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Revisions on Linear Algebra. I Revisions on Linear Algebra. II Denition: A linear function f is a mathematical function in which the variables appear only in the rst degree, are multiplied by constants, and are combined only by addition and subtraction. A linear equation is of the form f (x, y, · · · ) = 0 with f linear. Solving Linear equations: Find x such that: Find (x, y) such that: Tell me if those following functions or equations are linear. If not, do you know how they are called ? B. A. 1 2 1 f (x, y) = 3xy 2 2 x4 + 3x2 − 4 = 0 −4 3 3 x 4 3x + y − 2 = 0 f (x) = 3x − 4 f (x) = 3x − 4 3 f (x) = 4 f (x, y) = 3x + y − 2 3 x 1 Write both equations in the form ax + by = k 2 Make the coefcients of one of the variables the same in both equations. 3 Add or substract (depending on signs) on equation from the other to form a new equation in one variable. Denition: A matrix is a rectangular table of (real or complex) numbers. The order of a matrix gives the number of rows and columns it has. A matrix with m rows and n columns has order m × n. Write down the order of the following matrices: 1 4 Solve the new equation obtained in step 3. 5 Put the value obtained in step 4 into one of the given equations to nd the corresponding value of the other variables. 2 3 Revisions on Linear Algebra. V is not square, but B = 1 −1 2 5 Denition: The transpose of a matrix M , written M T is obtained by interchanging the rows and columns. For example, If M = −1 5 3 −2 ! −3 ! 1 −1 2 5 −5 8 2 A= B= C= ! Revisions on Linear Algebra. VI Denition: A square matrix has the same number!of rows and columns (order of the ! 2 x − 2y − 3 = 0 3x + y = −1 Revisions on Linear Algebra. IV Simultaneous linear equations are solved with the following steps −3 −4=0 Revisions on Linear Algebra. III form (m × m)). For instance, A = 3x = 4. , then M T = −1 3 5 −2 ! is. Compute 1 2 (M T )T MT with M = −1 5 3 −2 ! with M= 8 2 4 3 1 2 ! Revisions on Linear Algebra. VII Revisions on Linear Algebra. VIII Denitions: Example Matrices can only be added, or substracted, if they are of the same order. Simply add or substract corresponding elements. This gives a matrix of the same order. To multiply a matrix by a scalar (a number), multiply each element of the matrix by the number. Example 8 2 4 3 1 2 ! 2 = 16 4 8 6 2 4 ! 3 2 −1 6 2 4 2 With A = 1 A+B 2 A + CT 3 3B −3 8 4 ! ,B= 0 ! 5 + −1 2 5 1 −1 5 ! 2 4 0 + 1 2 ! 1 6 and C = ! Matrix addition is commutative. Let A and B be two matrices of the same order then: A+B=B+A Matrix addition is associative. Let A, B and c be three matrices of the same order then: −5 1 2 8 4 , 8 , compute (if you can): Example 1 0 1 4 1 5 6 8 7 9 0 9 2 = 5 3 4 13 17 7 9 6 8 5 3 2 Revisions on Linear Algebra. XII The determinant of a 2 × 2 matrix A = 2 9 is not dened. Denitions: ! 1 Memory aid: Row by Column Revisions on Linear Algebra. XI −5 6 ! matrix equals the number of rows in the second. 1 1 0 Multiplication of Matrices: Multiply each row of the rst matrix by each column of the second matrix. Two matrices can only be multiplied if the number of columns in the rst A + (B + C) = (A + B) + C 2 4 Revisions on Linear Algebra. X Denitions: A= 8 . Revisions on Linear Algebra. IX Let = 3 B= 6 2 −8 4 0 5 1 9 0 2 0 Find AB and BA, if possible. Is the multiplication of matrices commutative? ad − bc = det(A). a c b d ! is the number If det(A) = 0, A is called a singular matrix. Compute the determinant of the matrices: 1 2 A= B= 1 2 3 4 ! 3 1 −2 4 . ! Denition: A square matrix with 1s on the diagonal and 0s elsewhere is called the identity matrix and denoted by I . Revisions on Linear Algebra. XIV Revisions on Linear Algebra. XIII Denition: If A is a square matrix, and there exists a matrix B so that If A is invertible, we denote the inverse of A by A−1 , so A−1 A = AA−1 = I. AB = BA = I, plays the rôle of the reciprocal or multiplicative inverse in ordinary multiplication, since A−1 we say that A is invertible, and B is called an inverse of A. It can be shown that an invertible matrix has exactly one inverse, so we refer to the inverse of an invertible matrix. aa−1 = a−1 a = 1, for nonzero a ∈ R. Show that if A is not square, it cannot have an inverse. Revisions on Linear Algebra. XV If A = a c b d ! Revisions on Linear Algebra. XVI then A−1 = 1 det(A) d −c −b a ! 1 Exchange the elements of the diagonal. 2 Change the signs of the elements of the other diagonal. 3 Multiply by 1 det(A) . Prove the above result (i.e. equal to I ). compute AA−1 and A−1 A, and check Revisions on Linear Algebra. XVII If M = 1 2 3 3 5 2 3 ! Let M = and N = −1 3 −1 2 ! , Compute (MN)−1 . Compute N −1 M−1 Any comment on (MN)−1 and N −1 M −1 ? 1 −2 3 1 ! and A = 7 10 21 23 ! . 1 Compute M−1 A. 2 If MB = 2M + A, express B in matrix form. they are Revisions on Linear Algebra. XVIII Lets come back to solving a system of linear equations: x − 2y − 3 = 0 3x + y = −1 Solve this system by matrix methods. Revisions on Linear Algebra. XIX Revisions on Linear Algebra. XX Denitions: A vector (or column vector) is a matrix with a single column. A matrix with a single row is a row vector. The entries of a vector are its components. Solve Solving simultaneous equations of the form: 2x + 3y = 4 10x + 4y = 9 Ax = b with A a square matrix and x unknown vector and b vector of constants, is given by x = A−1 b when A is invertible. Revisions on Linear Algebra. XXI Revisions on Linear Algebra. XXII Solutions of linear equations in two variables correspond to intersection points of lines. Three ways in which lines can intersect: Three possibilities for solutions of LS in two unknowns (1) A unique solution (0) No solution (∞) Innitely many solutions We call a linear system with no solutions INCONSISTENT. Solve: 1 2 Intersect at unique point Parallel lines - no intersection point Lines coincide 3 x+y=1 2x − y = 2 2x − y = −1 2x − y = 0 y−x=2 3y − 3x = 6 Revisions on Linear Algebra. XXIV Revisions on Linear Algebra. XXIII Denitions: A scalar λ is called an eigenvalue of a square matrix A if it satises: Ax = λx with x a vector non null. x is then called an eigenvector. The eigenvalue λ is found by solving the characteristic equation: det(A − λI) = 0 Find the eigenvalues of A= 2 0 1 3 ! Find the eigenvectors corresponding to λ = 2 and λ = 3.