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.