Contents 1 Matrices 1 1.1 Introduction................................... 1 1. Determinant of a matrix............................ 1.3 Inverse of a matrix............................... 4 1.4 Eigenvalues and eigenvectors......................... 5 1.5 Diagonalisation of matrices.......................... 7 1.6 Solutions to systems of equations....................... 7 Vector Calculus 11.1 Introduction................................... 11. Equations of lines................................ 1.3 Equations of planes............................... 1.4 Vector functions................................. 1.5 Tangent lines and normal plane to space curves.............. 13.6 Functions with several variables....................... 14.6.1 Composite functions.......................... 14.6. Partial differentiation of vectors................... 15.7 Application: Scalar and vector fields..................... 15.8 Directional derivatives of scalar field..................... 15.9 Gradient of a scalar field............................ 16.10 Vector field.................................... 16.11 Divergence of a Vector field.......................... 17.1 Curl of a Vector field.............................. 17.13 Tangent plane to a surface at a point..................... 17.14 Normal line to a surface at a point...................... 17 1
CONTENTS
Chapter 1 Matrices 1.1 Introduction A matrix is an array of numbers in rows and columns. It is named by an alphabetic letter and a size. A matrix size or order is the number of rows number of columns. An m n matrix A is denoted by A m n = [a ij ] m n for i = 1,,...m and j = 1,,...n. A square matrix is one in which the number of rows is equal to the number of columns. Matrices A and B are equal if and only if the entries of matrix A, [a ij ] are equal to those of matrix B [b ij ] for all i and j. Matrix operations Addition and subtraction of matrices is only possible for matrices of the same order. Scalar multiplication is when all entries of a matrix are multiplied by same scalar, k. Multiplication of matrices is possible if the number of columns of the first matrix is equal to the number of rows of the second matrix. The (ij) th entry of XY is equal to the [ ] i th row of X j th column of Y = x i1 x i... x in y 1j y j.. y nj. Types of Matrices 1. Triangular Matrix This is a matrix in which a ij = 0 for all i > j or i < j but not both e.g. This is called a lower triangular matrix.. Singular Matrix This is a matrix whose determinant is zero. 3. Square Matrix 1 0 0 4 0 3 1.
Matrices A square matrix is a matrix in which the number of rows is equal to the number of columns i.e. A mn = A nn = A mm since m = n. 4. Diagonal Matrix This is a square matrix in which a ij = 0 for all i j e.g. [ 1 0 0 4 ] 5. Symmetric Matrix It is a square matrix in which the (a ij ) th entry is equal to the (a ji ) th entry for all 1 1 4 i and j e.g. 1 6 14 4 14 5 6. Skew-symmetric Matrix It is a square matrix in which the (a ij ) th entry is equal to the (a ji ) th entry for 1 1 64 all i and j e.g. 1 6 11 Transpose of a Matrix 64 11 5 The transpose of a matrix A is a matrix B such that the (a ij ) th entry of A is the (a ji ) th 4 1 4 entry of B. The transpose of A is denoted by A t. Take matrix A = 1 0 and B = A t = 4 6 1 1 1 4 0 5 Trace of a Matrix 6 1 5 The trace of a matrix is the sum of the leading diagonal elements. Taking matrix A above, its trace = tr(a) = 4 + 1 + 5 = 10. 1. Determinant of a matrix For a square matrix A = [a ij ], there is a unique number associated with matrix A called the determinant of A denoted by either deta or A. A lot of methods are available to calculate determinants of order 3 and above which are Sarrus, pivotal condensation or cofactor approach. [ ] a b a b Determinant of second order matrix, B = its determinant B = c d c d = ad bc Determinant of matrices of order 3 and above is calculated using:
Matrices 3 i. Determinant by Sarrus method The method is only applicable to square matrices of order 3. It requires you to augment your matrix by adding first columns of a 3 3 matrix in their order. Take a b c D = d e f. In this case we will add first columns of D to make a new matrix g h i a b c a b d e f d e g h i g h ii. Determinant by Pivotal condensation method iii. Determinant by Cofactor method The determinant is calculated by choosing any row or column that you want, then multiply each entry in the selected row or column by its cofactor and finally finding the sum of the resulting products. The cofactor of an entry is the determinant of a lower order matrix produced by deleting the row and column containing the entry times ( 1) i+j where ij are the subscripts of the entry i.e. det(d) = a i1 A i1 + a i A i +... + a in A in = 1 k=n k=1 a ika ik where A ik are the cofactors of the entries a ik. Give D = 3 1 4 1 then the det(d) is computed by: 1 D = 3 1 = ( 1) 1+1 1 4 1 1 + 3 1( 1)1+ 4 1 + 3 1 ( 1)1+3 4 = ( 1 + 4) 1(3 8) + ( 6 + 8) = 15 5 1 4 Find the determinant of B = 3 0 3 1 3 1 4 0 6 iv. Determination by reducing matrix to upper triangular form We reduce the matrix into triangular form by using property 9 below. The determinant is then the product of the entries in the leading diagonal. 1 3 1 3 1 3 B = 4 4 6 r = r 1 + r 0 0 r 3 = r 1 + r 0 0 = 1 9 1 9 0 0 6 ( )(6) = 4
4 Matrices Properties of determinants 1. det(ab) = detadetb. A = A t 3. AB = BA 4. Interchanging rows or columns changes sign of the determinant. 5. Multiplying matrix by scalar, k multiplies determinant by k n where n is the number of rows i.e. ka = k n A 6. Multiplying row or column by k multiplies determinant by k. 7. One row or column proportional to another gives a determinant of zero. 8. One row or column being equal results in determinant of zero. 9. Addingor subtracting scalar multiple of one row or column to another does not affect the determinant. 1.3 Inverse of a matrix Inverse by Adjoint method The inverse of a matrix can ( be calculated ) using the( adjoint or ) elementary row operations. For a x matrix A =, A 1 = 1 d b a b ad bc where ad bc 0. If c d c a A is an nxn matrix for n >, then its inverse is given by A 1 = 1 (T ranspose of Matrix of cofactors of A) A = 1 A AdjointA The minor of an entry is the determinant of a lower order matrix produced by deleting the row and column containing the entry. The cofactor is then this minor times ( 1) i+j where i and j are respectively the row and column number of the entry. Inverse by elementary row operations method The method requires you to augment the matrix with an identity matrix of the same order. It is put on the right side of the matrix. Perform row operations until the identity matrix is on the left side of the matrix and the matrix on the right is the inverse of the original matrix. Elementary row operations are used to reduce the matrix into the required form. The row operations are:
Matrices 5 1. Interchanging rows i and j of a matrix. Multiplying row by a non-zero scalar 3. Replace row by the sum of that row and scalar multiple of another row. Example Determine the inverse of matrix A = ( 4 3 ) using row operations. Solution( ) ( ) ( ) 4 1 0 (AI) = r 1 = 1 1 1 1 1 4 3 0 1 r 1 1 4 0 1 r = r 3r 1 4 0 1 3 0 1 0 3 4 1 ( ) 1 0 1 1 r = r, r 1 = r 1 r 0 1 3. Hence the inverse of matrix A is the square matrix of order two on the right side of this matrix. A 1 = ( 1 1 3 ) Properties of inverses 1. If matrix A is invertible then A 0. If A = 0 then A is not invertible. 3. If A is invertible then (A 1 ) 1 = A 4. If A has an inverse then that inverse is unique. 5. If A is invertible then A t is also invertible i.e. (A t ) 1 = (A 1 ) t 6. If A is invertible then A k is also invertible i.e. (A k ) 1 = (A 1 ) k 7. If A is invertible then ka is also invertible i.e. (ka) 1 = 1 k A 1 8. If A and B are invertible then their product is also invertible i.e. (AB) 1 = B 1 A 1 1.4 Eigenvalues and eigenvectors An eigenvector of a square matrix A is a non-zero column vector X such that AX = λx where λ is a scalar called the corresponding eigenvalue. e.g 0 1 1 3 4 = 3 4 1 0
6 Matrices Here 4 is the eigenvector and 3 is the correspoding eigenvalue. Note: A scalar multiple of an eigenvector is also an eigenvector of the same matrix with the eigenvalue unaltered. Finding eigenvalues and vectors of a matrix AX = λx (A λi)x = 0 = A λi = 0 A λi = 0 is the characteristic ( ) equation whose roots are the eigen values of A. 1 Find the eigen values of 3 0 Solution ( ) ( ) 1 λ 0 3 0 0 λ = λ 1 3 λ = 0 λ λ 3 = 0 λ = 1 or 3 So the corresponding eigen vectors are got by solving for matrix X from the equation x that (A λi)x = 0 where matrix X= y. So taking the first eigen value λ = 1 ( z ( 1) 1 3 0 ( 1) ) ( x y ) = 0 3x + y = 0 = y = 3x 3x + y = 0 ( So, taking x = 1, y = 3. The eigen vector corresponding to λ = 1 is X = Exercise 1 3 ) 1. Find the eigenvectors of the matrix after verifying that its eigenvalues are λ = 1, and 3. 0 1 1 3 1 0
Matrices 7 1.5 Diagonalisation of matrices A square matrix A = [a ij ] n n is said to be diagonalisable if it can be written in such a way that a ij = 0 for all i j. The entries in the leading diagonal are the eigen values of the original matrix A. A matrix P which diagolises A is the matrix of eigen vectors of A written in the order corresponding to that of the eigen values in the diagonal form D. To diagonalise a matrix we therefore need the eigen vectors of that matrix. If D is the diagonal form ( of A then ) AP = P D and we say D is similar to A e.g. Find the 5 3 diagonal form of 6 Linear independence A set of vectors v 1, v,...v n is said to be linearly dependent if there exists scalars a 1, a,...a n not all zero such that a 1 v 1 + a v +...a n v n = 0. If a 1 v 1 + a v +...a n v n = 0 = a i = 0 for all i then v 1, v...v n are said to be linearly independent. We show that vectors are linearly dependent or not by writing the vectors as rows of a matrix and reducing the matrix to row echelon form. If the resultant matrix has at least one zero row then the vectors are linearly dependent andif there is no zero row the vectors are linearly independent. Determine whether or not the vectors (1, 1,, 1), (1,, 3, 3) and (, 1, 5, 5) are linearly independent. 1.6 Solutions to systems of equations A system of n equations each with at most n unknowns is said to be consistent if it has at least one solution. A system which does not have any solution on the other hand is said to be inconsistent. Every linear system of equations has either a unique solution, an infinite number of solutions or no solutions at all. Solutions of systems of equation can be found using matrix methods that are i) Row operations, ii) Cramer s rule and iii) matrix inverse method Row echelon form of matrix An n n matrix is said to be in row echelon form if: 1. the leading entry in each non-zero row is 1.. the leading entry in each non-zero row appears to the right of the leading entry of all rows above it. 3. if zero rows exist, they should be at the bottom. If in each column that has a leading entry all other entries are zeros, the matrix is said to be in reduced row echelon form. Examples of matrices in row echelon and reduced row echelon form.
8 Matrices Gauss-Jordan elimination method This is a sequence of steps followed to reduce a matrix to row echelon form. The steps are: 1. Locate row with left most entry not equal to zero and move it to the top.. Change leading entry of top to 1. 3. Change all entries in column one below a 11 to zeros using row operations. 4. Locate (from the remaining rows) a row with left most entry not equal to zero and make it row. 5. Repeat step and 3. 6. Repeat steps 4, and 3. 7. To get reduced row echelon form start from bottom and move upwards adding suitable multiples of every row to those above. Example Solve the following system of equations by reducing the augmented matrix to row echelon form. Solution The augmented matrix is: 3 0 6 0 1 1 3 r 1 = r 1 3 4 0 1 1 0 0 r 3 = r 3 4r 0 1 0 3 0 0 4 13 3x 6z = 0 x + y + z = 3 x + 4y = 1 1 0 0 1 1 3 4 0 1 r 3 = r 3 4 r 1 = r 4 + r r = r + r 1 r 3 = r 3 r 1 1 0 0 13 0 1 0 3 0 0 1 13 4 From the final matrix, the solution to the system of equations is: Cramer s rule x = 13 13, y = 3, z = 4 1 0 0 0 1 0 3 0 4 4 1 This is another matrix method used to solve systems of equation given in the form AX = B. The method requires knowledge of calculating determinants of matrices. The following steps should be followed when using this method. 1. Identify matrices A and B from the equations.
Matrices 9. Calculate the determinant of matrix A. 3. Introduce matrices A i by replacing colum i of matrix A by matrix B. 4. Calculate the determinants of matrices A i. 5. Solutions to the system of equations, x i is given by x i = A i A for all i from 1 to n, where n is the number of variables. Inverse matrix method The inverse method assumes knowledge of inverses of matrices. The equation AX = B is solved by the following steps. 1. Calculate the inverse of matrix A.. Pre-multiply the equation AX = B by A 1 and you get A 1 AX = A 1 B. 3. This results in X = A 1 B. 4. Interpret your solutions from the relations that the column matrix X = [x i ] Exercise 1. Solve the system of equation: 3y + 4z = x 5y + z = 9 x + y 4z = 7
10 Matrices
Chapter Vector Calculus.1 Introduction A vector is a quantity that has both magnitude and direction. Examples of vector quantities are velocity, acceleration and force. vectors are denoted by bold or underlined letters, directed lines, column matrices of the i, j, k notation e.g. i-5j+4k where i, j and k are unit vectors in the x, y, and z axes respectively. A unit vector is a vector of magnitude 1. The modulus of a vector r is the length of a line segment that represents the vector. If r = xi + yj + zk the modulus of vector r is r = x + y + z Addition and Subtraction Correspnding compents are added or subtracted from the given vectors. Scalar Multiplication All vector components are multiplied by the scalar. Vector multiplication If vectors r 1 = x 1 i + y 1 j + z 1 k and r = x i + y j + z k then the scalar or dot product of r 1 and r denoted by r 1 r = x 1 x + y 1 y + z 1 z which is a scalar. The cross or vector product of two vectors r 1 and r denoted by r 1 r = i j k x 1 y 1 z 1 x y z which is a vector whose modulus denotes the area of the parrallelogram made by the two vectors. If r 3 = x 3 i + y 3 j + z 3 k then the scalar tripple product is given as below, whose
1 Vector Calculus modulus is the volume of the parallepiped with the three vectors as sides. x 1 y 1 z 1 r 1 r r 3 = x y z x 3 y 3 z 3 Orthogonal vectors have a dot product of zero. Orthogonal projection of one vector onto another ( a, b) is the length of the shadow of a along b when the light is vertically above a and b is horizontal. The projection of a onto b is = a cos θ. Equations of lines Vector equation of a line is r = a + ut or can be written in cartesian form as x x 0 a = y y 0 b = z z 0 c Nature of lines Lines in space can be intersecting, parallel, skew or coinciding. To discuss equations of lines that intersect, are parallel, skew and coincides..3 Equations of planes If a point r 0 and a normal to the plane are given, then the equation is: ( r r 0 ).N = 0 If three points are given as A, B, C and let R be a general point in the plane, then the normal to the plane is AB AC. The equation then takes the same form above. The cartesian equation takes the form ax + by + cz + d = 0 where ai + bj + ck is the vector normal to plane..4 Vector functions A vector function F of a single scalar value t is a function that assigns to each scalar value t 0 a unique vector F (t) called the value of F at t e.g. F (t) = (t +t)i+(1 t )j+tk. Each value of t corresponds to a point or vector whose initial point is the origin. A vector function like this describes a space curve.
Vector Calculus 13 Ordinary differentiation of a vector function F (t + t) F (t) A vector function is said to be differentiable at t if lim exists. This t 0 t limit is denoted by F (t) or df dt and is called the derivative of F. If F represents displacement then F represents velocity. For the function above F (t) = (t + )i tj + k Limit and continuity of a vector function A constant vector A is called the limit of a vector function F(t) as t t 0 written as lim F (t) = A if and only if for every ɛ > 0 there exists a > 0 such that F (t) A < ɛ t t 0 whenever 0 < t t 0 <. Show that the limit of F (t) = 3i + tj + tk as t 5 is 3i + 5j + 10k. Solution We show that F (t) (3i + 5j + 10k) 0 as t 0. Now F (t) (3i + 5j + 10k) = 3i + tj + tk (3i + 5j + 10k) = (t 5)j + t 10)k = (t 5)j + (t 5)k which approaches zero as (t 5) 0. Therefore lim t 5 (3i + tj + tk) = 3i + 5j + 10k. Find the lim( sin t i + cos t 1 j + e t k). To find the limit, show that a given constant t 0 t t vector is the limit of some vector function, one may consider the limits component by component and treat each component the way limits of scalars are treated..5 Tangent lines and normal plane to space curves Given that F (t) = (t cos t)i + (3 + sin t)j + (1 + cos 3t)k. Find the equation of the tangent line to this curve at t = π. Find also the equation of the normal plane through this point. Solution F (t) = (1 + sin t)i + ( cos t)j + ( 3 sin 3t)k, when t = π, F (t) = i j + 3k and F (t) = π i + 3j + k. Therefore equation of tangent is and equation of plane normal to curve is r(t) = ( π i + 3j + k) + λ(i j + 3k) [ r(t) ( π i + 3j + k)].(i j + 3k) = 0
14 Vector Calculus A unit vector tangent to a space curve is given by T = df i dt i + df dt j + df 3 dt k ( df 1 dt ) + ( df dt ) + ( df 3 dt ).6 Functions with several variables A function can have two or more independent variables e.g. W = f(x, y) = x + xy y and P = f(x, y, z) = x yz z. Partial derivatives are odrinary derivatives of a function of several variables with respect to one of the independent variables, keeping other independent variables constant. Partial derivatives of f(x, y) with respect to x and y are denoted by f x, f x or f x (x, y) and f y, f y or f y (x, y) respectively. Given a function f(x, y) = x 3 + 3xy then f x = 6x + 3y and f y = 6xy f xx = x ( f x ) = 1x, and f yy = y ( f y ) = 6x f xy = y ( f x ) = y (6x + 3y ) = 6y f yx = x ( f y ) = (6xy) = 6y x.6.1 Composite functions Let z = f(x, y) where x = g(r, s) and y = h(r, s), then z r = z x. x r + z y. y r and z s = z x. x s + z y. y s Example Given x = r s and y = r + s find u r in terms of r and s if u = x y 3. Solution u r = u x. x r + u y. y r = 4xy3.+6x y.1 = 8xy 3 +6x y = 8(r s)(r+s) 3 +6(r s) (r+s) Exercise Find z t at t = π given z = exy, x = t cos t and y = t sin t
Vector Calculus 15.6. Partial differentiation of vectors The partial derivative of a vector function F (x, y, z) = F 1 i + F j + F 3 k with respect to x where F 1, F, F 3 are function in x, y, z is F (x + x, y, z) F (x, y, z) F x = lim x 0 x provided it exists. F y and F z are defined similarly. If F (x, y, z) = (x 3 y + z )i + ye z j + (z + cos x)k then F x = 6x i sin xk..7 Application: Scalar and vector fields A field is a space or portion of space where the value of some quantity is defined. If the quantity in question is a scalar, the field is said to be scalar and if the quantity is a vector then the field is a vector. Examples of scaler fields are the density of atmospheric air and the temperature at each point in an insulated wall. F = x + y + 3z is an example of a scalar function. If x + y + 3z = k where k is constant, we say a level surface has been defined. There is a collection of all points at which the scalar field has the same value k..8 Directional derivatives of scalar field It is derivative of a scalar field / function along a given curve in a given direction. It measures the rate at which the scalar of interest in changing. It is defined by φ = φ φ φ cos α + cos β + u P o x y z cos γ where P o is the point of interest, u is the directional vector of interest, cos φ, cos β, and cos γ are the directional cosines of u and φ x, φ φ y and z are the partial derivatives with respect to each of the variables x, y and z. Example Find the directional derivative of φ(x, y, z) = xyz at P o (1,, 1) in the direction P ( 1, 1, ). Solution u = 1 1 1 1 = i + j + k cos α = 6, cos β = 1 6, cos φ = 1 6 u = ( ) + ( 1) + (1 ) = 6
16 Vector Calculus φ φ φ x = zy, y = xz, z = xy, and at point P o. φ φ φ x =, y =, z = 1, the directional derivative of the field at P o in the direction of P is φ u = ( 6 ) + ( )( 1 6 ) + ( 1)( 1 6 ) = 3 6 = 6 suggesting that the scalar is reducing along P. This value can also be got by making use of the vector differential operator del / nabla..9 Gradient of a scalar field It is a vector or vector function that gives the vector perpendicular to a given scalar field at a given point. It is defined by grad φ φ = i φ x + j φ y + k φ z where φ is the function and is the vector differential operator = i x + j y + k z. Example Find the gradient of φ(x, y, z) = x + y + z Solution φ = i (x+y+z) x + j (x+y+z) y + k (x+y+z) z = i + j + k which is normal to the family of planes x + y + z = k. Example Find the directional derivative of φ(x, y, z) = xyz at P (1,, 1) in the direction of P ( 1, 1, ). Solution The vector in the direction required is ( i + j + k) (i + j + k) = i j + k and the unit vector in this direction is 1 6 ( i j + k). φ = yzi + xzj + xyk. The directional derivative is therefore φ.n = (yzi + xzj + xyk).( 1 ( yz xz + xy) ( i j + k)) = 6 6 and at (1,, 1) this simplifies to 6.10 Vector field Thisis a space or portion of a space where a vector is defined at every point of the space. An example is the gravitational field of the earth. F (x, y, z) = x i + xyj + (x + y) 3 k and F = xi + yj are examples of vector functions.
Vector Calculus 17.11 Divergence of a Vector field This is a scalar which tells the extent to which a vector field explodes. It is given by: For F = xi + yj,. F = (i x + j y + k z ).(F 1i + F j + F 3 k). F = (i x + j y + k ).(xi + yj + ok) = 1 + 1 =. z.1 Curl of a Vector field It tells how a vector field rotates. It is given by: F i j k = x y z F 1 F F 3 Then for F = yi + xj the curl of F is: F = i j k x y z y x 0 = k.13 Tangent plane to a surface at a point Let r = (x, y, z) be a general point in a plane, r o = (x o, y o, z o ) be the point of interest and F be a scalar field of concern (surface). The tangent plane to the surface is given by: (r r o ).N = (r r o ). F Po = 0.14 Normal line to a surface at a point Let r = (x, y, z) be a general point on the line, r o = (x o, y o, z o ) be the point of interest and F be a scalar field. The equation of the line is given by x x o F x = y y 0 F P o y P o = z z o F z P o Example Find the equation of a) tangent plane b) normal line to the surface F (x, y, z) = x yz + 3y xz + 8z at (1,, 1).
18 Vector Calculus Solution A normal to the surface at (1,, 1) is.f (1,, 1) = (x y z z )i + (x z + 6y)j + (8 4xz + x y)k = 6i + 11j + 14k The equation of the tangent plane to the surface at (1,, 1) is: [r (i + j k)].( 6i + 11j + 14k) = 0 This reduces to: 6x + 11y + 14z = 0 The equation of the normal to surface at (1,, 1) is: This reduces to: Example r = (i + j k) + t( 6i + 11j + 14k) x 1 6 = y = z + 1 11 14 Find the equations of the a) tangent line b) normal plane to a curve R = (t cos t)i + (3 + sin t)j + (1 + cos 3t)k at t = π. Solution A vector tangent to the curve at a point where t = π is T o = dr dt t= π = (1 + sin t)i + cos tj 3 sin 3tk t= π = i j +3k and at t = π, R = π i+3j +k. The required equation is: x π = y 3 = z 1 3 A vector tangent to the curve is normal to the required plane and ( π, 3, 1) is on the plane, so the required equation is: Exercises [r ( π i + 3j + k)].(i j + 3k) = 0