determinants, properties and imt

Announcements

Ï Please bring any grade related questions regarding exam 1without delay.

Ï Homework set for exam 2 has been uploaded. Please check itoften, I may make small inclusions/exclusions.

Ï Last day to drop this class with grade "W" is Feb 4.

Ï No class tomorrow (Thurs, Feb 4).

Last Class

1. Saw how to compute 3×3 determinants.

2. Determinants of triangular matrices (just the product of thenumbers along the main diagonal)

3. Determinants of nice larger matrices:- Take advantage ofrows/columns which are predominantly zeros

Important Facts about Determinants

What is

A=∣∣∣∣ 1 23 4

∣∣∣∣?It is (1)(4)− (3)(2)=−2.

What about

A=∣∣∣∣ 3 41 2

∣∣∣∣?It is (3)(2)− (1)(4)= 2

Interchanged the rows and we have the same value but oppositesign


What about

A=∣∣∣∣ 12 161 2

∣∣∣∣?It is (12)(2)− (1)(16)= 24−16= 8.

Multiplied ONE row by 4 and the answer gets multiplied by 4

Note: If you multiply the entire determinant by 4, then the answergets multiplied by 4.4=16 not just by 4. Try it yourself.


Let A=∣∣∣∣ 3 41 2

∣∣∣∣ and B =∣∣∣∣ 3 44 6

∣∣∣∣

Here B is obtained by adding row 1 to row 2 of A.

What is B? It is (3)(6)− (4)(4)= 18−16= 2. The same!!


Let A=∣∣∣∣ 3 41 2

∣∣∣∣ and B =∣∣∣∣ 3 44 6

∣∣∣∣Here B is obtained by adding row 1 to row 2 of A.



Let A=∣∣∣∣ 3 41 2

∣∣∣∣ and B =∣∣∣∣ 3 47 10

∣∣∣∣

Here B is obtained by adding 2R1 to R2 of A.



Let A=∣∣∣∣ 3 41 2

∣∣∣∣ and B =∣∣∣∣ 3 47 10

∣∣∣∣Here B is obtained by adding 2R1 to R2 of A.


Theorem

Theorem

1. The value of determinant DOES NOT change if you add a

multiple of a row to another row.

2. The value of determinant only changes sign if you interchange

any 2 rows.

3. If you multiply any row of a determinant by a number k, the

value of the determinant gets multiplied by k.

Any uses?

1. Use these properties e�ectively and appropriately to reduce adeterminant to echelon form.

2. Then the determinant is just the product of the main diagonalentries. (It is a triangular matrix)

3. Do not forget to change sign if you interchange rows.

4. You can factor out any number common to all entries in a rowto make your determinant easier to work with.

Example 6, section 3.2

Find the determinant by row reduction to the echelon form.∣∣∣∣∣∣1 5 −33 −3 32 13 −7

∣∣∣∣∣∣

Solution: Basic row operations∣∣∣∣∣∣∣∣∣∣∣∣∣

1 5 −3

3 −3 3

2 13 −7

∣∣∣∣∣∣∣∣∣∣∣∣∣R2-3R1

R3-2R1


Find the determinant by row reduction to the echelon form.∣∣∣∣∣∣1 5 −33 −3 32 13 −7

∣∣∣∣∣∣Solution: Basic row operations∣∣∣∣∣∣∣∣∣∣∣∣∣

1 5 −3

3 −3 3

2 13 −7

∣∣∣∣∣∣∣∣∣∣∣∣∣R2-3R1

R3-2R1

∣∣∣∣∣∣∣∣∣∣∣∣∣

1 5 −3

0 −18 12

0 3 −1

∣∣∣∣∣∣∣∣∣∣∣∣∣= 6

∣∣∣∣∣∣∣∣∣∣∣∣∣

1 5 −3

0 −3 2

0 3 −1

∣∣∣∣∣∣∣∣∣∣∣∣∣ R2+R3

= 6

∣∣∣∣∣∣1 5 −30 −3 20 0 1

∣∣∣∣∣∣︸︷︷︸echelon

= 6(1)(−3)(1)=−18


Find the determinant by row reduction to the echelon form.∣∣∣∣∣∣∣∣∣1 3 3 −40 1 2 −52 5 4 −3−3 −7 −5 2

∣∣∣∣∣∣∣∣∣

Solution: Basic row operations∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 3 3 −4

0 1 2 −5

2 5 4 −3

−3 −7 −5 2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

R3-2R1R4+3R1


Find the determinant by row reduction to the echelon form.∣∣∣∣∣∣∣∣∣1 3 3 −40 1 2 −52 5 4 −3−3 −7 −5 2

∣∣∣∣∣∣∣∣∣Solution: Basic row operations∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 3 3 −4

0 1 2 −5

2 5 4 −3

−3 −7 −5 2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

R3-2R1R4+3R1


∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 3 3 −4

0 1 2 −5

0 −1 −2 5

0 2 4 −10

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣R2+R3

R4+2R3

∣∣∣∣∣∣∣∣∣1 3 3 −40 1 2 −50 0 0 00 0 0 0

∣∣∣∣∣∣∣∣∣

= (1)(1)(0)(0)= 0


∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 3 3 −4

0 1 2 −5

0 −1 −2 5

0 2 4 −10

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣R2+R3

R4+2R3

∣∣∣∣∣∣∣∣∣1 3 3 −40 1 2 −50 0 0 00 0 0 0

∣∣∣∣∣∣∣∣∣ = (1)(1)(0)(0)= 0

Important

1. If a determinant has a row (or column) full of zeros, then it isequal to 0. ALWAYS!

2. Even before reaching the "all-zero" stage if you see two (ormore) rows (or columns) being exactly the same, then thedeterminant is 0.

3. Same when any row (or respectively column) is a multiple ofany other row (or respectively column). The determinant is 0.

4. Sometimes you may have to do row reductions �rst and thendo co-factor expansion. (An echelon form may not happenalways)


Combine row reduction and cofactor expansion∣∣∣∣∣∣∣∣∣−1 2 3 03 4 3 05 4 6 64 2 4 3

∣∣∣∣∣∣∣∣∣

Solution: Basic row operations∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

−1 2 3 0

3 4 3 0

5 4 6 6

4 2 4 3

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

R2+3R1R3+5R1

R4+4R1


Combine row reduction and cofactor expansion∣∣∣∣∣∣∣∣∣−1 2 3 03 4 3 05 4 6 64 2 4 3

∣∣∣∣∣∣∣∣∣Solution: Basic row operations∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

−1 2 3 0

3 4 3 0

5 4 6 6

4 2 4 3

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

R2+3R1R3+5R1

R4+4R1


∣∣∣∣∣∣∣∣∣−1 2 3 00 10 12 00 14 21 60 10 16 3

∣∣∣∣∣∣∣∣∣−1 2 3 0

0 10 12 0

0 14 21 6

0 10 16 3

=⇒−1∣∣∣∣∣∣10 12 014 21 610 16 3

∣∣∣∣∣∣


Factor 2 from �rst row, =⇒−1(2)∣∣∣∣∣∣5 6 014 21 610 16 3

∣∣∣∣∣∣ .

Do cofactor expansion along the �rst row

5 6 0

14 21 6

10 16 3

5 6 0

14 21 6

10 16 3

5 6 0

14 21 6

10 16 3

detA= 5

∣∣∣∣ 21 616 3

∣∣∣∣︸︷︷︸63−96=−33

−6

∣∣∣∣ 14 610 3

∣∣∣∣︸︷︷︸42−60=−18

+0

∣∣∣∣ 14 2110 16

∣∣∣∣︸︷︷︸14

=−165+108+0=−57Don't forget to multiply the -2 we had. So the answer is 114.

Example 16, 18, 20 section 3.2

If

∣∣∣∣∣∣a b c

d e f

g h i

∣∣∣∣∣∣= 7, �nd the following.

1.

∣∣∣∣∣∣a b c

3d 3e 3fg h i

∣∣∣∣∣∣ = 21

2.

∣∣∣∣∣∣g h i

a b c

d e f

∣∣∣∣∣∣ = 7 (Two row interchanges, �rst between R1 and

R3 making it -7 and then between R2 and new R3 making it-(-7)=7)

3.

∣∣∣∣∣∣a+d b+e c + f

d e f

g h i

∣∣∣∣∣∣ = 7 (Adding 2 rows gives same value)


If

∣∣∣∣∣∣a b c

d e f

g h i


1.

∣∣∣∣∣∣a b c

3d 3e 3fg h i

∣∣∣∣∣∣

= 21

2.

∣∣∣∣∣∣g h i

a b c

d e f



3.

∣∣∣∣∣∣a+d b+e c + f

d e f

g h i



If

∣∣∣∣∣∣a b c

d e f

g h i


1.

∣∣∣∣∣∣a b c

3d 3e 3fg h i

∣∣∣∣∣∣ = 21

2.

∣∣∣∣∣∣g h i

a b c

d e f



3.

∣∣∣∣∣∣a+d b+e c + f

d e f

g h i



If

∣∣∣∣∣∣a b c

d e f

g h i


1.

∣∣∣∣∣∣a b c

3d 3e 3fg h i

∣∣∣∣∣∣ = 21

2.

∣∣∣∣∣∣g h i

a b c

d e f

∣∣∣∣∣∣

= 7 (Two row interchanges, �rst between R1 and


3.

∣∣∣∣∣∣a+d b+e c + f

d e f

g h i



If

∣∣∣∣∣∣a b c

d e f

g h i


1.

∣∣∣∣∣∣a b c

3d 3e 3fg h i

∣∣∣∣∣∣ = 21

2.

∣∣∣∣∣∣g h i

a b c

d e f

∣∣∣∣∣∣ = 7

(Two row interchanges, �rst between R1 and


3.

∣∣∣∣∣∣a+d b+e c + f

d e f

g h i



If

∣∣∣∣∣∣a b c

d e f

g h i


1.

∣∣∣∣∣∣a b c

3d 3e 3fg h i

∣∣∣∣∣∣ = 21

2.

∣∣∣∣∣∣g h i

a b c

d e f



3.

∣∣∣∣∣∣a+d b+e c + f

d e f

g h i



If

∣∣∣∣∣∣a b c

d e f

g h i


1.

∣∣∣∣∣∣a b c

3d 3e 3fg h i

∣∣∣∣∣∣ = 21

2.

∣∣∣∣∣∣g h i

a b c

d e f



3.

∣∣∣∣∣∣a+d b+e c + f

d e f

g h i

∣∣∣∣∣∣

= 7 (Adding 2 rows gives same value)


If

∣∣∣∣∣∣a b c

d e f

g h i


1.

∣∣∣∣∣∣a b c

3d 3e 3fg h i

∣∣∣∣∣∣ = 21

2.

∣∣∣∣∣∣g h i

a b c

d e f



3.

∣∣∣∣∣∣a+d b+e c + f

d e f

g h i


Back to Matrix Inverses

Theorem

A square matrix A is invertible if and only if detA6= 0

Invertible Matrix Theorem.. Again

Remember the following from the invertible matrix theorem?

1. If the columns of the matrix are linearly dependent whichmeans

2. There are non-pivot columns (or free variables)

3. The matrix IS NOT invertible

This just means that the determinant of the matrix is 0.

Invertible Matrix Theorem.. Again

If the determinant of a matrix is 0 then one of the following couldbe true.

1. The columns of the matrix are linearly dependent. (samecolumns or one column being multiple of another)

2. The rows of the matrix are linearly dependent. (same rows orone row being multiple of another)

3. Row or column full of zeros

Example 22, section 3.2Use determinant to �nd out if the matrix is invertible. 5 0 −1

1 −3 −20 5 3

5 0 −1

1 −3 −2

0 5 3

5 0 −1

1 −3 −2

0 5 3

5 0 −1

1 −3 −2

0 5 3

detA= 5

∣∣∣∣ −3 −25 3

∣∣∣∣︸︷︷︸1

−0

∣∣∣∣ 1 −20 3

∣∣∣∣︸︷︷︸3

−1

∣∣∣∣ 1 −30 5

∣∣∣∣︸︷︷︸5

= 5−0−5= 0

The matrix is not invertible.

Column Operations

1. You could do the same operations you do with the rows to thecolumns as well

2. Not usually done, just to avoid confusion.

Theorem

If A is an n×n matrix, then detA= detAT

Determinants and Matrix Products

Theorem

If A and B are n×n matrices, then detAB = (detA)(detB)

Also,

If A is an n×n matrix, then detAk = (detA)k for any number k .

(Use this in problem 29 in your homework)

However in general,If A and B are n×n matrices, then det(A+B) 6= detA+detB

Example 40 section 3.2

Let A and B be 4×4 matrices with det A =-1 and det B=2.Compute

1. detAB =−22. detB5 = 25 = 32

3. det2A=−164. detATA= (detA)(detA)= 1

5. detB−1AB = (detB−1)(detA)(detB)= 1

2(−1)(2)=−1

Do you really believe here that detB−1 = 1

2?Yes! Think why and do

homework problem 31. It is simple.



1. detAB

=−22. detB5 = 25 = 32



2(−1)(2)=−1






1. detAB =−2

2. detB5 = 25 = 32



2(−1)(2)=−1






1. detAB =−22. detB5

= 25 = 32



2(−1)(2)=−1






1. detAB =−22. detB5 = 25 = 32



2(−1)(2)=−1






1. detAB =−22. detB5 = 25 = 32

3. det2A

=−164. detATA= (detA)(detA)= 1


2(−1)(2)=−1






1. detAB =−22. detB5 = 25 = 32

3. det2A=−16

4. detATA= (detA)(detA)= 1


2(−1)(2)=−1






1. detAB =−22. detB5 = 25 = 32

3. det2A=−164. detATA

= (detA)(detA)= 1


2(−1)(2)=−1






1. detAB =−22. detB5 = 25 = 32



2(−1)(2)=−1






1. detAB =−22. detB5 = 25 = 32


5. detB−1AB

= (detB−1)(detA)(detB)= 1

2(−1)(2)=−1






1. detAB =−22. detB5 = 25 = 32



2(−1)(2)=−1






1. detAB =−22. detB5 = 25 = 32



2(−1)(2)=−1


2?

Yes! Think why and dohomework problem 31. It is simple.



1. detAB =−22. detB5 = 25 = 32



2(−1)(2)=−1




Example

Let A=[9 24 1

]. Find det A9.

Should you multiply the matrix A 9 times?

NO!!!!!!! Just rememberthat det A9 = (detA)9. Here detA= 9−8= 1. So (detA)9 = 19 = 1

Example

Let A=[9 24 1

]. Find det A9.

Should you multiply the matrix A 9 times? NO!!!!!!! Just rememberthat det A9 = (detA)9. Here detA= 9−8= 1. So (detA)9 = 19 = 1

determinants, properties and imt

Education

value of determinant

entire determinant

oppositesignimportant

zerosimportant facts

product of thenumbers

determinantswhat abouta

main diagonal3

main diagonalentries