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An in-depth exploration of various algebraic concepts, including expanding brackets, factorising, vectors, notation, subtraction, multiplication by a scalar, multiplication of two matrices, matrix-vector multiplication, inverse matrix, linear systems of equations, 3×3 determinants, eigenvectors, and symmetric matrices. Additionally, it covers the concept of gradient in multivariable calculus.
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This document contains notes on basic mathematics. There are links to the corresponding Leeds University Library skills@Leeds page , in which there are subject notes, videos and examples. If you require more in-depth explanations of these concepts, you can visit the Wolfram Math- world website: ⇒ Wolfram link ( http://mathworld.wolfram.com/ )
There are practice equations available online to accompany these notes.
ample: the top left element in matrix A , equal to 1, is in row 1 and column 1 and can be labelled as element a 11 ; the element in the 2nd^ column of row 1, equal to 3, is labelled as a 12. A general element aij is located in row i and column j (see equation 4 for a further example). 2.1.2 Notation There are different types of notation for matrices and vectors that you may encounter in text books. Below are some examples: Matrix A italics A bold, italics A double underline, italics Vector x italics →x top arrow, italics x x single underline, italics bold 2.1.3 Addition ⇒ Wolfram link ( http://mathworld.wolfram.com/MatrixAddition.html ) ⇒ Video link ( http://www.youtube.com/watch?v=FX4C-JpTFgY ) Two matrices (or vectors) of the same size ( n × m ) may be added together, element by element. For instance, if we have two matrices A and B : A = a 11 a 12 a 21 a 22
b 11 b 12 b 21 b 22
then, 2.1.4 Subtraction
a 11 + b 11 a 12 + b 12 a 21 + b 21 a 22 + b 22
Similar to addition, corresponding elements in A and B are subtracted from each other: a 11 − b 11 a 12 − b 12 a 21 − b 21 a 22 − b 22
2.1.5 Multiplication by a scalar If λ is a number (i.e. a scalar) and A is a matrix, then λA is also a matrix with entries λa 11 λa 12 λa 21 λa 22
2.1.6 Multiplication of two matrices ⇒ Wolfram link ( http://mathworld.wolfram.com/MatrixMultiplication.html ) This is non-trivial and is governed by a special rule. Two matrices A , where A is of size n × m , and B of size p × q , can only be multiplied if m = p , i.e. the number of columns in A must match the number of rows in B. The matrix produced has size n × q , with each entry being the dot (or scalar) product (see section 2.1.10) of a whole row in A by a whole column in B.
2.1.7 Motivation for matrix-matrix multiplication To understand why we may need to perform matrix-matrix multiplication, consider two cus- tomers of a repair garage, Peter and Alex, who require a number of car parts for each of their vehicles. Peter requires 1 × 3 litre engine and 2 doors, whereas Alex requires 1 × 5 litre engine and 4 doors. All the parts require a certain number of screws and bolts. But how many total screws and bolts do Peter and Alex need? We can present the quantity of each car part that Peter and Alex need in a table: 3 litre engine 5 litre engine Doors Peter 1 0 2 Alex 0 1 4 or as the matrix, A : A =
The number of screws and bolts for each car part are expressed in the table: bolts screws 3 litre 3 4 5 litre 1 8 doors 2 1 or can be expressed as matrix, B : 3 4 2 1
Using simple addition we can find out how many screws and bolts are needed.
2.1.8 Matrix-vector multiplication Since a vector is a special case of a matrix, this is simply a special case of the matrix-matrix multiplication we have already discussed. Consider multiplying a column vector of length m by a matrix of size n × m , e.g. 1 3 6 2 23 0 − 1 4 3
which results in a column vector of length n and in this case n = 2. 2.1.9 Special Matrices Identity Matrix, I : The identity matrix, I , of size n × n , is defined in equation 12. aij = 1 if i = j 0 if i /= j
Clearly when x = 0 this breaks down and x has no inverse — this is also true when dealing with some matrices. 2.1.10 Scalar products and orthogonality The scalar product (or dot product, or inner product) of two column vectors of length n , where x = ( x 1 , x 2 , x 3 ,... xn ) and y = ( y 1 , y 2 , y 3 ,... , yn ), is n x · y = xi yi. i = This can also be written as xT^ y ; that is, the product of a row vector of length n with a column vector of length n. Two vectors are said to be orthogonal if their scalar product is zero.
⇒ Wolfram link ( http://mathworld.wolfram.com/LinearSystemofEquations.html ) ⇒ Video link ( http://www.youtube.com/watch?v=ZK3O402wf1c ) A linear system of equations such as 5 x + 3 y + z = 3 2 x − 10 y − 3 z = − 1 4 y + 5 z = 7 (17) can be written as 5 3 1 x 3 2 − 10 − 3 y −^1 as can be verified by multiplying out the left hand side. When solving the linear system Ax = b , (where A is a matrix, x is the vector ( x, y, z,... ) and b is a vector of numbers) two cases can arise: i) A −^1 exists. Then A −^1 Ax = A −^1 b ⇒ Ix = A −^1 b so x = A −^1 b ii) A −^1 doesn’t exist. There is then no solution in general. 0 4 5 z
A = (^) a 21 a 22 a 23 = 1 = 21 − 24 = − 3 The determinant of the coefficient matrix is 1 2 = 4 4 = 0 (19) 2 4 Therefore, A has no inverse and so no solution exists. This can also be seen in the fact that it is not possible that x + 2 y simultaneously equal both 2 and 3. 3 × 3 determinants: Determinants can be generalised to n × n matrices. For the 3 × 3 matrix A , a 11 a 12 a 13 a 31 a 32 a 33 the determinant of A is: | A |= a a^22 a^23 − a a^21 a^23 + a a^21 a^22 or equivalently, 11 a 32 a 33 12 a 31 a 33 13 a 31 a 32 | A |= a a^22 a^23 − a a^12 a^13 + a a^12 a^13 e.g. 11 a 32 a 33 21 a 32 a 33 31 a 22 a 23 2 3 4 6 7
1
or 1 0 0 5 6 7 Do whichever is easier! 2.3.1 Using determinants to invert a 2 × 2 matrix The determinant can be used in finding the inverse of a 2 × 2 matrix. a b For A = ,^ the^ inverse^ can^ be^ found^ using^ the^ formula c d A −^1 = (^1) d − b =
d − b (20) det A (^) − c a ad − bc (^) − c a For example: find the inverse of matrix B =
− 1 B −^1 = =
10 − 10 x = 0 ⇒ x=y x (^) 1 1 1 y (^) 1 or 14 − λ − 10 = 0 or (14 − λ )(− 1 − λ ) − (−10)(5) = 0 2 5 − 1 − λ ⇒ λ − 13 λ + 36 = 0 ⇒ ( λ − 4)( λ − 9) = 0 so λ = 4 or λ = 9 ii) Now to find the eigenvectors, v : If λ = 4 then ( A − 4 I ) x = 0 or y 5 − 5 y Hence the eigenvector is: x^ = x^1 , or y^1 , or just 1 , (see below). If λ = 9 , then 5 − 10 x = (^0) so x = 2 y 5 − 10 y 0 giving the eigenvector of 2 y^ or just 2. The eigenvectors provide a direction and so we can ignore common factors. This is because if v is an eigenvector, then so is μv for any value of μ. i.e. Av = λv ⇒ A ( μv ) = λ ( μv ).
For example: if λ = 4 above, then 1 ,^2 ,^4 are all eigenvectors. We typically choose the 1 1 simplest! 1 2 4 Matrix diagonalisation Suppose A is a n × n matrix. Form a new matrix V , each column of which is an eigenvector of A. If V −^1 exists then, V −^1 AV = Λ or A = V Λ V −^1 where Λ is a diagonal matrix with ( λ 1 , λ 2 ,... ), the eigenvalues of A on the diagonal. E.g. A = 14 − 10 then^ V^ =^
NB - Eigenvector 1 is: 1 , eigenvector 2 is: 2 It turns out that V −^1 exists and is: V −^1 =
Check: V V −^1 =
So then the matrix product