Chapter 2: Matrices and Linear Systems

Chapter 2: Matrices and Linear Systems Paul Pearson

Outline Matrices Linear systems Row operations Inverses Determinants

Matrices Definition An m n matrix A = (a ij ) is a rectangular array of real numbers with m rows and n columns: a 11 a 12 a 1n a 21 a 22 a 2n A = (a ij ) =...... a m1 a m2 a mn The row index of a ij is i, and the column index of a ij is j. The set of all m n matrices with real entries is denoted M m,n (R). When m = n, this notation is shortened to M n (R).

Matrix addition and scalar mult. Definition Let A = (a ij ) and B = (b ij ) be m n matrices, and let α R. The operations of matrix addition and scalar multiplication are defined by adding corresponding entries and scaling all entries as follows: 1. A + B = (a ij + b ij ) 2. αa = (αa ij ) Example [ ] [ ] 1 2 3 10 20 30 1. + = 4 5 6 40 50 60 1 4 10 40 2. 10 2 5 = 20 50 3 6 30 60 [ 11 22 33 44 55 66 ]

Matrix addition and scalar mult. Theorem Let A, B, C M m,n (R) and α, β R. Let 0 denote the m n zero matrix. 1. 0 A = 0 2. 1 A = A 3. α 0 = 0 4. A + B = B + A 5. (A + B) + C = A + (B + C) 6. (α + β)a = αa + βa 7. α(a + B) = αa + αb 8. (αβ)a = α(βa)

Matrix multiplication Definition Let A M m,n (R) and B M n,k (R). The product matrix AB is the m k matrix whose (i, j)-entry is the dot product (Row i of A) (Column j of B) Example Compute by hand: A B AB ] 1 1 1 0 [ 0 3 2 0 = 2 3 3 0 [ 1 3 2 2 1 4 5 14 1 0 6 17 0 2 rows, 3 cols 3 rows, 4 cols 2 rows, 4 cols ]

Matrix multiplication Example Does AB = BA? [ ] [ ] 1 2 2 0 3 4 0 2 [ ] [ ] 2 0 1 2 0 2 3 4 = = [ ] [ ] 1 2 0 1 3 4 1 0 [ ] [ ] 0 1 1 2 1 0 3 4 = = [ ] [ ] 1 2 1 3 4 1 [ ] [ ] 1 1 2 1 3 4 = =

Matrix multiplication Example Does AB = BA? Not always, so we write AB BA. [ ] [ ] 1 2 2 0 3 4 0 2 [ ] [ ] 2 0 1 2 0 2 3 4 = = [ ] 2 4. 6 8 [ ] 2 4. 6 8 [ ] [ ] 1 2 0 1 3 4 1 0 [ ] [ ] 0 1 1 2 1 0 3 4 = = [ ] 2 1. 4 3 [ ] 3 4. 1 1 [ ] [ ] 1 2 1 3 4 1 [ ] [ ] 1 1 2 1 3 4 = [ ] 3. 7 = DNE

Matrix operations Definition I n is the n n identity matrix with all diagonal entries equal to 1 and all other entries equal to 0. Theorem Let A M m,n (R), B M n,k (R) and α R. 1. A0 n k = 0 m k 2. I m A = A and AI n = A. In particular, if A is an n n matrix, then I n A = AI n = A. 3. If C M k,l (R), then A(BC) = (AB)C. 4. If C M n,k (R), then A(B + C) = AB + AC. 5. A(αB) = α(ab) = (αa)b.

Matrix operations If A = [ 1 2 3 4 5 6 then by hand computation [ ] [ 1 0 1 2 3 I 2 A = 0 1 4 5 6 AI 3 = A0 3 2 = [ 1 2 3 4 5 6 ] [ 1 2 3 4 5 6 AI 2 = 1 0 0 0 1 0 0 0 1 ] [ 1 2 3 4 5 6 0 0 0 0 0 0 ] = = = ] [ 1 0 0 1 ], [ 1 2 3 4 5 6 [ 1 2 3 4 5 6 [ 0 0 0 0 ] = DNE ] = A, ] = A, ] = 0 2 2,

Matrix operations If [ ] 1 2 3 A =, B = 4 5 6 then by hand computation A(B + C) = = = AB + AC = = = 2 1 2 1, C = 2 1 [ ] 2 1 1 2 3 2 1 + 4 5 6 2 1 [ ] 3 3 1 2 3 4 4 4 5 6 [ ] 5 5 26 26 62 62 [ 30 15 ] 26 26 62 62 1 4 2 5 3 6 [ ] 2 1 [ ] 1 2 3 2 1 1 2 3 + 4 5 6 4 5 6 [ ] 2 [ 1 ] 12 6 14 32 + 32 77 1 4 2 5, 3 6 1 4 2 5 3 6

Matrix operations Suppose Then: A = [ 1 2 3 4 ] [ 1, b 1 = 1 ] [ 2, b 2 = 2 Ab 1 = Ab 2 = Ab 3 = A(b 1 b 2 b 3 ) = A((b 1 0 0) + (0 b 2 0) + (0 0 b 3 )) = ] [ 3, b 3 = 3 ].

Matrix operations Suppose A = [ 1 2 3 4 ] [ 1, b 1 = 1 ] [ 2, b 2 = 2 Then: [ ] [ ] 3 6 Ab 1 =, Ab 7 2 =, Ab 14 3 = [ ] 3 6 9 A(b 1 b 2 b 3 ) =. 7 14 21 [ 9 21 A((b 1 0 0) + (0 b 2 0) + (0 0 b 3 )) = In general, ] [ 3, b 3 = 3 ]. [ 3 6 9 7 14 21 AB = A(b 1 b n ) = (Ab 1 Ab n ). ]. ].

Matrix transpose Definition 1. The transpose of the matrix A = (a ij ) is the matrix A T = (a ji ), i.e., rows and columns swap: T = T and = 2. A square matrix A such that A T = A is called a symmetric matrix. 3. A square matrix A such that A T = A is called a skew-symmetric matrix. Theorem Let A M m,n (R) and α R. 1. (A T ) T = A. 2. If B M m,n (R), then (A + B) T = A T + B T. 3. (αa) T = αa T.

Matrix transpose Theorem If A M m,n (R) and B M n,k (R), then (AB) T = B T A T. Proof. The (i, j) entry of (AB) T is (ab) T i,j = (ab) j,i = (row j of A ) (column i of B). The (i, j) entry of B T A T is (row i of B T ) (column j of A T ) = (column i of B ) (row j of A). Remark AB BA, but (AB) T = B T A T is always true.

Matrix transpose Example Suppose Then A = [ 1 2 3 4 5 6 ], B = 10 20 30 40 50 60. (AB) T = B T A T =

Matrix transpose Example Suppose A = [ 1 2 3 4 5 6 ], B = 10 20 30 40 50 60. Then B T A T = (AB) T = [ 220 280 490 640 [ 10 30 50 20 40 60 ] ] T = 1 4 2 5 3 6 [ 220 490 280 640 = ]. [ 220 490 280 640 ].

Triangular and diagonal matrices Suppose A = (a ij ) is an n n square matrix. The diagonal entries of A are a 11, a 22,..., a nn. The matrix A is diagonal if all non-diagonal entries are zero. upper triangular if all entries below the diagonal are zero. lower triangular if all entries above the diagonal are zero. Example 1 0 0 0 2 0 0 0 3, 1 2 3 0 4 5 0 0 6, 1 0 0 2 4 0 3 5 6 1. If a matrix is both lower and upper triangular it is 2. The transpose of a lower triangular matrix is

Outline Matrices Linear systems Row operations Inverses Determinants

Systems of Linear Equations Definition An m n system of linear equations in variables x 1, x 2,..., x n is a list of m equations of the form a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.. a m1 x 1 + a m2 x 2 + + a mn x n = b m Any point (x 1, x 2,..., x n ) which satisfies all the equations in the system is called a solution of the system. A system that has at least one solution is called consistent, while a system with no solutions is called inconsistent.

Systems of linear equations Each graph below is the graph of a system of three linear equations in three unknowns. Determine which systems are consistent and inconsistent and the dimension of the solution set.

Systems of Linear Equations Definition We can represent a linear system as Ax = b: a 11 a 12 a 1n a 21 a 22 a 2n..... } a m1 a m2 {{ a mn } A x 1 x 2. x n }{{} x = b 1 b 2. b m } {{ } b Or, more succinctly, we can write the augmented matrix (A b): a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2...... a m1 a m2 a mn b m

Systems of linear equations Exercise 1. Find an equation for the plane P 1 in R 3 passing through the origin and parallel to 6y 2x + 3z = 100. Find an equation for the plane P 2 parallel to both 2i + 3j and 0, 1, 0 passing through the point (0, 0, 2). 2. Write the equations for P 1 and P 2 as a linear system and as an augmented matrix. 3. Solve the linear system.

Outline Matrices Linear systems Row operations Inverses Determinants

Solving linear systems The linear system { x + y = 1, x + y = 4. can be solved with three operations that leave the solution set unchanged: 1. Multiplying a row by a nonzero constant (e.g., x + y = 4 2x + 2y = 8) 2. Swapping rows (order doesn t matter) 3. Adding rows together (if (x, y) satisfies both x + y = 1 and x + y = 4, then it satisfies ( x + y) + (x + y) = 1 + 4, which reduces to 2y = 5) Since each of these operations affect only the coefficients, we can store all the pertinent information about the system in an array of numbers called a matrix and record the effects of these operations in matrices as we progress toward a solution.

Elementary row operations Definition Elementary Row Operations (EROs) 1. αr i R i 2. R i R j 3. R i + αr j R i Definition Inverses of Elementary Row Operations 1. e : 1 α R i R i 2. e : R i R j 3. e : R i αr j R i Remark Solution sets to linear systems are unchanged by EROs.

Solving linear systems { 2x + y = 6 3x + 4y = 12 y 4 3 2 1 L1 L2 x 1 2 3 4

Solving linear systems y { 2x + y = 6 3x + 4y = 12 3 2 1 L2 4 L1 x 1 2 3 4 y 3R 1 R 1 2R 2 R 2 { 6x + 3y = 18 6x 8y = 24 3 2 1 L2 4 L1 x 1 2 3 4

Solving linear systems y { 2x + y = 6 3x + 4y = 12 3 2 1 L2 4 L1 x 1 2 3 4 y 3R 1 R 1 2R 2 R 2 { 6x + 3y = 18 6x 8y = 24 3 2 1 L2 4 L1 x 1 2 3 4 y R 1 + R 2 R 2 { 6x + 3y = 18 0x 5y = 6 3 2 1 L2 L3 4 L1 x 1 2 3 4

Solving linear systems 1 3 R 1 R 1 { 2x + y = 6 0x 5y = 6 y 4 3 2 1 L1 L2 L3 x 1 2 3 4

Solving linear systems 1 3 R 1 R 1 { 2x + y = 6 0x 5y = 6 y 4 3 2 1 L1 L2 L3 x 1 2 3 4 y 4 L1 L4 5R 1 + R 2 R 1 { 10x + 0y = 24 0x 5y = 6 3 2 1 L2 L3 x 1 2 3 4

Solving linear systems 1 3 R 1 R 1 { 2x + y = 6 0x 5y = 6 y 4 3 2 1 L1 L2 L3 x 1 2 3 4 y 4 L1 L4 5R 1 + R 2 R 1 { 10x + 0y = 24 0x 5y = 6 3 2 1 L2 L3 x 1 2 3 4 y 4 L1 L4 1 10 R 1 R 1 1 5 R 2 R 2 { 1x + 0y = 2.4 0x + 1y = 1.2 3 2 1 L2 L3 x 1 2 3 4

Solving linear systems ( 2 1 6 3 4 12 )

Solving linear systems ( 2 1 6 ) 3R 1 R 1 2R 2 R 2 3 4 12 ( 6 3 18 6 8 24 )

Solving linear systems ( 2 1 6 ) 3R 1 R 1 2R 2 R 2 R 1 + R 2 R 2 3 4 12 ( 6 3 ) 18 6 8 24 ( 6 3 ) 18 0 5 6

Solving linear systems ( 2 1 6 ) 3R 1 R 1 2R 2 R 2 R 1 + R 2 R 2 3 4 12 ( 6 3 18 6 8 24 ( 6 3 ) 18 0 5 6 1 3 R 1 R 1 ( 2 1 6 0 5 6 ) )

Solving linear systems ( 2 1 6 3R 1 R 1 2R 2 R 2 R 1 + R 2 R 2 ) 3 4 12 ( 6 3 ) 18 6 8 24 ( 6 3 ) 18 0 5 6 1 3 R 1 R 1 ( 2 1 6 0 5 6 5R 1 + R 2 R 1 ( 10 0 24 0 5 6 ) )

Solving linear systems ( 2 1 6 3R 1 R 1 2R 2 R 2 R 1 + R 2 R 2 ) 3 4 12 ( 6 3 ) 18 6 8 24 ( 6 3 ) 18 0 5 6 1 3 R 1 R 1 ( 2 1 6 0 5 6 5R 1 + R 2 R 1 ( 10 0 ) 24 0 5 6 1 10 R 1 R 1 ( 1 0 ) 2.4 1 5 R 2 R 2 0 1 1.2 )

EROs and elementary matrices Definition Any ERO e may be represented as a matrix E = e(i) called an elementary matrix. The result of applying the ERO e to the matrix A is the same as the matrix product EA. Example The ERO e : R 1 + 2R 2 R 1 applied to A = is the same as EA, where 1 3 2 8 0 0 1 3 0 0 0 0 E = e(i 3 ) = 1 2 0 0 1 0 0 0 1.

Reduced row echelon form Definition A matrix A M m,n (R) is in reduced row echelon form (rref) if: 1. The leading entry in each nonzero row is 1. 2. All rows of zeros occur at the bottom of the matrix. 3. Every leading 1 occurs farther to the right of the leading 1 in the previous row. 4. Every entry above and below a leading 1 is equal to 0. Example Suppose represents any real number. RREF could look like: 1 0 0 0 1 0 0 0 0, 1 0 0 0 1 0 0 0 0 1

Gauss-Jordan Elimination Given a matrix A, its reduced row echelon form can be constructed by: 1. Starting with i = 1, find the row R j with i j whose leading entry is farthest to the left. If j i, do R i R j. 2. If the leading entry of R i is α 1, do (1/α)R i R i. This establishes a leading 1 in R i. 3. Make all entries above and below a leading 1 equal to 0 by doing EROs. 4. Go to the next row and repeat the process until the matrix is in rref.

Gauss-Jordan Elimination 1 2 1 2 3 1 3 5 0

Gauss-Jordan Elimination 1 2 1 2 3 1 3 5 0 2R 1 + R 2 R 2 1 2 1 0 1 3 3 5 0

Gauss-Jordan Elimination 1 2 1 2 3 1 3 5 0 2R 1 + R 2 R 2 3R 1 + R 3 R 3 1 2 1 0 1 3 3 5 0 1 2 1 0 1 3 0 1 3

Gauss-Jordan Elimination 1 2 1 2 3 1 3 5 0 2R 1 + R 2 R 2 3R 1 + R 3 R 3 R 2 + R 3 R 3 1 2 1 0 1 3 3 5 0 1 2 1 0 1 3 0 1 3 1 2 1 0 1 3 0 0 0

Gauss-Jordan Elimination 1 2 1 2 3 1 3 5 0 2R 1 + R 2 R 2 3R 1 + R 3 R 3 R 2 + R 3 R 3 R 1 2R 2 R 1 1 2 1 0 1 3 3 5 0 1 2 1 0 1 3 0 1 3 1 2 1 0 1 3 0 0 0 1 0 5 0 1 3 0 0 0

Gauss-Jordan Elimination If possible, solve the system x + 2y = 1, 2x 3y = 1, 3x + 5y = 0.

Gauss-Jordan Elimination If possible, solve the system x + 2y = 1, 2x 3y = 1, 3x + 5y = 0. Since rref 1 2 1 2 3 1 3 5 0 = 1 0 5 0 1 3 0 0 0 the system is consistent and the three given lines in R 2 intersect at the point (x, y) = ( 5, 3). There is only one solution (it is the unique solution) and it is zero dimensional.

Gauss-Jordan Elimination If possible, solve the system x + 2y + z = 0, 2x 3y + z = 0, 3x + 5y = 0.

Gauss-Jordan Elimination If possible, solve the system x + 2y + z = 0, 2x 3y + z = 0, 3x + 5y = 0. Since rref 1 2 1 0 2 3 1 0 3 5 0 0 = 1 0 5 0 0 1 3 0 0 0 0 0 the system is consistent and the three given planes in R 3 intersect. The first two rows say x 5z = 0 and y + 3z = 0. Since there is no pivot in the third column, z is free and z = z. Thus, the solution set is x = 5z, y = 3z, and z = z, i.e., the 1D line thru the origin (x, y, z) = (5z, 3z, z) = z 5, 3, 1 where z is any real number.

Reduced Row Echelon Form 1. A leading 1 in a row is also called a pivot. 2. A column without a pivot is called a free column and the variable it represents is called a free variable. The number of free variables determines the dimension of the solution set. 3. If a pivot occurs in the augmentation column, the system has no solution and is said to be inconsistent. Exercise 1. Solve Ax = 0 given 0 1 2 A = 2 6 4 EROs rref(a) = 2 7 2 1 0 1 1 2. B = 2 2 2 1 EROs rref(b) = 5 1 3 5 1 0 8 0 1 2 0 0 0 1 0 1 0 0 1 2 0 0 0 0 1

RREF and Gauss-Jordan Elimination Definition 1. A system of linear equations Ax = 0 is called a homogeneous system. 2. The solution set X H = {x Ax = 0} of a homogeneous system is called the homogeneous solution set. Theorem 1. 0 X H 2. For all x 1, x 2 X H, x 1 + x 2 X H 3. For all x X H and all α R, αx X H Remark These properties say that X H is a subspace. We ll study subspaces in chapter 3.

Solution sets Suppose A M m,n (R) and b R n. The solution set for a consistent homogeneous linear system Ax = 0 is X H = {x Ax = 0}. The solution set for a consistent non-homogeneous linear system Ax = b is X G = p + X H = {x Ax = b}, a translation of X H by a particular solution p (i.e., Ap = b). What happens if p X H? z p + XH = {x Ax = b} z p + XH = {x Ax = b} p + tv p p XH = {x Ax = 0} tv XH = {x Ax = 0} v y y x x

Systems of Linear Equations Write the system of linear equations 3x + y 3z = 2 x + y z = 0 x z = 1 as an augmented matrix (A b), solve the system by row reduction, and write the solution in the form p + X H.

Solving linear systems 3 1 3 2 1 1 1 0 1 0 1 1

Solving linear systems 3 1 3 2 1 1 1 0 1 0 1 1 R 1 R 3 1 0 1 1 1 1 1 0 3 1 3 2

Solving linear systems 3 1 3 2 1 1 1 0 1 0 1 1 R 1 R 3 R 1 + R 2 R 2 1 0 1 1 1 1 1 0 3 1 3 2 1 0 1 1 0 1 0 1 3 1 3 2

Solving linear systems 3 1 3 2 1 1 1 0 1 0 1 1 R 1 R 3 R 1 + R 2 R 2 3R 1 + R 3 R 3 1 0 1 1 1 1 1 0 3 1 3 2 1 0 1 1 0 1 0 1 3 1 3 2 1 0 1 1 0 1 0 1 0 1 0 1

Solving linear systems 3 1 3 2 1 1 1 0 1 0 1 1 R 1 R 3 R 1 + R 2 R 2 3R 1 + R 3 R 3 R 2 + R 3 R 3 1 0 1 1 1 1 1 0 3 1 3 2 1 0 1 1 0 1 0 1 3 1 3 2 1 0 1 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1 0 0 0 0

Outline Matrices Linear systems Row operations Inverses Determinants

Finding inverses 1 0 4 3 1 12 2 0 7 = A

Finding inverses 3R 1 + R 2 R 2 1 0 4 3 1 12 2 0 7 1 0 4 0 1 0 2 0 7 = A = E 1 A

Finding inverses 3R 1 + R 2 R 2 2R 1 + R 3 R 3 1 0 4 3 1 12 2 0 7 1 0 4 0 1 0 2 0 7 1 0 4 0 1 0 0 0 1 = A = E 1 A = E 2 E 1 A

Finding inverses 3R 1 + R 2 R 2 2R 1 + R 3 R 3 R 1 4R 3 R 1 1 0 4 3 1 12 2 0 7 1 0 4 0 1 0 2 0 7 1 0 4 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1 = A = E 1 A = E 2 E 1 A = E 3 E 2 E 1 A = EA

Finding inverses 3R 1 + R 2 R 2 2R 1 + R 3 R 3 R 1 4R 3 R 1 1 0 4 3 1 12 2 0 7 1 0 4 0 1 0 2 0 7 1 0 4 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1 = A = E 1 A = E 2 E 1 A = E 3 E 2 E 1 A = EA Therefore, I = EA. How do we find E = E 3 E 2 E 1?

Finding inverses E is the result of how the row operations transform I. 1 0 0 0 1 0 = I 0 0 1

Finding inverses E is the result of how the row operations transform I. 1 0 0 0 1 0 = I 0 0 1 1 0 0 3R 1 + R 2 R 2 3 1 0 = E 1 0 0 1

Finding inverses E is the result of how the row operations transform I. 1 0 0 0 1 0 = I 0 0 1 1 0 0 3R 1 + R 2 R 2 3 1 0 = E 1 0 0 1 1 0 0 2R 1 + R 3 R 3 3 1 0 = E 2 E 1 2 0 1

Finding inverses E is the result of how the row operations transform I. 1 0 0 0 1 0 = I 0 0 1 1 0 0 3R 1 + R 2 R 2 3 1 0 = E 1 0 0 1 1 0 0 2R 1 + R 3 R 3 3 1 0 = E 2 E 1 2 0 1 7 0 4 R 1 4R 3 R 1 3 1 0 = E 3 E 2 E 1 = E 2 0 1

Finding Inverses Definition Given an n n matrix A with real entries, we say that A is invertible if there exists an n n matrix E with real entries such that EA = I, in which case we call E an inverse of A and write E = A 1. Theorem Given A M n,n (R), A 1 exists if there is some sequence of row operations that reduces A to the identity matrix (i.e., if rref(a) = I). Applying those same row operations to I yields E = A 1, hence applying those row operations simultaneously to A and I gives a method for finding A 1 : (A I) EROs (I A 1 ). If rref(a) I, then A 1 does not exist.

Using inverses Exercise 1. Given E and A below, verify that E is an inverse of A. E = 7 0 4 3 1 0 2 0 1, A = 2. Find a general formula for the dot product (row i of E) (column j of A). 3. Find AE. Is A an inverse of E? 4. Find a general formula for the dot product (row i of A) (column j of E). 1 0 4 3 1 12 2 0 7.

Using inverses Exercise Let A = 1 0 4 3 1 12 2 0 7, and E = A 1 = 7 0 4 3 1 0 2 0 1 1. Use E = A 1 to find the intersection of the planes x + 4z = 2, 3x + y + 12z = 1, and 2x + 7z = 3. 2. If b = 2, 1, 3, use E = A 1 to solve Ax = b for x. 3. If b is any vector in R 3, find the general solution to the linear system Ax = b. 2 20 4. If B = 1 10, solve AX = B. 3 30. 5. If B is any matrix in M 3,k (R), find the general solution to AX = B.

Using inverses Exercise 1. If possible, find the inverse of A = 1 1 0 2 3 1 0 2 2 2. What is the solution set X H to the homogeneous linear system Ax = 0? 0 1 2 3. Is B = 1 4 2 invertible? If so, find its inverse. 1 2 1 4. What is the solution set X H to the homogeneous linear system Bx = 0? 5. Conjecture a relationship between invertible matrices and homogeneous solution sets..

Computing inverses Theorem ( ) a b When ad bc 0, the matrix A = is invertible and its c d inverse is ( ) A 1 1 d b =. ad bc c a Example The inverse of A = A 1 = 1 5 ( 3 1 2 1 ( 1 1 2 3 ) is ) = ( 0.2 0.2 0.4 0.6 ).

Properties of inverses Theorem Suppose the matrices below are all n n invertible matrices with real entries. 1. (A 1 ) 1 = A 2. (AB) 1 = B 1 A 1 3. (A 1 A 2 A k ) 1 = A 1 k A 1 2 A 1 1 Theorem Let A M n (R) be invertible. 1. If B M n,k (R) and AB = 0, then B = 0 2. If C M m,n (R) and CA = 0, then C = 0 3. If B, C M n,k (R) and AB = AC, then B = C 4. If B, C M m,n (R) and BA = CA, then B = C

Using inverses Exercise Solve A T + XB = C, where A = ( 1 2 3 4 5 6 ) ( 1 1, B = 3 2 ), C = 1 7 0 1 1 3.

Using inverses Exercise Solve A T + XB = C, where A = ( 1 2 3 4 5 6 ) ( 1 1, B = 3 2 ), C = 1 7 0 1 1 3 Since XB = C A T, it follows that X = (C A T )B 1 and thus X = 9 3 8 2 5 1.

Outline Matrices Linear systems Row operations Inverses Determinants

Computing determinants Definition ( a b Let A = c d ). Then det(a) = A = ad bc. Theorem Let A = (a ij ), and let A ij denote the ij-minor of A obtained by deleting the ith row and jth column of A. Then det(a) = ( 1) i+j a ij det A ij where the sum is taken from j = 1 to j = n if expanding along row i and from i = 1 to i = n if expanding along column j. The quantities det(a ij ) are (n 1) (n 1) determinants, and this process can be repeated until only determinants of 2 2 matrices remain.

Computing determinants Alternating signs: Minors of A = + + + + + + + +.. a 1,1 a 1,2 a 1,3 a 2,1 a 2,2 a 2,3 a 3,1 a 3,2 a 3,3 a det(a)=+a 2,2 a 2,3 1,1 a 3,2 a 3,3 a 1,2 a 2,1 a 2,3 a 3,1 a 3,3 + a 1,3 a 2,1 a 2,2 a 3,1 a 3,2 = a 1,2 a 2,1 a 2,3 a 3,1 a 3,3..... + a 2,2 a 1,1 a 1,3 a 3,1 a 3,3 a 3,2 a 1,1 a 1,3 a 2,1 a 2,3

Computing determinants Theorem Let A M n,n (R). Then the value of det(a) is the same regardless of the choice of row or column to expand over at each stage of the algorithm. Theorem Given A M n,n (R), det(a) = det(a T ). Proof. (Rough sketch:) Calculating det(a) by expanding along the top row of A is exactly the same as calculating det(a T ) by expanding along the leftmost column of A T.

Computing determinants Example Compute det 1 2 1 0 0 1 0 1 1 2 3 1 3 2 1 2 Exercise Compute det 1 2 1 3 2 2 1 3 2

Computing determinants Example Compute det 1 2 1 0 0 1 0 1 1 2 3 1 3 2 1 2 = 20 Exercise Compute det 1 2 1 3 2 2 1 3 2 = 3

Determinants geometrically x 2 x 2 2 0 Ae 2 e A = 2 x 1 0 3 x 1 e 1 Ae 1 ( ) 0 0 = ( 0 ) 1 e 1 = ( 0 ) 0 e 2 = 1( ) 1 e 1 + e 2 = 1 ( 2 0 det 0 3 ) = 6 ( ) 0 A0 = ( 0 ) 2 Ae 1 = ( 0 ) 0 Ae 2 = 3( ) 2 A(e 1 + e 2) = 3

Determinants geometrically x 2 x 2 2 0 e A = 2 x 1 0 3 Ae 1 x 1 e 1 Ae 2 ( ) 0 0 = ( 0 ) 1 e 1 = ( 0 ) 0 e 2 = 1( ) 1 e 1 + e 2 = 1 ( 2 0 det 0 3 ) = 6 ( ) 0 A0 = ( 0 ) 2 Ae 1 = ( 0 ) 0 Ae 2 = 3 ( ) 2 A(e 1 + e 2) = 3

Determinants geometrically e 2 e 1 A = a c = (a b d 1 a 2 ) Ae2 = a2 = ( ) c d n = ( ) b a p Ae1 = a1 = ( ) a b Note that n = b, a is perpendicular to a 1 = a, b. The signed length of p is the component of a 2 along n, comp n a 2 = b, a c, d b, a so the signed area of the parallelogram is = ad bc b, a, ad bc Signed area = (base)(height) = a, b = ad bc. b, a

Determinants geometrically z z x e 3 A = (a 1 a 2 a 3 ) e 1 e 2 y = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 x Ae 3 = a 3 Ae 2 = a 2 y Ae 1 = a 1 Using geometry, projections, and properties of, we showed that the triple scalar product a 1 (a 2 a 3 ) is the volume the parallelepiped formed by a 1, a 2, and a 3. It is easy to verify that det(a) = a 1 (a 2 a 3 ), so Signed volume = det(a) = a 1 (a 2 a 3 ). The linear function f : R 3 R 3 defined by f (x) = Ax has magnification factor det(a), i.e., one cubic unit of volume gets mapped to det(a) cubic units of signed volume.

Determinants geometrically More generally, a matrix A M n,n (R) defines a linear function f : R n R n by f (x) = Ax, and the determinant of A is the signed magnification factor for f : det(a) = signed area (or volume) of output region signed area (or volume) of input region, where we use area when n = 2, volume when n = 3, and hypervolume when n 4. The sign of det(a) is determined by the number of reflections in the linear transformation (e.g., if a right-handed coordinate system in the input space gets mapped to a left-handed coordinate system in the output space, then the sign of det(a) is negative).

Geometry of determinants x 2 x 2 Ae1 x 2 A 2 e 2 e 2 x 1 A x 1 e 1 A A 2 e 1 x 1 Ae 2 Suppose f (x) = Ax. 1. Find a formula for f (x). 2. Describe how f and f 2 transform the unit square (e.g., dilation, reflection, rotation, shear, projection, etc.). 3. Find the signed magnification factor for f, f 2, f 3, and f k. 4. Find a formula for f 1, if possible. 5. Find the signed magnification factor for f 1 and f k. What is det(a 1 )?

Geometry of determinants x 2 x 2 x 2 Ae 2 A 2 e 2 e 2 x 1 A x 1 A x 1 e 1 Ae 1 A 2 e 1 Suppose f (x) = Ax. 1. Find a formula for f (x). 2. Describe how f and f 2 transform the unit square (e.g., dilation, reflection, rotation, shear, projection, etc.). 3. Find the signed magnification factor for f, f 2, f 3, and f k. 4. Find a formula for f 1, if possible. 5. Find the signed magnification factor for f 1 and f k. What is det(a 1 )?

Geometry of determinants 1. Find a nonzero 2 2 matrix A = (a 1 a 2 ) with real entries such that det(a) = 0. What is the geometric relationship between the column vectors a 1 and a 2? 2. If A = (a 1 a 2 a 3 ) is in M 3,3 (R) and det(a) = 0, what is the geometric relationship among the column vectors a 1, a 2, and a 3? 3. Suppose a linear function f : R 3 R 3 defined by f (x) = Ax for some A M 3,3 (R) maps a sphere of radius 3 in the input space to an ellipsoid with volume 7 in the output space. What can you say about the determinant of A?

Geometry of determinants 1. Suppose A = 1 2 1 3 2 2 1 3 2, B = 1 10 1 3 10 2 1 15 2. Since det(a) = 3, what is det(b)? 2. Suppose A = (a 1 a 2 a 3 ) is in M 3,3 (R) has det(a) = 3. What are det(a 2 a 1 a 3 ), det(a 2 a 3 a 1 ), det(2a 1 5a 2 a 3 ), and det(5a)? 3. Find the area of the hexagon with side length 1.

Properties of determinants Theorem Let A, B M n,n (R). 1. det(a) = det(a T ). 2. det(ab) = (deta)(detb). 3. A is invertible if and only if det(a) 0, in which case det(a 1 ) = 1 det(a). 4. If A = (a ij ) is an upper-triangular, lower-triangular, or diagonal matrix, then det(a) = a 11 a 22 a nn. 5. If A has a row or column of zeros, then det(a) = 0. 6. For all n 1, det(i n ) = 1.