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 Planning to do parts of chapters 3, 5 and 6 for exam 2.

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

Section 3.1 Introduction to Determinants

1. A 2×2 matrix A is invertible if and only if det A 6= 0.

2. We can now extend this idea to a 3×3 or larger matrices.

3. Determinants exist only for square matrices.

Notation The notation aij means the element in the i-th row andj-th column of a matrix.

So a23 means the element in the second row, third column of agiven matrix.

Section 3.1 Introduction to Determinants

1. A 2×2 matrix A is invertible if and only if det A 6= 0.

2. We can now extend this idea to a 3×3 or larger matrices.

3. Determinants exist only for square matrices.

Notation The notation aij means the element in the i-th row andj-th column of a matrix.

So a23 means the element in the second row, third column of agiven matrix.

Section 3.1 Introduction to Determinants

1. A 2×2 matrix A is invertible if and only if det A 6= 0.

2. We can now extend this idea to a 3×3 or larger matrices.

3. Determinants exist only for square matrices.

Notation The notation aij means the element in the i-th row andj-th column of a matrix.

So a23 means the element in the second row, third column of agiven matrix.

Section 3.1 Introduction to Determinants

1. A 2×2 matrix A is invertible if and only if det A 6= 0.

2. We can now extend this idea to a 3×3 or larger matrices.

3. Determinants exist only for square matrices.

Notation The notation aij means the element in the i-th row andj-th column of a matrix.

So a23 means the element in the second row, third column of agiven matrix.

Determinant of a 2×2 matrix

You know this one!!

If

A=[

a11 a12a21 a22

]Here, det A =a11a22−a21a12. It is a number.

Determinant of a 2×2 matrix

You know this one!!

If

A=[

a11 a12a21 a22

]Here, det A =a11a22−a21a12. It is a number.

What about a 3×3 matrix?

If

A= a11 a12 a13

a21 a22 a23a31 a32 a33

We have to break this down to multiple 2×2 determinants.

What about a 3×3 matrix?

If

A= a11 a12 a13

a21 a22 a23a31 a32 a33

We have to break this down to multiple 2×2 determinants.

What about a 3×3 matrix?

You can start the computation using any row or column as ananchor.

Suppose you choose the �rst row.

Ï Each entry of the �rst row will give one term each as follows.

Ï Add the terms at the end to get det A.

Ï To get the �rst term of det A, cover the row and columncorresponding to a11.

a11 a12 a13

a21 a22 a23

a31 a32 a33

What about a 3×3 matrix?

You can start the computation using any row or column as ananchor.

Suppose you choose the �rst row.

Ï Each entry of the �rst row will give one term each as follows.

Ï Add the terms at the end to get det A.

Ï To get the �rst term of det A, cover the row and columncorresponding to a11.

a11 a12 a13

a21 a22 a23

a31 a32 a33

What about a 3×3 matrix?

You can start the computation using any row or column as ananchor.

Suppose you choose the �rst row.

Ï Each entry of the �rst row will give one term each as follows.

Ï Add the terms at the end to get det A.

Ï To get the �rst term of det A, cover the row and columncorresponding to a11.

a11 a12 a13

a21 a22 a23

a31 a32 a33

What about a 3×3 matrix?

You can start the computation using any row or column as ananchor.

Suppose you choose the �rst row.

Ï Each entry of the �rst row will give one term each as follows.

Ï Add the terms at the end to get det A.

Ï To get the �rst term of det A, cover the row and columncorresponding to a11.

a11 a12 a13

a21 a22 a23

a31 a32 a33

What about a 3×3 matrix?

You can start the computation using any row or column as ananchor.

Suppose you choose the �rst row.

Ï Each entry of the �rst row will give one term each as follows.

Ï Add the terms at the end to get det A.

Ï To get the �rst term of det A, cover the row and columncorresponding to a11.

a11 a12 a13

a21 a22 a23

a31 a32 a33

Ï Multiply a11 with the determinant of the remaining matrix[a22 a23a32 a33

]

Ï Thus the �rst term is a11(a22a33−a32a23).Ï To get the second term of det A, cover the row and columncorresponding to a12.

a11 a12 a13

a21 a22 a23

a31 a32 a33

Ï Multiply the negative of a12 with the determinant of theremaining matrix [

a21 a23a31 a33

]

Ï Multiply a11 with the determinant of the remaining matrix[a22 a23a32 a33

]Ï Thus the �rst term is a11(a22a33−a32a23).

Ï To get the second term of det A, cover the row and columncorresponding to a12.

a11 a12 a13

a21 a22 a23

a31 a32 a33

Ï Multiply the negative of a12 with the determinant of theremaining matrix [

a21 a23a31 a33

]

Ï Multiply a11 with the determinant of the remaining matrix[a22 a23a32 a33

]Ï Thus the �rst term is a11(a22a33−a32a23).Ï To get the second term of det A, cover the row and columncorresponding to a12.

a11 a12 a13

a21 a22 a23

a31 a32 a33

Ï Multiply the negative of a12 with the determinant of theremaining matrix [

a21 a23a31 a33

]

Ï Multiply a11 with the determinant of the remaining matrix[a22 a23a32 a33

]Ï Thus the �rst term is a11(a22a33−a32a23).Ï To get the second term of det A, cover the row and columncorresponding to a12.

a11 a12 a13

a21 a22 a23

a31 a32 a33

Ï Multiply the negative of a12 with the determinant of theremaining matrix [

a21 a23a31 a33

]

Ï Thus the second term is −a12(a21a33−a31a23).

Ï To get the third term of det A, cover the row and columncorresponding to a13.

a11 a12 a13

a21 a22 a23

a31 a32 a33

Ï Multiply a13 with the determinant of the remaining matrix[a21 a22a31 a32

]

Ï Thus the second term is a13(a21a32−a31a22).

Ï Thus the second term is −a12(a21a33−a31a23).

Ï To get the third term of det A, cover the row and columncorresponding to a13.

a11 a12 a13

a21 a22 a23

a31 a32 a33

Ï Multiply a13 with the determinant of the remaining matrix[a21 a22a31 a32

]

Ï Thus the second term is a13(a21a32−a31a22).

Ï Thus the second term is −a12(a21a33−a31a23).

Ï To get the third term of det A, cover the row and columncorresponding to a13.

a11 a12 a13

a21 a22 a23

a31 a32 a33

Ï Multiply a13 with the determinant of the remaining matrix[a21 a22a31 a32

]

Ï Thus the second term is a13(a21a32−a31a22).

Ï Thus the second term is −a12(a21a33−a31a23).

Ï To get the third term of det A, cover the row and columncorresponding to a13.

a11 a12 a13

a21 a22 a23

a31 a32 a33

Ï Multiply a13 with the determinant of the remaining matrix[a21 a22a31 a32

]

Ï Thus the second term is a13(a21a32−a31a22).

Ï Add the 3 terms you obtained above

Ï a11(a22a33−a32a23)−a12(a21a33−a31a23)+a13(a21a32−a31a22)

Ï This is det A for a 3×3 matrix A.

Ï DO NOT try to memorize this as a formula

Ï Remember the steps (all the covering and multiplying games)!!

Ï To �nd determinant of a 4×4 matrix A, break it down intofour 3×3 determinants using the same idea. (more work).

Ï This method works for a square matrix of any size.

Ï Add the 3 terms you obtained above

Ï a11(a22a33−a32a23)−a12(a21a33−a31a23)+a13(a21a32−a31a22)

Ï This is det A for a 3×3 matrix A.

Ï DO NOT try to memorize this as a formula

Ï Remember the steps (all the covering and multiplying games)!!

Ï To �nd determinant of a 4×4 matrix A, break it down intofour 3×3 determinants using the same idea. (more work).

Ï This method works for a square matrix of any size.

Ï Add the 3 terms you obtained above

Ï a11(a22a33−a32a23)−a12(a21a33−a31a23)+a13(a21a32−a31a22)

Ï This is det A for a 3×3 matrix A.

Ï DO NOT try to memorize this as a formula

Ï Remember the steps (all the covering and multiplying games)!!

Ï To �nd determinant of a 4×4 matrix A, break it down intofour 3×3 determinants using the same idea. (more work).

Ï This method works for a square matrix of any size.

Ï Add the 3 terms you obtained above

Ï a11(a22a33−a32a23)−a12(a21a33−a31a23)+a13(a21a32−a31a22)

Ï This is det A for a 3×3 matrix A.

Ï DO NOT try to memorize this as a formula

Ï Remember the steps (all the covering and multiplying games)!!

Ï To �nd determinant of a 4×4 matrix A, break it down intofour 3×3 determinants using the same idea. (more work).

Ï This method works for a square matrix of any size.

FAQs

Ï Which row to choose for anchor? Any row (or column)!!

Ï Any caveats?? Yes!! Need to make sure that you do propersign alternating depending on which row or column youchoose. Keep the following in mind.

A= + − +

− + −+ − +

Ï So, if you decide to use second column, the �rst term will benegative, the second positive and the third negative. (withproper covering and multiplying)

Ï Choose a row or column with as many zeros as possible.

FAQs

Ï Which row to choose for anchor? Any row (or column)!!

Ï Any caveats?? Yes!! Need to make sure that you do propersign alternating depending on which row or column youchoose. Keep the following in mind.

A= + − +

− + −+ − +

Ï So, if you decide to use second column, the �rst term will benegative, the second positive and the third negative. (withproper covering and multiplying)

Ï Choose a row or column with as many zeros as possible.

FAQs

Ï Which row to choose for anchor? Any row (or column)!!

Ï Any caveats?? Yes!! Need to make sure that you do propersign alternating depending on which row or column youchoose. Keep the following in mind.

A= + − +

− + −+ − +

Ï So, if you decide to use second column, the �rst term will benegative, the second positive and the third negative. (withproper covering and multiplying)

Ï Choose a row or column with as many zeros as possible.

FAQs

Ï Which row to choose for anchor? Any row (or column)!!

Ï Any caveats?? Yes!! Need to make sure that you do propersign alternating depending on which row or column youchoose. Keep the following in mind.

A= + − +

− + −+ − +

Ï So, if you decide to use second column, the �rst term will benegative, the second positive and the third negative. (withproper covering and multiplying)

Ï Choose a row or column with as many zeros as possible.

FAQs

Ï Which row to choose for anchor? Any row (or column)!!

Ï Any caveats?? Yes!! Need to make sure that you do propersign alternating depending on which row or column youchoose. Keep the following in mind.

A= + − +

− + −+ − +

Ï So, if you decide to use second column, the �rst term will benegative, the second positive and the third negative. (withproper covering and multiplying)

Ï Choose a row or column with as many zeros as possible.

Before we go further..

Notation: Use a pair of vertical lines for determinants.

Example

If

A= 1 2 3

4 5 67 8 9

then

detA=∣∣∣∣∣∣1 2 34 5 67 8 9

∣∣∣∣∣∣

Before we go further..

Notation: Use a pair of vertical lines for determinants.

Example

If

A= 1 2 3

4 5 67 8 9

then

detA=∣∣∣∣∣∣1 2 34 5 67 8 9

∣∣∣∣∣∣

Going back to our 3×3 matrix

A= a11 a12 a13

a21 a22 a23a31 a32 a33

,

we can write

detA= a11

∣∣∣∣ a22 a23a32 a33

∣∣∣∣︸ ︷︷ ︸C11

−a12∣∣∣∣ a21 a23a31 a33

∣∣∣∣︸ ︷︷ ︸C12

+a13∣∣∣∣ a21 a22a31 a32

∣∣∣∣︸ ︷︷ ︸C13

Here C11, C12 and C13 are called the cofactors of A.

This method of computing determinants is called cofactor

expansion across �rst row.

Going back to our 3×3 matrix

A= a11 a12 a13

a21 a22 a23a31 a32 a33

,

we can write

detA= a11

∣∣∣∣ a22 a23a32 a33

∣∣∣∣︸ ︷︷ ︸C11

−a12∣∣∣∣ a21 a23a31 a33

∣∣∣∣︸ ︷︷ ︸C12

+a13∣∣∣∣ a21 a22a31 a32

∣∣∣∣︸ ︷︷ ︸C13

Here C11, C12 and C13 are called the cofactors of A.

This method of computing determinants is called cofactor

expansion across �rst row.

Going back to our 3×3 matrix

A= a11 a12 a13

a21 a22 a23a31 a32 a33

,

we can write

detA= a11

∣∣∣∣ a22 a23a32 a33

∣∣∣∣︸ ︷︷ ︸C11

−a12∣∣∣∣ a21 a23a31 a33

∣∣∣∣︸ ︷︷ ︸C12

+a13∣∣∣∣ a21 a22a31 a32

∣∣∣∣︸ ︷︷ ︸C13

Here C11, C12 and C13 are called the cofactors of A.

This method of computing determinants is called cofactor

expansion across �rst row.

Going back to our 3×3 matrix

A= a11 a12 a13

a21 a22 a23a31 a32 a33

,

we can write

detA= a11

∣∣∣∣ a22 a23a32 a33

∣∣∣∣︸ ︷︷ ︸C11

−a12∣∣∣∣ a21 a23a31 a33

∣∣∣∣︸ ︷︷ ︸C12

+a13∣∣∣∣ a21 a22a31 a32

∣∣∣∣︸ ︷︷ ︸C13

Here C11, C12 and C13 are called the cofactors of A.

This method of computing determinants is called cofactor

expansion across �rst row.

In General..

Theorem

1. The determinant of an n×n matrix A can be computed by

cofactor expansion along any row or column.

2. Expansion across the ith row will be

detA= ai1Ci1+ai2Ci2+ . . .+ainCin.

Don't forget to take care of proper sign alternations depending

on the row.

3. Expansion across the jth column will be

detA= a1jC1j +a2jC2j + . . .+anjCnj .

Don't forget to take care of proper sign alternations depending

on the column.

In General..

Theorem

1. The determinant of an n×n matrix A can be computed by

cofactor expansion along any row or column.

2. Expansion across the ith row will be

detA= ai1Ci1+ai2Ci2+ . . .+ainCin.

Don't forget to take care of proper sign alternations depending

on the row.

3. Expansion across the jth column will be

detA= a1jC1j +a2jC2j + . . .+anjCnj .

Don't forget to take care of proper sign alternations depending

on the column.

In General..

Theorem

1. The determinant of an n×n matrix A can be computed by

cofactor expansion along any row or column.

2. Expansion across the ith row will be

detA= ai1Ci1+ai2Ci2+ . . .+ainCin.

Don't forget to take care of proper sign alternations depending

on the row.

3. Expansion across the jth column will be

detA= a1jC1j +a2jC2j + . . .+anjCnj .

Don't forget to take care of proper sign alternations depending

on the column.

In General..

Theorem

1. The determinant of an n×n matrix A can be computed by

cofactor expansion along any row or column.

2. Expansion across the ith row will be

detA= ai1Ci1+ai2Ci2+ . . .+ainCin.

Don't forget to take care of proper sign alternations depending

on the row.

3. Expansion across the jth column will be

detA= a1jC1j +a2jC2j + . . .+anjCnj .

Don't forget to take care of proper sign alternations depending

on the column.

Example 2, section 3.1

Compute using cofactor expansion along �rst row.∣∣∣∣∣∣0 5 14 −3 02 4 1

∣∣∣∣∣∣

Solution:

detA= 0

∣∣∣∣ −3 04 1

∣∣∣∣︸ ︷︷ ︸−3

−5

∣∣∣∣ 4 02 1

∣∣∣∣︸ ︷︷ ︸4

+1

∣∣∣∣ 4 −32 4

∣∣∣∣︸ ︷︷ ︸22

= 0−20+22= 2

Example 2, section 3.1

Compute using cofactor expansion along �rst row.∣∣∣∣∣∣0 5 14 −3 02 4 1

∣∣∣∣∣∣Solution:

detA= 0

∣∣∣∣ −3 04 1

∣∣∣∣︸ ︷︷ ︸−3

−5

∣∣∣∣ 4 02 1

∣∣∣∣︸ ︷︷ ︸4

+1

∣∣∣∣ 4 −32 4

∣∣∣∣︸ ︷︷ ︸22

= 0−20+22= 2

Example 2, section 3.1

Compute using cofactor expansion along �rst row.∣∣∣∣∣∣0 5 14 −3 02 4 1

∣∣∣∣∣∣Solution:

detA= 0

∣∣∣∣ −3 04 1

∣∣∣∣︸ ︷︷ ︸−3

−5

∣∣∣∣ 4 02 1

∣∣∣∣︸ ︷︷ ︸4

+1

∣∣∣∣ 4 −32 4

∣∣∣∣︸ ︷︷ ︸22

= 0−20+22= 2

Example 2, section 3.1

Compute using cofactor expansion along �rst row.∣∣∣∣∣∣0 5 14 −3 02 4 1

∣∣∣∣∣∣Solution:

detA= 0

∣∣∣∣ −3 04 1

∣∣∣∣︸ ︷︷ ︸−3

−5

∣∣∣∣ 4 02 1

∣∣∣∣︸ ︷︷ ︸4

+1

∣∣∣∣ 4 −32 4

∣∣∣∣︸ ︷︷ ︸22

= 0−20+22= 2

Example 2, section 3.1

Compute using cofactor expansion along �rst row.∣∣∣∣∣∣0 5 14 −3 02 4 1

∣∣∣∣∣∣Solution:

detA= 0

∣∣∣∣ −3 04 1

∣∣∣∣︸ ︷︷ ︸−3

−5

∣∣∣∣ 4 02 1

∣∣∣∣︸ ︷︷ ︸4

+1

∣∣∣∣ 4 −32 4

∣∣∣∣︸ ︷︷ ︸22

= 0−20+22= 2

Example 2, section 3.1

Compute using cofactor expansion down the second column.∣∣∣∣∣∣0 5 14 −3 02 4 1

∣∣∣∣∣∣

Solution:

0 5 1

4 −3 0

2 4 1

=⇒−5∣∣∣∣ 4 02 1

∣∣∣∣︸ ︷︷ ︸4

=−20

Example 2, section 3.1

Compute using cofactor expansion down the second column.∣∣∣∣∣∣0 5 14 −3 02 4 1

∣∣∣∣∣∣

Solution:

0 5 1

4 −3 0

2 4 1

=⇒−5∣∣∣∣ 4 02 1

∣∣∣∣︸ ︷︷ ︸4

=−20

Example 2, section 3.1

Compute using cofactor expansion down the second column.∣∣∣∣∣∣0 5 14 −3 02 4 1

∣∣∣∣∣∣

Solution:

0 5 1

4 −3 0

2 4 1

=⇒−5∣∣∣∣ 4 02 1

∣∣∣∣︸ ︷︷ ︸4

=−20

Example 2, section 3.1

0 5 1

4 −3 0

2 4 1

=⇒−3∣∣∣∣ 0 12 1

∣∣∣∣︸ ︷︷ ︸−2

= 6

0 5 1

4 −3 0

2 4 1

=⇒−4∣∣∣∣ 0 14 0

∣∣∣∣︸ ︷︷ ︸−4

= 16.

Add these terms, -20+6+16=2.

Example 2, section 3.1

0 5 1

4 −3 0

2 4 1

=⇒−3∣∣∣∣ 0 12 1

∣∣∣∣︸ ︷︷ ︸−2

= 6

0 5 1

4 −3 0

2 4 1

=⇒−4∣∣∣∣ 0 14 0

∣∣∣∣︸ ︷︷ ︸−4

= 16.

Add these terms, -20+6+16=2.

Example 2, section 3.1

0 5 1

4 −3 0

2 4 1

=⇒−3∣∣∣∣ 0 12 1

∣∣∣∣︸ ︷︷ ︸−2

= 6

0 5 1

4 −3 0

2 4 1

=⇒−4∣∣∣∣ 0 14 0

∣∣∣∣︸ ︷︷ ︸−4

= 16.

Add these terms, -20+6+16=2.

Example 2, section 3.1

0 5 1

4 −3 0

2 4 1

=⇒−3∣∣∣∣ 0 12 1

∣∣∣∣︸ ︷︷ ︸−2

= 6

0 5 1

4 −3 0

2 4 1

=⇒−4∣∣∣∣ 0 14 0

∣∣∣∣︸ ︷︷ ︸−4

= 16.

Add these terms, -20+6+16=2.

Example 2, section 3.1

0 5 1

4 −3 0

2 4 1

=⇒−3∣∣∣∣ 0 12 1

∣∣∣∣︸ ︷︷ ︸−2

= 6

0 5 1

4 −3 0

2 4 1

=⇒−4∣∣∣∣ 0 14 0

∣∣∣∣︸ ︷︷ ︸−4

= 16.

Add these terms, -20+6+16=2.

Comments

1. Again, be careful with the alternating signs.

2. If you are expanding down the second column, the �rst termwill be negative, second positive (but already we have a -3)and the third negative.

Comments

1. Again, be careful with the alternating signs.

2. If you are expanding down the second column, the �rst termwill be negative, second positive (but already we have a -3)and the third negative.

Comments

1. Again, be careful with the alternating signs.

2. If you are expanding down the second column, the �rst termwill be negative, second positive (but already we have a -3)and the third negative.

Example 8, section 3.1Compute using cofactor expansion along �rst row.∣∣∣∣∣∣

8 1 64 0 33 −2 5

∣∣∣∣∣∣

8 1 6

4 0 3

3 −2 5

8 1 6

4 0 3

3 −2 5

8 1 6

4 0 3

3 −2 5

detA= 8

∣∣∣∣ 0 3−2 5

∣∣∣∣︸ ︷︷ ︸6

−1

∣∣∣∣ 4 33 5

∣∣∣∣︸ ︷︷ ︸11

+6

∣∣∣∣ 4 03 −2

∣∣∣∣︸ ︷︷ ︸−8

= 48−11−48=−11

Example 8, section 3.1Compute using cofactor expansion along �rst row.∣∣∣∣∣∣

8 1 64 0 33 −2 5

∣∣∣∣∣∣8 1 6

4 0 3

3 −2 5

8 1 6

4 0 3

3 −2 5

8 1 6

4 0 3

3 −2 5

detA=

8

∣∣∣∣ 0 3−2 5

∣∣∣∣︸ ︷︷ ︸6

−1

∣∣∣∣ 4 33 5

∣∣∣∣︸ ︷︷ ︸11

+6

∣∣∣∣ 4 03 −2

∣∣∣∣︸ ︷︷ ︸−8

= 48−11−48=−11

Example 8, section 3.1Compute using cofactor expansion along �rst row.∣∣∣∣∣∣

8 1 64 0 33 −2 5

∣∣∣∣∣∣8 1 6

4 0 3

3 −2 5

8 1 6

4 0 3

3 −2 5

8 1 6

4 0 3

3 −2 5

detA= 8

∣∣∣∣ 0 3−2 5

∣∣∣∣︸ ︷︷ ︸6

−1

∣∣∣∣ 4 33 5

∣∣∣∣︸ ︷︷ ︸11

+6

∣∣∣∣ 4 03 −2

∣∣∣∣︸ ︷︷ ︸−8

= 48−11−48=−11

Example 8, section 3.1Compute using cofactor expansion along �rst row.∣∣∣∣∣∣

8 1 64 0 33 −2 5

∣∣∣∣∣∣8 1 6

4 0 3

3 −2 5

8 1 6

4 0 3

3 −2 5

8 1 6

4 0 3

3 −2 5

detA= 8

∣∣∣∣ 0 3−2 5

∣∣∣∣︸ ︷︷ ︸6

−1

∣∣∣∣ 4 33 5

∣∣∣∣︸ ︷︷ ︸11

+6

∣∣∣∣ 4 03 −2

∣∣∣∣︸ ︷︷ ︸−8

= 48−11−48=−11

Example 8, section 3.1Compute using cofactor expansion along �rst row.∣∣∣∣∣∣

8 1 64 0 33 −2 5

∣∣∣∣∣∣8 1 6

4 0 3

3 −2 5

8 1 6

4 0 3

3 −2 5

8 1 6

4 0 3

3 −2 5

detA= 8

∣∣∣∣ 0 3−2 5

∣∣∣∣︸ ︷︷ ︸6

−1

∣∣∣∣ 4 33 5

∣∣∣∣︸ ︷︷ ︸11

+6

∣∣∣∣ 4 03 −2

∣∣∣∣︸ ︷︷ ︸−8

= 48−11−48=−11

De�nitionA square matrix A is a Triangular matrix if the entries above OR

below the main diagonal are ALL zeros

TheoremIf A is a triangular matrix, then det A is the product of entries on

the main diagonal of A.

De�nitionA square matrix A is a Triangular matrix if the entries above OR

below the main diagonal are ALL zeros

TheoremIf A is a triangular matrix, then det A is the product of entries on

the main diagonal of A.

Example

If

A=

1 2 377 4 514 60 5 69 77 81 90 0 2 2321 45 880 0 0 1 45 76.670 0 0 0 2 81.630 0 0 0 0 1

detA= (1)(5)(2)(1)(2)(1)= 20.

Example

If

A=

1 2 377 4 514 60 5 69 77 81 90 0 2 2321 45 880 0 0 1 45 76.670 0 0 0 2 81.630 0 0 0 0 1

detA= (1)(5)(2)(1)(2)(1)= 20.

Larger "Convenient" Matrices

1. If you have a 4×4 or larger matrix with a row or columnmostly zeros, use that row(column) as the anchor.

2. Be careful with the sign alterations.

3. Have a sign template of proper size handy.

Larger "Convenient" Matrices

1. If you have a 4×4 or larger matrix with a row or columnmostly zeros, use that row(column) as the anchor.

2. Be careful with the sign alterations.

3. Have a sign template of proper size handy.

Larger "Convenient" Matrices

1. If you have a 4×4 or larger matrix with a row or columnmostly zeros, use that row(column) as the anchor.

2. Be careful with the sign alterations.

3. Have a sign template of proper size handy.

Example 10, section 3.1

Compute the following determinant using least amount ofcomputation. ∣∣∣∣∣∣∣∣∣

1 −2 5 20 0 3 02 −6 −7 55 0 4 4

∣∣∣∣∣∣∣∣∣

Use row 2 as the anchor. To be sure about the signs use thefollowing ∣∣∣∣∣∣∣∣∣

+ − + −− + − ++ − + −− + − +

∣∣∣∣∣∣∣∣∣Only the cofactor of 3 matters here. It will be negative. Others areall zero.

Example 10, section 3.1

Compute the following determinant using least amount ofcomputation. ∣∣∣∣∣∣∣∣∣

1 −2 5 20 0 3 02 −6 −7 55 0 4 4

∣∣∣∣∣∣∣∣∣Use row 2 as the anchor. To be sure about the signs use thefollowing ∣∣∣∣∣∣∣∣∣

+ − + −− + − ++ − + −− + − +

∣∣∣∣∣∣∣∣∣

Only the cofactor of 3 matters here. It will be negative. Others areall zero.

Example 10, section 3.1

Compute the following determinant using least amount ofcomputation. ∣∣∣∣∣∣∣∣∣

1 −2 5 20 0 3 02 −6 −7 55 0 4 4

∣∣∣∣∣∣∣∣∣Use row 2 as the anchor. To be sure about the signs use thefollowing ∣∣∣∣∣∣∣∣∣

+ − + −− + − ++ − + −− + − +

∣∣∣∣∣∣∣∣∣Only the cofactor of 3 matters here. It will be negative. Others areall zero.

Slide corrected on Feb 2, 12.00pm

1 −2 5 2

0 0 3 0

2 −6 −7 5

5 0 4 4

=⇒−3∣∣∣∣∣∣1 −2 22 −6 55 0 4

∣∣∣∣∣∣

We can expand along the last row. To be safe, keep the signtemplate ∣∣∣∣∣∣

+ − +− + −+ − +

∣∣∣∣∣∣

Slide corrected on Feb 2, 12.00pm

1 −2 5 2

0 0 3 0

2 −6 −7 5

5 0 4 4

=⇒−3∣∣∣∣∣∣1 −2 22 −6 55 0 4

∣∣∣∣∣∣

We can expand along the last row. To be safe, keep the signtemplate ∣∣∣∣∣∣

+ − +− + −+ − +

∣∣∣∣∣∣

Slide corrected on Feb 2, 12.00pm

1 −2 5 2

0 0 3 0

2 −6 −7 5

5 0 4 4

=⇒−3∣∣∣∣∣∣1 −2 22 −6 55 0 4

∣∣∣∣∣∣

We can expand along the last row. To be safe, keep the signtemplate ∣∣∣∣∣∣

+ − +− + −+ − +

∣∣∣∣∣∣

1 −2 2

2 −6 5

5 0 4

1 −2 2

2 −6 5

5 0 4

1 −2 2

2 −6 5

5 0 4

detA= 5

∣∣∣∣ −2 2−6 5

∣∣∣∣︸ ︷︷ ︸2

−0

∣∣∣∣ 1 22 5

∣∣∣∣︸ ︷︷ ︸0

+4

∣∣∣∣ 1 −22 −6

∣∣∣∣︸ ︷︷ ︸−2

= 10+0+ (−8)= 2

Don't forget to multiply the -3 we had. So the answer is -6.

1 −2 2

2 −6 5

5 0 4

1 −2 2

2 −6 5

5 0 4

1 −2 2

2 −6 5

5 0 4

detA= 5

∣∣∣∣ −2 2−6 5

∣∣∣∣︸ ︷︷ ︸2

−0

∣∣∣∣ 1 22 5

∣∣∣∣︸ ︷︷ ︸0

+4

∣∣∣∣ 1 −22 −6

∣∣∣∣︸ ︷︷ ︸−2

= 10+0+ (−8)= 2

Don't forget to multiply the -3 we had. So the answer is -6.

1 −2 2

2 −6 5

5 0 4

1 −2 2

2 −6 5

5 0 4

1 −2 2

2 −6 5

5 0 4

detA= 5

∣∣∣∣ −2 2−6 5

∣∣∣∣︸ ︷︷ ︸2

−0

∣∣∣∣ 1 22 5

∣∣∣∣︸ ︷︷ ︸0

+4

∣∣∣∣ 1 −22 −6

∣∣∣∣︸ ︷︷ ︸−2

= 10+0+ (−8)= 2

Don't forget to multiply the -3 we had. So the answer is -6.

Sarrus' Mnemonic Rule

1. An easy to remember method for 3×3 matrices

2. DO NOT apply this method for larger matrices.

3. Make sure all rows and columns are properly aligned, otherwiseit becomes very confusing.

4. Start by repeating the �rst 2 rows immediately beneath thedeterminant.

Sarrus' Mnemonic Rule

1. An easy to remember method for 3×3 matrices

2. DO NOT apply this method for larger matrices.

3. Make sure all rows and columns are properly aligned, otherwiseit becomes very confusing.

4. Start by repeating the �rst 2 rows immediately beneath thedeterminant.

Sarrus' Mnemonic Rule

1. An easy to remember method for 3×3 matrices

2. DO NOT apply this method for larger matrices.

3. Make sure all rows and columns are properly aligned, otherwiseit becomes very confusing.

4. Start by repeating the �rst 2 rows immediately beneath thedeterminant.

Sarrus' Mnemonic Rule

a11 a12 a13

a21 a22 a23

a31 a32 a33

a11 a12 a13

a21 a22 a23

+

+

+

−

−

−

Sarrus' Mnemonic Rule

a11 a12 a13

a21 a22 a23

a31 a32 a33

a11 a12 a13

a21 a22 a23

+

+

+

−

−

−

Sarrus' Mnemonic Rule

a11 a12 a13

a21 a22 a23

a31 a32 a33

a11 a12 a13

a21 a22 a23

+

+

+

−

−

−

Sarrus' Mnemonic Rule

a11 a12 a13

a21 a22 a23

a31 a32 a33

a11 a12 a13

a21 a22 a23

+

+

+

−

−

−

Sarrus' Mnemonic Rule

a11 a12 a13

a21 a22 a23

a31 a32 a33

a11 a12 a13

a21 a22 a23

+

+

+

−

−

−

Sarrus' Mnemonic Rule

a11 a12 a13

a21 a22 a23

a31 a32 a33

a11 a12 a13

a21 a22 a23

+

+

+

−

−

−

Sarrus' Mnemonic Rule

a11 a12 a13

a21 a22 a23

a31 a32 a33

a11 a12 a13

a21 a22 a23

+

+

+

−

−

−