Announcements Monday, October 23 Webwork due next week, No quiz - PowerPoint PPT Presentation

Announcements Monday, October 23 ◮ Webwork due next week, ◮ No quiz this week.

Chapter 3 Determinants

Section 3.1 Introduction to Determinants

Orientation Recall: This course is about learning to: ◮ Solve the matrix equation Ax = b We’ve said most of what we’ll say about this topic now. ◮ Solve the matrix equation Ax = λ x (eigenvalue problem) We are now aiming at this. ◮ Almost solve the equation Ax = b This will happen later. The next topic is determinants . This is a completely magical function that takes a square matrix and gives you a number. It is a very complicated functionthe formula for the determinant of a 10 × 10 matrix has 3 , 628 , 800 summands! so we need efficient ways to compute it. Today is mostly about the computation of determinants; in the next lecture we will focus on the theory.

The Idea of Determinants Let A be an n × n matrix. Determinants are only for square matrices. The columns v 1 , v 2 , . . . , v n give you n vectors in R n . These determine a parallelepiped P . P P v 3 v 1 v 2 v 2 v 1 Observation: the volume of P is zero ⇐ ⇒ the columns are linearly dependent ( P is “flat”) ⇐ ⇒ the matrix A is not invertible. The determinant of A will be a number det( A ) whose absolute value is the volume of P .

Determinants of 2 × 2 Matrices Revisited We already have a formula in the 2 × 2 case: � a � b det = ad − bc . c d What does this have to do with volumes? The area of the parallelogram is always | ad − bc | . Note: this shows det( A ) � = 0 ⇐ ⇒ A is invertible in this case.

Determinants of 3 × 3 Matrices Here’s the formula:   a 11 a 12 a 13  = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 det a 21 a 22 a 23  − a 13 a 22 a 31 − a 11 a 23 a 32 − a 12 a 21 a 33 a 31 a 32 a 33 How to remember this? Draw a bigger matrix, repeating the first two columns to the right: � � � � a 11 a 12 a 13 a 11 a 12 a 11 a 12 a 13 a 11 a 12 � � � � � � � � + a 21 a 22 a 23 a 21 a 22 − a 21 a 22 a 23 a 21 a 22 � � � � � � � � a 31 a 32 a 33 a 31 a 32 a 31 a 32 a 33 a 31 a 32 � � � � For example,  5 1 0   = det − 1 3 2  4 0 − 1 What does this have to do with volumes ? Next time.

A Formula for the Determinant When n ≥ 4, the determinant is not that simple to describe . The formula is recursive: We need some notation. Let A be an n × n matrix. A ij = ij th minor of A = ( n − 1) × ( n − 1) matrix you get by deleting the ith row and jth column C ij = ij th cofactor of A = ( − 1) i + j det A ij The signs of the cofactors follow a checkerboard pattern :  + − + −  + + − − + + − − − + − + − − + + − − + +   ± in the ij entry is the sign of C ij   + + + − − − + + + − − −   − − − + + + − − − + + + Definition The determinant of an n × n matrix A is n � det( A ) = a 1 j C 1 j = a 11 C 11 + a 12 C 12 + · · · + a 1 n C 1 n . j =1 This formula is called cofactor expansion along the first row .

A Formula for the Determinant 1 × 1 Matrices This is the beginning of the recursion. det( a 11 ) = a 11 .

A Formula for the Determinant 2 × 2 Matrices � a 11 � a 12 A = a 21 a 22 The minors are: A 11 = A 12 = A 21 = A 22 = The cofactors are C 11 = C 12 = C 21 = C 22 = The determinant is det A = a 11 C 11 + a 12 C 12 = a 11 a 22 − a 12 a 21 .

A Formula for the Determinant 3 × 3 Matrices   a 11 a 12 a 13 A = a 21 a 22 a 23   a 31 a 32 a 33 The top row minors and cofactors are: A 11 = C 11 = A 12 = C 12 = A 13 = C 13 =

A Formula for the Determinant Example  5 1 0   = det − 1 3 2  4 0 − 1

Cofactor expasion: Specify point of reference... Recall: the formula n � det( A ) = a 1 j C 1 j = a 11 C 11 + a 12 C 12 + · · · + a 1 n C 1 n . j =1 is called cofactor expansion along the first row. Actually, you can expand cofactors along any row or column you like! n � det A = a ij C ij for any fixed i j =1 n � det A = a ij C ij for any fixed j i =1 Good trick: Use cofactor expansion along a row or a column with a lot of zeros .

Cofactor Expansion Example  2 1 0  A = 1 1 0   5 9 1 It looks easiest to expand along the third column: det A =

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The Determinant of an Upper-Triangular Matrix Trick: Expand along the first row This works for any matrix that is upper-triangular (all entries below the main diagonal are zero). Theorem The determinant of an upper-triangular matrix is the product of the diagonal entries:  · · ·  a 11 a 12 a 13 a 1 n 0 a 22 a 23 · · · a 2 n    0 0 a 33 · · · a 3 n  det = a 11 a 22 a 33 · · · a nn .    . . . .  ... . . . .   . . . .   0 0 0 · · · a nn The same is true for lower-triangular matrices. (Repeatedly expand along the first row.)

Extra: A Formula for the Inverse from § 3.3 For 2 × 2 matrices we had a nice formula for the inverse: � a � d � C 11 � � � 1 − b 1 b ⇒ A − 1 = C 21 A = = = . − c c d ad − bc a det A C 12 C 22 Theorem This last formula works for any n × n invertible matrix A :  C 11 C 21 C 31 · · · C n 1  (3 , 1) entry C 12 C 22 C 32 · · · C n 2   1 1 A − 1 =   C 13 C 23 C 33 · · · C n 3 � T � =   C ij det A det A  . . . .  ... . . . .   . . . .   · · · C 1 n C 2 n C 3 n C nn Note that the cofactors are “transposed”: the ( i , j ) entry of the matrix is C ji . The proof uses Cramer’s rule. See Dan Margalit’s notes on the website for a nice explanation.

A Formula for the Inverse Example  1 0 1  Compute A − 1 , where A = 0 1 1  .  1 1 0 The minors are: The cofactors are (don’t forget to multiply by ( − 1) i + j ): The determinant is (expanding along the first row): det A =

✧ Extra: A Formula for the Inverse Example, continued  1 0 1  Compute A − 1 , where A = 0 1 1  .  1 1 0 The inverse is A − 1 = Check: