1 7/2 −3/2 −1/2
Example of a 3×3 Matrix1 0 −2 1 0 0
0 −1 2 −3 1 0
0 0 1 7/2 −3/2 −1/2
R2-2R3
1 0 −2 1 0 0
0 −1 0 −10 4 1
0 0 1 7/2 −3/2 −1/2
Divide R2 by -1
1 0 −2 1 0 0
0 1 0 10 −4 −1
0 0 1 7/2 −3/2 −1/2
R1+2R2
Example of a 3×3 Matrix
1 0 0 8 −3 −1
0 1 0 10 −4 −1
0 0 1 7/2 −3/2 −1/2
=
[I A−1
]
Example of a 3×3 Matrix
1 0 0 8 −3 −1
0 1 0 10 −4 −1
0 0 1 7/2 −3/2 −1/2
=
[I A−1
]
Example 32, section 2.2
Let A= 1 −2 1
4 −7 3
−2 6 −4
. Find A−1 using the algorithm.
Solution: Start with the augmented matrix1 −2 1 1 0 0
4 −7 3 0 1 0
−2 6 −4 0 0 1
R2-4R1
1 −2 1 1 0 0
0 1 −1 −4 1 0
−2 6 −4 0 0 1
R3+2R1
Example 32, section 2.2
1 −2 1 1 0 0
0 1 −1 −4 1 0
0 2 −2 2 0 1
R3-2R2
1 −2 1 1 0 0
0 1 −1 −4 1 0
0 0 0 10 −2 1
R3-2R2
Example 32, section 2.2
1 −2 1 1 0 0
0 1 −1 −4 1 0
0 2 −2 2 0 1
R3-2R2
1 −2 1 1 0 0
0 1 −1 −4 1 0
0 0 0 10 −2 1
R3-2R2
Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
2. Here A does not have pivot in every column (there is a free
variable)
3. The equation Ax= 0 has Non-trivial solution
4. The columns of A are linearly Dependent
5. A cannot be row-reduced to the 3×3 identity matrix
Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
2. Here A does not have pivot in every column (there is a free
variable)
3. The equation Ax= 0 has Non-trivial solution
4. The columns of A are linearly Dependent
5. A cannot be row-reduced to the 3×3 identity matrix
Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
2. Here A does not have pivot in every column (there is a free
variable)
3. The equation Ax= 0 has Non-trivial solution
4. The columns of A are linearly Dependent
5. A cannot be row-reduced to the 3×3 identity matrix
Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
2. Here A does not have pivot in every column (there is a free
variable)
3. The equation Ax= 0 has Non-trivial solution
4. The columns of A are linearly
Dependent
5. A cannot be row-reduced to the 3×3 identity matrix
Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
2. Here A does not have pivot in every column (there is a free
variable)
3. The equation Ax= 0 has Non-trivial solution
4. The columns of A are linearly Dependent
5. A cannot be row-reduced to the 3×3 identity matrix
Very Important
Conclusion: You cannot reduce A to I because of the zero row and
so A−1 does not exist
1. Here A does not have a pivot in every row.
2. Here A does not have pivot in every column (there is a free
variable)
3. The equation Ax= 0 has Non-trivial solution
4. The columns of A are linearly Dependent
5. A cannot be row-reduced to the 3×3 identity matrix
Section 2.3 Invertible Matrices
If a square matrix A of size n×n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n×n identity matrix
3. The equation Ax= 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly independent
5. The equation Ax= b is consistent for every b in Rn
6. The columns of A span Rn (because every row has a pivot and
recall theorem 4, sec 1.4)
Section 2.3 Invertible Matrices
If a square matrix A of size n×n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n×n identity matrix
3. The equation Ax= 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly independent
5. The equation Ax= b is consistent for every b in Rn
6. The columns of A span Rn (because every row has a pivot and
recall theorem 4, sec 1.4)
Section 2.3 Invertible Matrices
If a square matrix A of size n×n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n×n identity matrix
3. The equation Ax= 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly independent
5. The equation Ax= b is consistent for every b in Rn
6. The columns of A span Rn (because every row has a pivot and
recall theorem 4, sec 1.4)
Section 2.3 Invertible Matrices
If a square matrix A of size n×n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n×n identity matrix
3. The equation Ax= 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly independent
5. The equation Ax= b is consistent for every b in Rn
6. The columns of A span Rn (because every row has a pivot and
recall theorem 4, sec 1.4)
Section 2.3 Invertible Matrices
If a square matrix A of size n×n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n×n identity matrix
3. The equation Ax= 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly
independent
5. The equation Ax= b is consistent for every b in Rn
6. The columns of A span Rn (because every row has a pivot and
recall theorem 4, sec 1.4)
Section 2.3 Invertible Matrices
If a square matrix A of size n×n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n×n identity matrix
3. The equation Ax= 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly independent
5. The equation Ax= b is consistent for every b in Rn
6. The columns of A span Rn (because every row has a pivot and
recall theorem 4, sec 1.4)
Section 2.3 Invertible Matrices
If a square matrix A of size n×n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n×n identity matrix
3. The equation Ax= 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly independent
5. The equation Ax= b is consistent for every b in Rn
6. The columns of A span Rn (because every row has a pivot and
recall theorem 4, sec 1.4)
Section 2.3 Invertible Matrices
If a square matrix A of size n×n is invertible the following is true
Theorem
The Invertible Matrix Theorem (IMT)
1. A has n pivot positions (n pivot rows and n pivot columns)
2. A can be row-reduced to the n×n identity matrix
3. The equation Ax= 0 has only the trivial solution (no free
variables)
4. The columns of A are linearly independent
5. The equation Ax= b is consistent for every b in Rn
6. The columns of A span Rn (because every row has a pivot and
recall theorem 4, sec 1.4)
Example 2, Section 2.3
Determine whether A=[ −4 6
6 −9]is invertible. Why(not)?
Solution: Clearly the determinant is (-4)(-9)-(6)(6)=36-36=0. So
not invertible. Checking invertibility of 2×2 matrices are thus easy.
Not so obvious fact: The second column is -1.5 times the �rst
column. Since we have linearly dependent columns, by IMT, A is
not invertible.
Example 2, Section 2.3
Determine whether A=[ −4 6
6 −9]is invertible. Why(not)?
Solution: Clearly the determinant is (-4)(-9)-(6)(6)=36-36=0. So
not invertible. Checking invertibility of 2×2 matrices are thus easy.
Not so obvious fact: The second column is -1.5 times the �rst
column. Since we have linearly dependent columns, by IMT, A is
not invertible.
Example 4, Section 2.3
Determine whether A= −7 0 4
3 0 −12 0 9
is invertible. Why(not)?
Solution: Remember what happens to a set of vectors if the zero
vector is present in that set?
The vectors are linearly dependent.
Since the second column of A is full of zeros, we have linearly
dependent columns, and so A is not invertible.
Example 4, Section 2.3
Determine whether A= −7 0 4
3 0 −12 0 9
is invertible. Why(not)?
Solution: Remember what happens to a set of vectors if the zero
vector is present in that set? The vectors are linearly dependent.
Since the second column of A is full of zeros, we have linearly
dependent columns, and so A is not invertible.
Example 6, Section 2.3
Determine whether A= 1 −5 −4
0 3 4
−3 6 0
is invertible. Why(not)?
1 −5 −4
0 3 4
−3 6 0
R3+3R1
1 −5 −4
0 3 4
0 −9 −12
R3+3R2
Example 6, Section 2.3
1 −5 −4
0 3 4
0 0 0
Since the third row (and the third column) does not have a pivot,
we have linearly dependent columns and by the IMT A is not
invertible.
Example 6, Section 2.3
1 −5 −4
0 3 4
0 0 0
Since the third row (and the third column) does not have a pivot,
we have linearly dependent columns and by the IMT A is not
invertible.
Example 8, Section 2.3
Determine whether A=
1 3 7 4
0 5 9 6
0 0 2 8
0 0 0 10
is invertible. Why(not)?
Solution: This matrix is already in the echelon form. There are 4
pivot rows and 4 pivot columns. So A is invertible.
Example 8, Section 2.3
Determine whether A=
1 3 7 4
0 5 9 6
0 0 2 8
0 0 0 10
is invertible. Why(not)?
Solution: This matrix is already in the echelon form. There are 4
pivot rows and 4 pivot columns. So A is invertible.