Brief Introduction to Vectors and Matrices

1 CHAPTER 1 Brief Introduction to Vectors and MatricesIn this chapter, we will discuss some needed concepts found in in-troductory course in linear algebra. We will introduce matrix, vector , vector -valued function, and linear independency of a group of vectorsand vector -valued Vectors and MatricesAmatrixis a group of numbers(elements) that are arranged inrows and columns. In general, anm nmatrix is a rectangular arrayofmnnumbers (or elements) arranged inmrows andncolumns. Ifm=nthe matrix is called a square matrix. For example a 2 2 matrixis[a11a12a21a22]and an 3 3 matrix is a11a12a13a21a22a23a31a32a33 Generally, we use bold phase letter, likeA, to denote a matrix,and lower case letters with subscripts, likeaij,to denote element ofa matrix.

2 Hereaijwould be the element atithrow an element at1strow and column. Sometime we use theabbreviationA= (aij)for a matrix with Special the zero matrix whose elementsare all zeroes. So2 2and3 3zero Matrices are[0 00 0]and 0 0 00 0 00 0 0 Another special matrix is the identity matrix, denoted byI, a iden-tity matrix is an matrix whose main diagonal elements are 1, and all121. Brief Introduction TO Vectors AND Matrices other elements are 0. So2 2and3 3zero Matrices are[1 00 1]and 1 0 00 1 00 0 1 A vector is a matrix with one row or one column.

3 In this chapter,a vector is always a matrix with one column as[x1x2]for a two-dimensional vector and x1x2x3 for a three dimensional vector . Here the element has only one indexthat denotes the row position (Sometimes we use different variable todenote number in different position such as using[xy]for a 2-dimensional vector ). We use bold lower case, such asv, todenote a Operations on number in rectangularfashion, as a matrix, itself is not something terribly interesting. Themost important advantage from that kind arrangement is that we candefine matrix addition, multiplication, and scalar (i)Equality:Two matrixA= (aij)andB= (bij)are equalif corresponding elements are equal, (ii)Addition:IfA= (aij)andB= (bij)and the sum ofAandBisA+B= (cij) =aij+bij.

4 (iii)Scalar Product:IfA= (aij)is matrix andkis num-ber(scalar), thekA= (kaij)is product the above definition, we see that, to multiply a matrix bya numberk, we simply multiply each of its entries byk;to add twomatrices we just add their corresponding entries;A B=A+( 1) [2 3 1 4]1. Vectors AND MATRICES3andB=[0 53 4],find (a)A+B, (b)3A, (c)4A (a)A+B=[2 3 1 4]+[0 53 4]=[2 + 0 3 + 5 1 + 3 4 + ( 4)]=[2 82 0](b)3A= 3[2 3 1 4]=[6 9 3 12](c)4A B=[8 12 4 16] [0 53 4]=[8 7 7 20]aThe following fact lists all properties of matrix addition and , bB,andCbe Matrices .

5 Leta, bbe scalars(numbers). We have(1)A+ 0 = 0 +A=A, A A= 0;(2)A+B=B+A(commutativity);(3)A+ (B+C) = (A+B) +C,(ab)A=a(bA)(associa-tivity);(4)a(A+B) =aA+aB,(a+b)A=aA+bA(distributivity)When we have a row vector and a column vector with the samenumber of elements, we can define the dot product Product: in 2-dimension:Letx=[x1x2]andy=[y1y2], thedot product ofxandyis,x y=x1y1+x2y241. Brief Introduction TO Vectors AND Matrices in 3-dimension:Letx=[x1x2x3]andy= y1y2y3 ,the dot product ofxandyis,x y=x1y1+x2y2+ productLetA= (aij)andB= (bij),if the number of columns ofAis the same as number of rows ofB, then the product ofAandBisgiven byAB= (cij)wherecijis dot product ofithrow ofAwithjthcolumn [2 3 1 4]andB=[0 53 4],findABSolutionAB=[2 3 1 4][0 53 4]=[2 (0) + 3 (3) 2 5 + 3 ( 4)( 1) (0) + 4 3 1 5 + 4 ( 4)]=[9 212 21]Notice, the first element ofABis2 (0) + 3 (3)

6 Which is the dotproduct of first row ofA,[2 3]and first column ofB,[03]aThe following fact gives properties of matrix product, , B, Cbe three Matrices andrbe a scalar,we have A(BC) = (AB)C, r(AB) =A(rB)(associativity) A(B+C) =AB+AC(distributivity)Notice, in generalAB6=BA, that is for most of the times,ABis not equal the matrix notation and matrix product, we can write thefollowing system of equations{ax1+bx2=y1cx1+dx2=y21. Vectors AND MATRICES5asAx=ywithA=[abc d],x=[x1x2],andy=[y1y2]. square ( 2or3 3) matrixAisinvertible if there is a matrixA-1such thatAA-1=A-1A= [abc d]be a2 2matrix, ifAis invertible, we haveA-1=1ad bc[d b c d]So ifAis invertible, to solveAx=y, we need to simply multiplyboth sides withA-1,that isx= the system of equation{3x1 4x2= 2 2x1+ 5x2= 7 SolutionThe equation can be rewriteAx=ywithA=[3 4 2 5],x=[x1x2],andy=[27].}}

7 So in matrix form the system ofequation is[3 4 2 5][x1x2]=[27].Now the inverse ofAisA-1=13(5) ( 2)( 4)[5 42 3],so the solution isx=A-1y=17[5 42 3][27]=[387257]a61. Brief Introduction TO Vectors AND the system of equation 3x1 4x2+ 5x3= 2 2x1+ 5x2= 7x1 5x2+ 8x3= 1 SolutionThe equation can be rewriteAx=ywithA= 3 4 5 2 5 01 5 8 ,x= x1x2x3 ,andy= 27 1 .So in matrix form the system ofequation,Ax=y,is 3 4 5 2 5 01 5 8 x1x2x3 = 27 1 .It is a little harder to compute the inverse of a3 3matrix, wewill use Mathcad to solve the equation.

8 Here is how to do it, TypeA:[Ctrl][M]at a blank area to bring up the matrixdefinition screen, put 3 in the both input boxes and click OK,you will get a3 3matrix place holder likeA:= Fill the entries ofAin the corresponding position, using [Tab]key to navigate among the place holders(or just click each one). Typeb:[Ctrl][M]in another blank area, the matrix defini-tion screen is up again. This time put 3 in the number of rowbox, and 1 in the number of column box and click OK. Youwill get b:= put the values ofyin the correspondingposition.

9 TypeA^-1 *b=you will get the solution, which is, 15481175818081 1. Vectors AND MATRICES7 Notice, by default, Mathcad will display the results as dec-imal, you can double click on the result vector to change itto fraction, after you double click the result you will have apopup menu such asFigure ResultaThe next example shows how we can determine the unknown con-stants typically found in the initial value problems of system of differ-ential (t) =C1et+C2e 2tandx2= 2C1et C2e (0) = 2, x2(0) = 3,findC1andC2 SolutionFromx1(t) =C1et+C2e 2t, sett= 0we havex1(0) =C1e0+C2e 2(0)=C1+ ,x2(t)

10 = 2C1et C2e 2t,givesx2(0) = 2C1e(0) C2e 2(0)= 2C1 withx1(0) = 2, x2(0) = 3we have the following systemof equations,{C1+C2= 12C1 C2= 3 Rewrite the equation in matrix form[1 12 1][C1C2]=[13],and using Mathcad we find the solution is[C1C2]=[43 13]81. Brief Introduction TO Vectors AND MATRICESSox1(t) =43et 13e 2tandx2=83et+13e are solution ofthe following system of differential equations,{x 1(t) = x1(t) +x2(t)x 2(t) = Eigenvalues and [abc d]the de-termined ofAis defined as|A|= a bc d =ad a3 3matrix a11a12a13a21a22a23a31a32a33 we can compute the matrix as a11a12a13a21a22a23a31a32a33 =a11 a22a23a32a33 a12 a21a23a31a33 +a13 a21a22a31a32.}}