Contents. 1 Vectors, Lines and Planes 1. 2 Gaussian Elimination Matrices Vector Spaces and Subspaces PDF Free Download

Contents 1 Vectors, Lines and Planes 1 11 Vectors in R 2 and R 3 1 12 Vector Addition and Scalar Multiplication 5 13 Linear Equations in Two Variables 10 14 Linear Equations in Three Variables 20 15 Problems 31 2 Gaussian Elimination 36 21 Row echelon form of a matrix 36 22 Linear systems of equations 46 23 Homogeneous Linear System of Equations 59 24 Reduced row echelon form of a matrix 64 25 Problems 68 3 Matrices 75 31 Matrix Arithmetic 75 32 Matrix Product 80 33 Some Special Matrices 90 34 Elementary Matrices and Matrix Inverse 98 35 Problems 115 4 Vector Spaces and Subspaces 124 41 Vector Spaces 124 42 Subspaces of a Vector Space 132 43 Problems 143 ii

Contents 5 Linear Independence, Basis and Dimension 149 51 Linear Combinations and Span 149 52 Linear Independence 153 53 Basis and Dimension 163 54 Problems 171 6 The four fundamental subspaces 174 61 Column and Row Spaces 174 62 Nullspace 183 63 Dimensions of the fundamental subspaces 190 64 Problems 196 7 Linear Transformations 200 71 Linear Transformations Associated to Matrices 200 72 The Matrix of a Linear Transformation 209 73 Problems 231 8 Determinants 236 81 Determinant of a 2 2 Matrix 236 82 Determinant of an n n Matrix 242 83 Computing Determinants by Row Reductions 261 84 Minors and Cofactors 264 85 Inverse of a Matrix using Determinants 275 86 Cramer s Rule 280 87 Problems 286 9 Diagonalization 291 91 Eigenvalues and Eigenvectors 291 92 Diagonal Form of a Matrix 314 93 Applications 336 931 Matrix Powers 336 932 Differential Equations 338 94 Problems 343 10 Solutions to Problems 347 101 Chapter1 Solutions 347 iii

Contents 102 Chapter2 Solutions 351 103 Chapter3 Solutions 356 104 Chapter4 Solutions 364 105 Chapter5 Solutions 369 106 Chapter6 Solutions 372 107 Chapter7 Solutions 377 108 Chapter8 Solutions 382 109 Chapter9 Solutions 387 iv

List of Figures 11 Vectors in R 2 2 12 Vectors in R 3 3 13 Parallelogram law of vector addition 6 14 Triangle law of vector addition 7 15 Vector joining two points 7 16 Unique solution 12 17 Inconsistent system 13 18 Parametric equation of a line 15 19 Vector w is a linear combination of u and v 18 110 Example 131 explained by linear combinations 18 111 The plane x + 2y + 3z = 6 20 112 Planes intersecting in a line 21 113 Planes with no common points of intersection 21 71 Rotation by angle θ 216 72 Rotation of the basis vectors e 1 and e 2 217 73 Reflection across the line y = x 218 74 Reflection across the x-axis 219 75 Projection onto the x-axis 221 76 Reflection across an arbitrary line in the plane 228 v

1 Vectors, Lines and Planes: Linear Equations in Two and Three Variables 11 Vectors in R 2 and R 3 The coordinate plane R 2 is the set of all ordered pairs of real numbers (a, b) We have a preferred point (0,0) called the origin, and each coordinate pair (a, b) represents a unique point in the plane Similarly the threedimensional coordinate space R 3 is the set of all ordered triplets of real numbers (a, b, c) More generally the n-dimensional coordinate space R n is the set of all ordered n-tuples of real numbers (x 1, x 2,, x n ) Recall from your physics courses that vectors are objects with magnitude and direction Parallel vectors having the same length are equal A vector can be represented by an arrow pointing out from the origin A point (a, b) in R 2 corresponds to a vector represented by a directed line connecting the origin (0,0) to (a, b) We will write this vector as a column vector v = [ ] a, (see Figure 11) b 1

1 Vectors, Lines and Planes y (a, b) v = [ a b ] = ai + bj j tan 1 ( b a ) O i x Figure 11: Vectors in R 2 Definition 111 The length of a vector is also known as its magnitude or norm A vector of norm 1 is called a unit vector It follows from the Pythagorean theorem that the length or norm of the vector v = [ a] b is v = a 2 + b 2 Its direction makes an angle tan 1 ( b a) with the positive x-axis In physics books a vector v is often written as v = [ ] a = ai + bj, b where i and j are unit vectors in the directions of the positive x- and y-axes respectively With our notation i = [ ] 1 and j = 0 [ ] 0 1 2

1 Vectors, Lines and Planes The above observations can be generalized to the three-dimensional coordinate space R 3 The point (a, b, c) in R 3 corresponds to the vector a v = b = ai + bj + ck, c where the unit vectors in the directions of the positive x-, y- and z-axes are respectively given by (see Figure 12) 1 0 0 i = 0, j = 1, and k = 0 0 0 1 Figure 12: Vectors in R 3 Again the Pythagorean theorem gives the norm of v as v = a 2 + b 2 + c 2 Analogously, where n is any positive integer, we find that the point (a 1, a 2,, a n ) 3

1 Vectors, Lines and Planes in R n corresponds to the vector a 1 a 2 a n v = = a 1e 1 + a 2 e 2 + + a n e n, where the unit vectors in the directions of the positive x 1, x 2,, x n -axes are respectively given by 1 0 0 0 e 1 =, e 1 2 =,, e 0 n = 0 0 1 The Pythagorean formula for the norm of the vector v in R n is v = a 2 1 + a2 2 + + a2 n Coordinate spaces with the usual notion of distance between points (Pythagorean formula), and a notion of angles between vectors, are also known as Euclidean spaces In later chapters we will study the vector space R n in detail Here, we merely review some familiar facts about two- and threedimensional spaces 4

1 Vectors, Lines and Planes 12 Vector Addition and Scalar Multiplication Definition 121: Vector Addition Vectors in R 2 can be added componentwise to give a resultant vector in R 2 Thus if [ ] [ ] a c u = and v = b d are two vectors in R 2, then u + v is a vector in R 2 given by u + v = [ ] a + c b + d Similarly vectors can be added component wise in R 3 If a 1 b 1 u = a 2 and v = b 2 a 3 b 3 are two vectors in R 3, then u + v is a vector in R 3 given by a 1 + b 1 u + v = a 2 + b 2 a 3 + b 3 Remark 122 We write 0 for the zero vector in R n Note this is not the scalar 0 For example, if v is a vector in R 3, then 0 v v = 0 = 0 0 5

1 Vectors, Lines and Planes Remark 123: Parallelogram Law of Vector Addition Vector addition follows the Parallelogram Law which is illustrated in Figure 13 The sum of the two vectors represented by the two adjacent sides of the parallelogram is given by the diagonal y (a + c, b + d) (a, b) u u + v (c, d) v Figure 13: Parallelogram law of vector addition x Remark 124: Triangle Law of Vector Addition We also have the triangle law of addition of vectors This is illustrated in Figure 14 The sum of two vectors represented by two sides of a triangle in the same order is given by the third side of the triangle in the reverse order Remark 125 Let P and Q be two points in R n corresponding to the vectors u = OP and v = OQ, where O is the origin of the coordinate system Then the vector from P to Q is given by PQ = v u This follows from the 6

1 Vectors, Lines and Planes y (a + c, b + d) u + v u (c, d) v Figure 14: Triangle law of vector addition x triangle law of vector addition as illustrated in Figure 15 y P v u Q u v O x Figure 15: Vector joining two points 7

1 Vectors, Lines and Planes Example 126 The vector pointing from P(1, 2,4) to Q( 2,3,5) in R 3 is given by PQ = OQ OP = 2 3 5 1 3 = 5 1 2 4 Definition 127: Scalar Multiplication Vectors can also be multiplied componentwise by scalars (real numbers) For example, if c is any scalar (real number) and if v = [ a1 is a vector in R 2 then cv is a vector in R 2 given by a 2 ] [ ] ca1 cv = ca 2 Similarly if a 1 v = a 2 a 3 is a vector in R 3, then cv is a vector in R 3 given by ca 1 cv = ca 2 ca 3 The above definintions can be generalized to vectors in R n Vectors in R n can be added componentwise and they satisfy the parallelogram and 8

1 Vectors, Lines and Planes triangle law of addition Scalar multiplication of vectors in R n is also defined componentwise We will see detailed discussions of these operations in R n in the following chapters Remark 128 Two nonzero vectors u and v are parallel if and only if there is a scalar c 0 such that u = cv Remark 129 If a 1 v = a 2 a 3 is a vector in R 3 and if c is a scalar, we have cv = c 2 a 2 1 + c2 a 2 2 + c2 a 2 3 = c v Recall that c 2 = c The norm (length) of a vector is a nonnegative number Example 1210 Let 1 v = 2, so that 2 3 v = 2 2 3 4 3 4 3 9

1 Vectors, Lines and Planes The norms of these vectors are v = 1 2 + ( 2) 2 + 2 2 = 3 and ( 2 3 v = 2 ) 2 ( ) 4 2 ( + + 4 ) 2 3 3 3 = 2 = 2 3 v Observe that 2 3v is a vector whose length is two thirds that of v and which points in a direction opposite to that of v 13 Linear Equations in Two Variables A system of m linear equations in n variables (unknowns) x 1, x 2,, x n can be written as 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, 10

1 Vectors, Lines and Planes where the coefficients a i j are real numbers We say x 1 c 1 x 2 = c 2 x n c n is a solution vector of this system if and only if x 1 = c 1, x 2 = c 2,, x n = c n satisfies all m equations of this system simultaneously solution vector is a vector in R n Note that the Much of linear algebra is about the solutions of systems of linear equations, as shown in the following chapters Here we do a review of some familiar topics from linear equations in two or three variables A linear equation ax + by = c, in two variables x and y, represents a line in the xy plane (assuming that a and b not both zero) If b 0, the equation is that of a line of slope a b and its y-intercept is c b If b = 0, the line is parallel to the y axis with x intercept ( c a,0) We have always been told that we need two equations to solve for two unknowns Let us look at some examples to see how this really works Example 131: Unique Solution Consider the linear system of equations 2x 3y = 5 4x + y = 3 (131) 11

1 Vectors, Lines and Planes We solve equations by the method of elimination To eliminate x we multiply the first equation by 2 and add the resulting equation to the second We arrive at the system 2x 3y = 5 7y = 7 Consequently we can read off y = 1 from the second equation Substituting y = 1 in the first equation we get x = 1 We therefore have a unique solution vector [ ] x = y [ ] 1 1 for the system The two equations represent two lines with different slopes Hence they intersect at a unique point which in this case is (1, 1) This is illustrated in Figure 16 y (1, 1) x Figure 16: Unique solution 12

1 Vectors, Lines and Planes Example 132: Inconsistent System Let us now try to solve the system 2x 3y = 5 4x 6y = 3 (132) We eliminate x as in the last example by multiplying the first equation by 2 and adding the resulting equation to the second We arrive at the system 2x 3y = 5 0 = 7 The last equality is absurd Hence the system of equations has no solution We say that the system of equations is inconsistent The two equations represent two lines having the same slope They are parallel but they pass through different points Hence they do not intersect This is illustrated in Figure 17 y x Figure 17: Inconsistent system 13

1 Vectors, Lines and Planes Example 133: Infinite Solutions As a final example, let us consider the system 2x 3y = 5 4x 6y = 10 (133) We eliminate x as in the last two examples by multiplying the first equation by 2 and adding the resulting equation to the second We arrive at the system 2x 3y = 5 0 = 0 There is no inconsistency in this case We are left with a single equation because the second equation in Equation (133) is a multiple of the first The two equations represent the same line Any point (x, y) on the line is a solution to the system It is instructive to write the solutions in parametric form Observe that we can write x in terms of y using 2x 3y = 5 We call y a free variable We assign any real number t to be the value of y, then for each such t we have y = t = x = 3 2 t + 5 2 Thus all vectors of the form [ ] [ x 3 = 2 t + 5 ] 2, y t where t is any real number, are solutions of the system The system has infinitely many solutions 14

1 Vectors, Lines and Planes Using the rules of vector addition and multiplication by scalars we write the solutions in the form [ ] x 3 ] [ 5 ] = t[ 2 + 2 = t y 1 0 2 [ ] [ 3 5 ] + 2, 2 0 where t is any real number This is the parametric equation of the line 2x 3y = 5 Equivalently, writing s = t 2 we could express the parametric equation of the line in a simpler form as [ ] [ ] [ x 3 5 ] = s + 2, < s < y 2 0 Observe that as s runs through all real numbers, the vector s [ ] 3 2 gives all points on the line through the origin parallel to the vector [ ] 3 [ 2 Adding the 5 ] constant vector 2 simply gives a parallel shift to the line to the point ( 5 0 2,0) This is illustrated in Figure 18 y (3, 2) [ x y ] = s [ 3 2 ] [ x y ] = s [ 3 2 ] + [ ] 5/2 0 ( 5 2, 0) x Figure 18: Parametric equation of a line 15

1 Vectors, Lines and Planes Remark 134: Parametric equation of a line The last example led us to the parametric equation of a line If [ a b ] is a fixed vector in R 2, then [ ] [ ] x a = t y b is the parametric equation of a line through the origin (0,0), parallel to the vector [ a b ], where the parameter t can take any real value The parametric equation of a parallel line through some other point (c, d) is given by [ ] x = t y [ ] a + b [ ] c, < t < d A second interpretation Let us revisit our first example Refer to the system Equation (131) in Example 131 above We will be studying matrices in the next chapter Let us jump ahead a bit and write the system in matrix notation We write the system as [ ][ ] 2 3 x 4 1 y = [ ] 5 3 A rectangular array of numbers such as [ 2 3 4 1] is known as a matrix Since this matrix has 2 rows and 2 columns, we say that it is a 2 2 matrix Similarly, since the column vectors in R 2 have 2 rows and 1 column, they are 2 1 matrices The following is a simple consequence of the rules of matrix multiplication (which we will discuss shortly): [ ][ ] 2 3 x 4 1 y = [ ] 2x 3y 4x + y 16

1 Vectors, Lines and Planes In other words the matrix notation is merely stating that [ ] [ ] 2x 3y 5 = 4x + y 3 Comparing rows we get the system Equation (131) that we started with When we look at the system of equations as rows, we get lines in the plane R 2, which in this example intersect at the point (1, 1) Now let us look at the columns in the system Note that we can also write the system as [ ] [ ] [ ] 2 3 5 x + y = 4 1 3 For real numbers x and y an expression such as x [ ] [ ] 2 3 + y 4 1 is called a linear combination of the two column vectors [ 2 4 ] and [ 31 ] Hence a solution to the system of equations exists, if we can find real numbers x and y such that the linear combination of the two vectors on the left gives us exactly the vector [ 5 3 ] on the right Now if two vectors u and v in the plane R 2 are not parallel, then given any other vector w we can always complete a parallelogram with sides parallel to u and v and with w as the diagonal This is illustrated in Figure 19 In other words we can find unique scalars x and y such that xu + yv = w This is exactly the situation in the first example The vectors [ ] [ 2 4 and 31 ] are not parallel (because one is not a scalar multiple of the other) Hence we can find unique scalars x = 1, y = 1 such that 1 [ 2 4 ] + ( 1) [ 3 1 ] = [ 5 3 ], 17

1 Vectors, Lines and Planes w xu u v yv Figure 19: Vector w is a linear combination of u and v see Figure 110 (2, 4) (5, 3) ( 3, 1) (0, 0) (3, 1) Figure 110: Example 131 explained by linear combinations Now let us look at the other examples What goes wrong in our second example Example 132? In terms of column vectors the system Equation (132) can be written as x [ ] 2 + y 4 [ ] 3 = 6 [ ] 5 3 18

1 Vectors, Lines and Planes You can check that the column vectors on the left are parallel, in fact both point in the same direction as the vector [ 1 2 ] Expressing each vector on the left as a multiple of [ 1 2 ] we have (2x 3y) [ ] 1 = 2 [ ] 5 3 This is clearly impossible since [ 5 3 ] is not parallel to (is not a multiple of) the vector [ 1 2 ] Finally the system of equations Equation (133) which appears in the third example (Example 133) can be written as x [ ] [ ] [ ] 2 3 5 + y = 4 6 10 The column vectors on the left are parallel, but they are also parallel to the constant vector on the right We have (2x 3y) [ ] [ ] 1 1 = 5 2 2 Hence 2x 3y = 5 and there are an infinite number of solutions When the column vectors of the matrix of coefficients are parallel we say that the matrix is singular The first example involved a nonsingular matrix and we had unique solutions The next two examples had a singular matrix as the matrix of coefficients We had no solutions in one case and infinitely many soutions in the other Convince yourself that these are all the possibilities with intersecting lines in the plane R 2 19

1 Vectors, Lines and Planes 14 Linear Equations in Three Variables Let us now turn to systems of linear equations in three variables A linear equation of the form ax + by + cz = d represents a plane in R 3 For example, the xy, yz and zx-planes are respectively given by z = 0, x = 0 and y = 0 The plane parallel to the yz-plane and containing the point (2,0,0) is Similarly x = 2 x + 2y + 3z = 6 represents a plane intersecting the coordinate axes at the points (6,0,0), (0,3,0), (0,0,2) respectively, as shown in Figure 111 Solving equations in three variables Figure 111: The plane x + 2y + 3z = 6 involves the intersection of planes in R 3 In this case there are several possibilities 20

1 Vectors, Lines and Planes Three planes may intersect at a single point in R 3 For example the three coordinate planes x=0, y=0, z=0 intersect at the origin (0,0,0) Three planes can intersect along a line, as shown in Figure 112 Three planes may have no common points of intersection at all Two examples are illustrated in Figure 113 Figure 112: Planes intersecting in a line Figure 113: Planes with no common points of intersection 21

1 Vectors, Lines and Planes Let us look at some examples of systems of linear equations in three variables and see how the solutions are related to the geometry of intersecting planes Example 141: Unique Solutions Consider the system of equations x + y + 2z = 9 2x + y + z = 3 3x + 2y + z = 4 (141) In matrix notation 1 1 2 x 9 2 1 1 y = 3 3 2 1 z 4 We follow the same process of elimination as in the earlier examples involving systems of equations in two variables First we eliminate x from the second and third equations We multiply the first equation by 2 and add the resulting equation to the second This eliminates the variable x from the second equation We also multiply the first equation by 3 and add the result to the third to eliminate x from the third equation We now have the system x + y + 2z = 9 y 3z = 15 y 5z = 23 22

1 Vectors, Lines and Planes Now we multiply the second equation by 1 and add the result to the third to eliminate y and get x + y + 2z = 9 y 3z = 15 2z = 8 From the last row we read off z = 4 Substituting z = 4 in the second row we get y = 3 Finally substituting the values of z and y in the first row we get x = 2 Hence there is a unique solution vector x 2 y = z 3 4 The system of equations Equation (141) represents three planes intersecting at a single point As we did previously with equations in two variables, let us now look at the column picture for the system Equation (141) We rewrite the system as 1 1 2 9 x 2 + y 1 + z 1 = 3 3 2 1 4 Note that the vectors on the left are the three column vectors of the coefficient matrix The system has a solution precisely if the vector on the right can be written as a linear combination of the three vectors on the left In this example, there is a unique triple x = 2, y = 3, z = 4 23

1 Vectors, Lines and Planes such that 1 1 2 9 2 2 + 3 1 + 4 1 = 3 3 2 1 4 Example 142: Inconsistent System Consider the system of equations 2x y + z = 3 4x 3y + 3z = 7 2x + y z = 2 (142) Once again we use the method of elimination First we eliminate x from the second and third equations To do this we multiply the first equation by 2 and add the resulting equation to the second We then multiply the first equation by 1 and add the result to the third We now have 2x y + z = 3 y + z = 1 2y 2z = 1 We then eliminate y from the third equation by multiplying the second equation by 2 and adding the result to the third We have 2x y + z = 3 y + z = 1 0 = 1 24

1 Vectors, Lines and Planes The last row says 0 = 1 which is absurd The system is inconsistent and has no solutions The three planes represented by Equation (142) [ do not have 31 ] a common point of intersection Equivalently, the vector cannot be written as a linear combination of the columns of the coefficient matrix 1 Example 143: Infinite Solutions 1 Consider the system of equations 2x y + z = 3 4x 3y + 3z = 7 2x + y z = 1 (143) To use the method of elimination we multiply the first equation by 2 and add the resulting equation to the second We then multiply the first equation by 1 and add the result to the third We now have 2x y + z = 3 y + z = 1 2y 2z = 2 We then eliminate y from the third equation to get 2x y + z = 3 y + z = 1 0 = 0 (144) The system is consistent However the three equations in Equation (143) are reduced to an equivalent system Equation (144), that consists of two equations in three variables Observe that we can find y in terms of z from 25

1 Vectors, Lines and Planes the second equation As we did previously for the two variable case, we call z a free variable Set Then from the second equation z = t, where t is any real number y = t 1 Substituting z = t and y = t 1 in the first equation we get x = 1 The solution vector is x 1 y = t 1, z t where t is any real number It is instructive to write the solution in the form x 0 1 y = t 1 + 1 z 1 0 This, as we discussed earlier, is the parametric equation of a line As t takes 01 ] on all real values, t[ traces the line in R 3 that passes through the origin 1 [ 01 ] (0,0,0) and is parallel to the vector In fact it is the line y = z on the 1 [ 1 ] yz-plane (x = 0) Adding the constant vector 1 merely shifts the line to a 0 parallel line passing through the point (1, 1, 0) The three planes intersect along a line and we have an infinite number of solutions in this case 26

1 Vectors, Lines and Planes Remark 144: Parametric equation of a line in R 3 The previous example led us to the parametric equation of a line in R 3 If u is a given vector in R 3, then x y = tu, < t < z is the parametric equation of a line through the origin (0,0,0) parallel to the vector u The parametric equation of a parallel line through some other point (a, b, c) is given by x a y = tu + b z c Example 145: Infinite Solutions 2 Finally let us consider the system of equations 2x 4y + z = 3 4x + 8y 2z = 6 8x 16y + 4z = 12 (145) To use the method of elimination we multiply the first equation by 2 and add the resulting equation to the second We also multiply the first equation 27

1 Vectors, Lines and Planes by 4 and add the resulting equation to the third We have 2x 4y + z = 3 0 = 0 0 = 0 The system is consistent The three equations are multiples of each other They represent the same plane All points on the plane 2x 4y + z = 3 are solutions Observe that we can express x in terms of the variables y and z Consequently we now have two free variables y and z We let z = t and y = s, where s and t are arbitrary real numbers Then substituting y = s, z = t and solving for x we get x = 2s 1 2 t + 3 2 There are infinitely many solution vectors of the form x 2s 1 2 t + 3 2 y = s, < s, t < z t Let us rewrite the solution vectors in the form x 2 y = s 1 + t 3 1 2 0 + 0 2 z 0 2 0 28

1 Vectors, Lines and Planes Writing r = t 2 we can express the solutions in the simpler form The vectors [ 21 0 3 x 2 1 2 y = s 1 + r 0 + 0, < s, r < (146) z 0 2 0 ] and [ 10 2 ] 21 ] [ 10 ] are not parallel (why?) Hence s[ and r give two nonparallel lines, say L 1 and L 2, passing through the origin It follows from the parallelogram law that as s and r take all real values the linear combination 2 1 s + r 1 0 gives all points on the plane passing through the origin and containing the two nonparallel lines L 1 and L 2 (convince ] yourself of this by drawing a picture) Adding the constant vector simply shifts this plane to a parallel plane passing through the point ( 3 2,0,0) This is precisely the plane 0 2 [ 3 2 0 0 2x 4y + z = 3, and Equation (146) above is the parametric equation of this plane 0 2 Remark 146: Parametric equation of a plane The previous example led us to the parametric equation of a plane Let u and v be two nonzero vectors in R 3 that are not parallel to each other Let L 1 and L 2 be the nonparallel lines in R 3 given by 29

1 Vectors, Lines and Planes the parametric equations x y = su and z x y = tv z respectively, where s and t are real numbers Then x y = su + tv z is the parametric equation of a plane through the origin (0,0,0), containing the lines L 1 and L 2 The parametric equation of a parallel plane through some other point (a, b, c) is given by x a y = su + tv + b z c 30

1 Vectors, Lines and Planes 15 Problems 11 Find a vector of norm 5 in the direction of the vector from the point (1, 2,1) to the point (2, 1, 2) 12 Find all vectors of norm 4 parallel to the line x 2y = 3 13 Solve the system of equations 3x + 5y = 1 2x 3y = 7 Give a geometric interpretation of your answer 14 Solve the following system of equations 2x + y = 1 3x + 15y = 15 Give a geometric interpretation of your answer 15 Find vectors u and v such that the parametric equation of the line y = mx + b can be expressed as [ x y ] = tu + v 16 The parametric equation of a line in R 2 is given by [ ] x = t y [ ] 2 + 2 [ ] 3 1 Write the equation of the line in the form ax + by = c 17 Can the vector [ 1 2 ] be written as a linear combination of the vectors [ 2 3] and [ 46 ]? Explicitly write down all such linear combinations when they exist 18 Can the vector [ 1 2 ] be written as a linear combination of the vectors [ 1 1] and [ 23 ]? Explicitly write down all such linear combinations when they exist 31

1 Vectors, Lines and Planes 19 Can the vector [ 5 [ 5] be written as a linear combination of the vectors 1 [ 1] and 22 ]? Explicitly write down all such linear combinations when they exist 110 Consider the following system of equations 2x 3y = c 4x + 6y = 3 For what value(s) of c does the system have no solutions, a unique solution, infinitely many solutions? Give the solutions explicitly, when they exist 111 Consider the following system of equations x 2y = 1 2x + cy = 3 For what value(s) of c does the system have no solutions, a unique solution, infinitely many solutions? Give the solutions explicitly, when they exist 112 Let L 1 and L 2 be two lines in R 2 whose parametric equations are given by [ ] [ ] [ ] x 3 2 = t + y 2 0 and [ ] x = t y [ ] 1 + 1 [ ] 1 0 respectively Find the point of intersection of L 1 and L 2 or show that they do not intersect ] [ 1 ] 113 Let u = and v = a+b be two vectors in R 3 For what values of a [ 3 9 6 a b and b are the vectors u and v parallel? 32

1 Vectors, Lines and Planes 114 Solve the system of equations x y + z = 4 2x y z = 1 x + 3y + 2z = 2 Give a geometric interpretation of your result 115 Solve the system of equations x y + 3z = 4 2x y + 5z = 1 x + 3y z = 2 Give a geometric interpretation of your result 116 Solve the system of equations x y + 3z = 1 2x y + 5z = 3 x + 3y z = 5 Give a geometric interpretation of your result 117 Find a parametric equation for the line of intersection of the planes 2x y + z = 1 and x + y + z = 2 [ 20 ] [ 11 ] [ 13 ] 118 Write the vector as a linear combination of the vectors, [ 3 1 3 13 ] and 6 119 Solve the system of equations x 2y + 3z = 1 2x 4y + 6z = 2 5x 10y + 15z = 5 Give a geometric interpretation of your result 33

1 Vectors, Lines and Planes 120 Consider the system of equations x y z = 1 x + 2y + 4z = 2 x + 2y + 8z = c For what value(s) of c does the system have no solutions, a unique solution, infinitely many solutions? Give the solutions explicitly, when they exist 121 Let x 1 1 1 y = s 1 + t 2 + 2 z 1 3 0 be the parametric equation of a plane in R 3 Write the equation of the plane in the form ax + by + cz = d [ 122 Let L be the line in R 3 1 ] that is parallel to the vector 2 and passes 3 through the point (1,0, 2) Find the point of intersection (if any) of the line L with the plane 2x 5y z = 1 123 Find the parametric form of the equation of the plane 2x y + 2z = 1 124 The parametric equation of a plane P 1 is given by x 1 1 2 y = s 2 + t 0 + 0 z 0 2 0 The parametric equation of a second plane P 2 is given by x 1 1 1 y = s 1 + t 0 + 0 z 0 1 0 Find the parametric equation of the line of intersection of P 1 and P 2 Otherwise show that the two planes are parallel and do not intersect 34

1 Vectors, Lines and Planes 125 Find the equation of the plane passing through the points (1,1,0), (1,2,1) and ( 1, 1,2) in the form ax + by + cz = d 126 Find the parametric form of the equation of the plane passing through the points (1,3, 1), (2,2,3) and ( 1,1,1) 127 Do the points ( 3,5,3), (5, 7,7) and ( 1,2,4) lie on a line in R 3 If your answer is yes, then find the parametric equation of the line If your answer is no, then find the equation of the unique plane that passes through these three points 128 Let L 1 and L 2 be two lines in R 3 whose parametric equations are respectively given by x 2t + 2 y = at 3 and z 3t + 5 x 4t + 1 y = 3t 4 z bt + 8 For what values of a and b are the two lines parallel? 129 Let L 1 and L 2 be two lines in R 3 whose parametric equations are respectively given by x t + 2 y = t + 3 and z t + 1 x t 1 y = 2t + 3 z t 4 Are L 1 and L 2 parallel? Find the points of intersection of L 1 and L 2 if any Otherwise explain why they do not intersect 130 Find four distinct planes which intersect along the line given by the parametric equations x = t, y = 2t, z = 3t 35

2 Gaussian Elimination 21 Row echelon form of a matrix We are going to describe an algorithm (Gaussian Elimination) for solving a system of linear equations The algorithm is a systematic implementation of the method of elimination that we employed in Chapter 1 to solve systems of equations in two or three variables Our goal is to find the solution vectors for a system of m linear equations in n variables x 1, x 2,, x n given by 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, (211) where the coefficients a i j and b i, 1 i m,1 j n are scalars (real numbers) Using matrix notation, we write the above system as a 11 a 12 a 1n x 1 b 1 a 21 a 22 a 2n x 2 = b 2 a m1 a m2 a mn x n b m 36

2 Gaussian Elimination More succinctly we write Ax = b, where a 11 a 12 a 1n x 1 b 1 a 21 a 22 a 2n A =, x = x 2 and b = b 2 a m1 a m2 a mn x n b m Definition 211 Recall, from the previous chapter, that a m n matrix is a rectangular array of numbers having m rows and n columns The m n matrix A described above, is called the matrix of coefficients for the system of equations (211) The column vector x is a n 1 matrix and its entries are the n variables or unknowns The column vector b is a m 1 matrix and its entries are the constants that appear on the right The i j-th element a i j of a matrix A is the element whose position is at the i-th row and j-th column of the matrix A All the information about the system of equations (211) is contained in the matrix A and the column vector b We combine them to form the augmented matrix for the system of equations given by [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 (212) 37

2 Gaussian Elimination Observe that the augmented matrix is a m (n + 1) matrix, and that the i-th row in the augmented matrix corresponds to the i-th equation in the system of equations Equation (211) We draw a vertical bar before the last column to make it clear that the rightmost column consists of the constants that appear on the right of the system of equations When we solve a system of equations, the solution set remains unchanged if we switch two equations of the system we multiply an equation by a nonzero constant we take the multiple of one equation of the system and add the result to another equation Note that each of these operations on the equations give rise to a corresponding operation on the rows of the augmented matrix These are called the elementary row operations These are the operations that we are allowed to perform on the augmented matrix when we solve a system of equations We formalize this in the following definition Definition 212: Elementary Row Operations 1 Interchanging i-th and j-th rows of a matrix For the augmented matrix (212), this corresponds to interchanging the i-th equation with the j-th equation in the system (211) We will denote this operation in short by R i R j (R i represents the i-th row of the augmented matrix) 2 Multiplying a row by a nonzero scalar This corresponds to multiplying an equation of the system by a nonzero scalar We will denote this operation by R i cr i (The i-th row is multiplied by the constant c 0) 38

2 Gaussian Elimination 3 Adding a multiple of one row to another This corresponds to adding the multiple of one equation in the system to another we will denote this operation by R j R j + cr i (The i-th row is multiplied by c and the result is added to the j-th row We also need the following Definition 213: Row equivalence A matrix A is row equivalent to a matrix B, if B can be obtained from A by elementary row operations Remark 214 It is easy to show that if A is row equivalent to B then B is row equivalent to A Why? Also if A is row equivalent to B and B is row equivalent to C then A is row equivalent to C Why? We can restate our previous discussions in the following proposition Proposition 215 Let Ax = b and A x = b be two systems of linear equations If the augmented matrices [A b] and [A b ] are row equivalent, then the two systems Ax = b and A x = b have the same solution vectors We need one more definition before discussing an algorithm for solving linear systems 39

2 Gaussian Elimination Definition 216: Row echelon form A matrix A is in row echelon form if the following conditions are satisfied All rows consisting entirely of zeros (if any) appear at the bottom of A The first (from the left) non-zero entry of a non-zero row is called a pivot In two successive rows the pivot entry of the lower row occurs to the right of the pivot entry for the higher row It follows from the definition that for a row-echelon matrix, all entries in a pivot column, that appear below a pivot, must be 0 Example 217 Let 0 2 3 0 1 2 3 0 2 3 0 1 2 3 0 0 0 3 0 3 5 A = 0 0 0 0 5 7 0 and B = 0 0 0 3 0 3 5 0 0 0 5 7 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Then A is a row-echelon matrix by definition The matrix B is not in row-echelon form because the pivot in the third row is not to the right of the pivot in the second row All entries below the pivot 3 in the second row, must be 0 The word echelon comes from the French échelon, which literally means rungs of a ladder (échelle is the French word for ladder) The step like pattern in a row-echelon matrix is shown below The pivots are the entries 40

2 Gaussian Elimination a 0, b 0 and c 0 B = a 0 0 0 0 b 0 0 0 0 0 c 0 0 0 0 0 0 Any matrix can be transformed to its echelon form using elementary row operations The algorithm which achieves this is called Gaussian Elimination We describe the method with the help of an example Example 218: Gaussian Elimination We want to find an echelon matrix that is row equivalent to the matrix 0 0 0 2 1 4 0 2 2 4 2 6 A = 0 3 1 1 1 3 0 2 4 3 4 5 Step 1 Locate the first non-zero column and the first non-zero entry in that column This is a pivot Interchange rows to move the pivot to the top In our example the first non-zero entry in the first non-zero column is a 22 = 2 Here a i j refers to the element in the i-th row and j-th column of the matrix A So we switch the first row and the second row to move the pivot to the top row We write this step as 0 0 0 2 1 4 0 2 2 4 2 6 0 3 1 1 1 3 0 2 4 3 4 5 R 1 R 2 0 2 2 4 2 6 0 0 0 2 1 4 0 3 1 1 1 3 0 2 4 3 4 5 41

2 Gaussian Elimination Step 2 We normalize the pivot to 1 by multiplying the first row by 1 2 This step is not strictly necessary We do it to simplify the calculations 0 2 2 4 2 6 0 0 0 2 1 4 0 3 1 1 1 3 0 2 4 3 4 5 R 1 1 2 R 1 0 1 1 2 1 3 0 0 0 2 1 4 0 3 1 1 1 3 0 2 4 3 4 5 Step 3 We now add suitable multiples of the pivot row to the rows below it, so that all the elements in the pivot column appearing below the pivot are 0 In our example we multiply the first row by 3 and add it to the third We also multiply the first row by 2 and add it to the fourth row Thus 0 1 1 2 1 3 0 0 0 2 1 4 0 3 1 1 1 3 0 2 4 3 4 5 R 3 R 3 3R 1 R 4 R 4 +2R 1 0 1 1 2 1 3 0 0 0 2 1 4 0 0 2 7 2 6 0 0 6 1 6 11 Step 4 We repeat the above process ignoring the pivot rows on the top We keep repeating these steps till there are no more pivots to consider For our example we ignore the first row which is the pivot row from the last step In the submatrix that is left we find that the first non-zero entry in the first non-zero column is a 33 = 2 (the pivot) Following step 1, we bring this entry to the top by switching rows 2 and 3 Remember we are ignoring the first pivot row Thus we have 0 1 1 2 1 3 0 0 0 2 1 4 0 0 2 7 2 6 0 0 6 1 6 11 R 2 R 3 0 1 1 2 1 3 0 0 2 7 2 6 0 0 0 2 1 4 0 0 6 1 6 11 42

2 Gaussian Elimination We now add multiples of the second pivot row to rows below it so that all the elements below the pivot are 0 Here we get 0 1 1 2 1 3 0 0 2 7 2 6 0 0 0 2 1 4 0 0 6 1 6 11 R 4 R 4 +3R 2 0 1 1 2 1 3 0 0 2 7 2 6 0 0 0 2 1 4 0 0 0 20 0 7 We now ignore the first two pivot rows and repeat the process The first non-zero entry in the first non-zero column is (ignoring the top two rows) is a 34 = 2 This is a pivot We want zeros below the pivot in the third row So 0 1 1 2 1 3 0 0 2 7 2 6 0 0 0 2 1 4 0 0 0 20 0 7 R 4 R 4 +10R 3 0 1 1 2 1 3 0 0 2 7 2 6 0 0 0 2 1 4 0 0 0 0 10 33 We are done There are no more pivots left The resulting matrix on the right is in row echelon formthe matrix 0 0 0 2 1 4 0 2 2 4 2 6 A = 0 3 1 1 1 3 0 2 4 3 4 5 is row equivalent to the echelon matrix 0 1 1 2 1 3 0 0 2 7 2 6 B = 0 0 0 2 1 4 0 0 0 0 10 33 Here is another example 43

2 Gaussian Elimination Example 219 We want an echelon matrix that is row equivalent to the matrix 2 1 1 4 A = 1 3 1 1 3 1 2 0 We first switch the two rows R 1 and R 2, to avoid multiplication by the fraction 1 2 This is not strictly necessary Here are the steps 2 1 1 4 1 3 1 1 1 3 1 1 R1 R 2 2 1 1 4 3 1 2 0 3 1 2 0 1 3 1 1 R 2 R 2 2R 1 R 3 R 3 3R 1 0 7 1 6 0 8 5 3 1 3 1 1 R 3 R 3 8 7 R 2 0 7 1 6 0 0 27 7 27 7 1 3 1 1 R 3 7 27 R 3 0 7 1 6 0 0 1 1 The matrix in the final step is in row echelon form In fact the matrix in the penultimate step is also in row echelon form We perform the last row operation to simplify the entries in the third row 44

2 Gaussian Elimination Remark 2110 Note that the echelon form of a matrix is not unique The same matrix A can be row equivalent to different echelon matrices This is true even if we demand that each pivot element should be normalized and made equal to 1 For example Let A = [ ] 1 2 1 3 Then we can row reduce A as [ ] [ ] 1 2 1 2 R 2 R 2 R 1 1 3 0 1 Thus A is row equivalent to the echelon matrix Given any scalar c we have [ ] 1 2 0 1 A = [ ] 1 2 0 1 [ ] 1 2 + c R 1 R 1 +cr 2 0 1 For each c, the matrix on the right is an echelon matrix Consequently we have infinitely many echelon matrices that are row equivalent to A Hence, although one sometimes says "the" echelon form of a matrix A, one should remember that this is an abuse of language and that one is in fact talking about one member of an infinite family of echelon matrices that are row equivalent to A Later we will define a special echelon matrix that is row equivalent to a given matrix A This is called the reduced row echelon form of a matrix A The reduced echelon form is a "canonical form" in the sense that it is unique for a given matrix A 45

2 Gaussian Elimination We state the following without proof The proof follows from the uniqueness of the reduced row echelon form of a matrix The proof will also be clear from our discussions of the column space of a matrix in a later chapter Theorem 2111 Let A and B be two row echelon matrices which are row equivalent to each other Then the i-th column of A is a pivot column if and only if the i-th column of B is also a pivot column 22 Linear systems of equations How does row reduction of a matrix to its echelon form help solve linear systems? Let us look at the following example Example 221 Find the solution vectors for the following system of linear equations: 2x 1 x 2 + x 3 = 4 x 1 + 3x 2 + x 3 = 1 3x 1 + x 2 2x 3 = 0 (221) The system can be written in matrix form as 2 1 1 x 1 4 1 3 1 x 2 = 1 3 1 2 x 3 0 46

2 Gaussian Elimination The augmented matrix for the system is 2 1 1 4 1 3 1 1 3 1 2 0 This is precisely the matrix we examined in Example 219 By the Gaussian elimination algorithm (as shown in Example 219), we find that this augmented matrix is row equivalent to the echelon matrix 1 3 1 1 0 7 1 6 0 0 1 1 By Proposition 215, the system of equations, Equation (221), has the same solution vectors as the system corresponding to the echelon matrix above We use the echelon form to solve the system by back substitution Notice that the pivots (leading non-zero entries) occur in columns 1,2 and 3 The corresponding variables x 1, x 2, x 3 are called basic variables From row 3, we get x 3 = 1 From row 2 we have 7x 2 x 3 = 6 Back substituting the value of x 3 we get x 2 = 1 From row 1 x 1 + 3x 2 + x 3 = 1 47

2 Gaussian Elimination Back substituting the values of x 2 and x 3 we get x 1 = 1 The system has a unique solution vector given by x 1 1 x 2 = 1 x 3 1 Remark 222 Observe that the number of pivots is the same as the number of unknowns in the above example This is precisely when the system has a unique solution Here is another example Example 223 Solve the system of equations 7y + 7z = 3 3x 2y + z = 4 2x + y + 3z = 1 (222) We modify the above algorithm slightly, multiplying rows by suitable integers to avoid computations with fractions This makes the calculations somewhat less tedious when performing row reductions by hand We start with the augmented matrix for the system and row reduce it to an echelon 48

2 Gaussian Elimination matrix as follows 0 7 7 3 3 2 1 4 2 1 3 1 R1 R 2 R 1 2R 1 R 3 3R 3 R 3 R 3 R 1 R 3 R 3 R 2 3 2 1 4 0 7 7 3 2 1 3 1 6 4 2 8 0 7 7 3 6 3 9 3 6 4 2 8 0 7 7 3 0 7 7 5 6 4 2 8 0 7 7 3 0 0 0 8 We now have an echelon matrix that is row equivalent the initial augmented matrix The last row says 0 = 8, which is impossible The system is inconsistent and has no solutions Remark 224 Note that in the example above, the column of constants on the very right of the echelon form of the augmented matrix contains a pivot This is precisely what makes the system inconsistent Let us look at another example Example 225 We want to find the intersection of two planes x + 2y + 3z = 4 and 2x + 4y + 5z = 6 49

2 Gaussian Elimination in R 3 In other words, we have to solve the system of equations x + 2y + 3z = 4 2x + 4y + 5z = 6 (223) We start with the augmented matrix for the system and row reduce it to an echelon matrix as follows [ 1 2 3 4 2 4 5 6 ] R 2 R 2 2R 1 [ 1 2 3 4 0 0-1 2 We have an echelon matrix that is row equivalent to the initial augmented matrix We have enclosed the pivots in boxes The pivots are in the first and third columns corresponding to the variables x and z These are the so called basic variables The variables corresponding to the non-pivot columns are called free variables Thus y is a free variable A more detailed discussion of this is given below Here, let us focus on the solution set for this problem We assign arbitrary values to the free variables The basic variables can then be computed in terms of the values of the free variables Specifically, for the problem at hand we let From the last row we read off y = t, where t is any scalar z = 2 ] From the first row x + 2y + 3z = 4 Back substituting the values of y and z we get x = 2t 2 50

2 Gaussian Elimination The solution vectors are of the form x 2t 2 y = z t 2 We rewrite the solutions as x 2 2 y = t z 1 + 0 0 2 There are infinitely many solutions The solution vectors are given by the parametric equation of the line of intersection[ of the two planes in R 3 The 21 ] line of intersection is parallel to the vector and passes through the point ( 2,0,2) 0 Remark 226 Observe that the system of equations in the above example is consistent, and the number of pivots is less than the number of unknowns This is precisely when we have free variables and consequently an infinite number of solutions Definition 227: Free and basic variables We want to solve the system of linear equations Ax = b, 51

2 Gaussian Elimination see (211), which can be explicitly written as 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 (224) We have a system of m equations in n unknowns x 1, x 2,, x n The variable x i corresponds to the i-th column of the matrix A By Gaussian elimination we can row reduce the augmented matrix [A b] to an echelon matrix Let us denote the echelon form of the matrix [A b] by [A b ] We call the variables corresponding to the pivot columns of the echelon matrix A, basic variables The variables corresponding to columns without pivots are the free variables Note that from Theorem 2111, the free and basic variables are the same for any echelon form of the matrix A Remark 228 Using the same notation as in the above definition, let us denote an echelon form of the matrix [A b] by [A b ] Observe that if there are r pivots in A, there are r basic variables and n r free variables The number r is also known as the rank of the matrix A, written as rank(a) The rank is a unique nonnegative integer associated with the matrix A Clearly we have r = rank(a) n Also, since each row can have at most one pivot r = rank(a) m 52

2 Gaussian Elimination Now we turn to the following fundamental questions: When is the system Equation (224) inconsistent (ie does not have a solution)? If the system is consistent, how many solutions does it have? If r < m (the number of pivots is strictly less than the number of equations), then the last m r rows of A consist of zeros If the last m r entries of the column vector b are not zero then the system is inconsistent and there are no solutions Note that the system is inconsistent if the column of constants b, on the right of the echelon form of the augmented matrix, is a pivot column This was the case in example Example 223 above If r = m or if r < m and the last m r rows of the column vector b consist of zeros, then the system is consistent and always has solutions For a consistent system if r < n, then there are n r free variables We assign arbitrary values to the n r free variables and the r basic variables are computed in terms of the free variables by back substitution In such a case there are an infinite number of solutions This is illustrated by the example Example 225 above For a consistent system if r = n, then there are no free variables and the system has a unique solution This is what we had in the system of equations Equation (221) above We summarize our discussions in the following: Theorem 229: Existence and uniqueness of solutions Let Ax = b be a linear system of equations and let [A b ] be an echelon form of the augmented matrix [A b] 53

2 Gaussian Elimination The system is consistent if and only if the column b of the echelon matrix is not a pivot column Otherwise the system is inconsistent and has no solutions If the system is consistent and there are no free variables, then the system has a unique solution This happens when the number of pivots r = rank(a) = n, where n is the number of variables or the number of columns of A If the system is consistent and there are free variables, then it has infinitely many solutions This happens when the number of pivots r = rank(a) < n, where n is the number of variables or the number of columns of A Let us look at how all this works in the following examples Example 2210 Let us find the solutions of the system 3x 1 2x 2 + 8x 3 = 4 x 1 x 2 + 2x 3 + x 4 = 2 x 1 + 4x 3 x 4 = 1 We row reduce the augmented matrix for the system to its echelon form as 54

2 Gaussian Elimination follows 3 2 8 0 4 1 1 2 1 2 1 0 4 1 1 R1 R 2 R 2 R 2 3R 1 R 3 R 3 R 1 R 3 R 3 R 2 1 1 2 1 2 3 2 8 0 4 1 0 4 1 1 1 1 2 1 2 0 1 2 3 2 0 1 2 2 1 1 1 2 1 2 0 1 2 3 2 0 0 0 1 1 The final matrix on the right is in echelon form The pivots are enclosed in boxes The first, second and fourth columns are pivot columns Hence x 1, x 2 and x 4 are basic variables and x 3 is the only free variable We let x 3 = t, where t can take any real value From the third row we see that x 4 = 1 By back substituting the values of x 3 and x 4 into the second row, we get x 2 + 2t 3 = 2, so that x 2 = 1 2t Finally by back substitution again we see from the first row that x 1 = 2 4t 55

2 Gaussian Elimination There are infinitely many solutions and the solution vectors are of the form x 1 4 2 x 2 x 3 = t 2 1 + 1 0, 0 1 x 4 where t is any real number The solutions are given by a line in R 4 The number of pivots (or the rank of the matrix of coefficients) is r = 3 The number of variables or the number of columns of the coefficient matrix is n = 4 In this particular example we have a consistent system with r < n Hence we have infinitely many solutions Example 2211 Let us find the solutions of the system x 1 + 2x 2 + x 3 = 1 3x 1 + 6x 2 + 2x 3 x 4 = 4 x 1 + 2x 2 + 2x 3 + x 4 = 0 The augmented matrix of the system is 1 2 1 0 1 3 6 2 1 4 1 2 2 1 0 Using Gaussian elimination it is easy to check that the augmented matrix 56

2 Gaussian Elimination above is row equivalent to the echelon matrix 1 2 1 0 1 0 0 1 1 1 0 0 0 0 0 The pivots are once again enclosed in boxes The basic variables are x 1 and x 3 and they correspond to the pivot columns The variables x 2 and x 4 are free We let x 2 = s and x 4 = t, where s and t can take arbitrary real values The third row gives 0 = 0, so the system is consistent From the second row, substituting for x 2 and x 4 we get x 3 = 1 t Back substituting, from the first row we get x 1 = 2 2s + t There are infinitely many solutions which can be written in the form x 1 x 2 x 3 x 4 2 1 2 = s 1 0 + t 0 1 + 0 1 0 1 0 The solution set represents a plane in R 4 Why? As in the previous example, here too we have a consistent system The number of pivots (or rank of the matrix of coefficients) is r = 2 The rank r is strictly less than the number of unknowns n = 4 Hence there are infinitely many solutions Here is one more example 57

Contents. 1 Vectors, Lines and Planes 1. 2 Gaussian Elimination Matrices Vector Spaces and Subspaces 124