Solutions Review # Math 7 Instructions: Use the following problems to study for Exam # which will be held Wednesday Sept For a set of nonzero vectors u v w} in R n use words and/or math expressions to complete the following statements (a) A linear combination of u v and w is any vector of the form ru + sv + tw for scalars r s t A vector b R n is in spanu v w} if if b is a linear combination of u v w ie if we can find scalars r s t so that (c) A vector b R n is not in spanu v w} if ru + sv + tw b if it is not a linear combination of u v w ie if ru + sv + tw b is impossible no matter which scalars r s t are used (d) The set u v w} is linearly dependent if if can be expressed as a nontrivial linear combination of u v w ie if we can find scalars r s t (not all zero) so that (e) The set u v w} is linearly independent if ru + sv + tw if cannot be written as a nontrivial linear combination of u v w ie if the only way to make is by taking u + v + w (a) Give two vectors in R that are linearly dependent Any nonzero vector u and any multiple of u will be linearly dependent for example [ [ u and v u 4 8 Give two vectors in R that are linearly independent for example Any nonzero vectors u v which are not multiples of each other will be independent u [ 4 and v [ 3 3 (c) Give three vectors in R that are linearly dependent Any triple of vectors will work since the maximum number of independent vectors in R is two Here is an example: [ [ [ 3 5 u v and w satisfy 7u + 6v 5w 4 3 (d) Give three vectors in R that are linearly independent This is impossible; the maximum number of linearly independent vectors in R n is n
(e) Give three vectors in R 3 that are linearly dependent One way is to make the third vector a linear combination of the first two for example u v and 3 w u + v (f) Give three vectors in R 3 that are linearly independent If you pick 3 random vectors chances are high that they are independent but you would have to check Here is a fairly simple but nontrivial example: 5 u v 3 and 4 w 3 If we have ru + sv + tw then 5r + s + 4t 3s + 3t t The third component guarantees that t ; plugging in t to the second component we see that s ; and plugging in s t into the first component we see that r Another way to see this is to make the augmented matrix 5 4 3 3 Since there are 3 pivots and 3 vectors there is a unique solution namely the solution with coefficients all equal to 3 Find all solutions (if any) to the following systems of equations x 3y 9 (a) 4x + 6y 8 The augmented system reduces to [ 9 R [ / 9/ 4 6 8 R + R o there are infinitely many solutions where y is a free variable; if we pick y t the solution set is x 9 + 3 t and y t By the way this line is the intersection of two lines so both original lines are actually the same line and the solution is just a parametrization of the line x + 5y 7 x + y [ [ 5 7 so there is a unique solution x and y This is the intersection point of the two lines in R
(c) x 3y x + y x 5y 4 3 5 4 (d) so there is a unique solution x 3 and y This is the intersection point of the three lines in R x y + 3z 5 3x + y 5z [ 3 5 3 5 [ 8 so there are infinitely many solutions Since z is a free variable we can set z t The equation x z 8 becomes x 8 + t and the equation y z becomes y + t So the solution set is x 8 + t y + t and z t This is the parametric equation of the line which is the intersection of the two planes in R 3 4 Find all solutions (if any) to the following linear combination problems If infinitely many solutions exist write down 3 of the linear combinations that work [ [ [ 4 3 (a) Determine if is a linear combination of and 8 4 4 The equation we want to solve is [ [ 3 x + x 4 4 [ 4 8 The associated augmented system eventually reduces to [ [ 3 4 5 4 4 8 3 so there is a unique solution x 5 and x 3 In other words there is exactly one linear combination that works: [ [ [ 3 4 5 + 3 4 4 8 Determine if 5 3 6 is a linear combination of and 4 The equation we want to solve is 3 5 x + x 6 4
(c) (d) The associated augmented system eventually reduces to 3 5 4 6 4 so there is a unique solution x 4 and x In other words there is exactly one linear combination that works: 3 5 4 6 4 Determine if is a linear combination of and 4 3 The equation we want to solve is x + x 3 4 The associated augmented system eventually reduces to 3 4 3 so there is no solution (since the last equation says x + x 3 which is impossible) In other words there is no linear combination that works so the vector on the right-hand side is not in the span of the vectors on the left [ [ [ Determine if is a linear combination of 3 3 is [ [ [ 5 x + x 3 + x The associated augmented system eventually reduces to [ 5 3 3 and [ 3 [ 3 [ 5 The equation we want to solve so there are infinitely many solutions Since x 3 is a free variable we can set x 3 t The equation x + x 3 3 becomes x 3 t and the equation x x 3 becomes x + t For example here are the linear combinations corresponding to t 3 t : 5 [ 3 [ t : 4 + 3 [ t : 3 + 3 [ t : + 4 3 [ t : + 6 3 [ t 3 : + 8 3 [ [ [ [ [ [ + + + + 3 [ [ 5 3 [ [ 5 3 [ [ 5 3 [ [ 5 3 [ [ 5 3 [ [ 5 3
5 Find all solutions (if any) to the following matrix equations 4 6 (a) 3 x 5 3 8 4 4 6 5 3 3 5 3 8 4 so there is no solution since the last row says that 4 x 3 4 3 so there are infinitely many solutions We have x 4 is a free variable so we set x 4 t The first equation x + x 4 becomes x t; the second equation x + x 4 becomes x t; the third equation x 3 x 4 becomes x 3 t In other words the solution set is x t t t t + t (c) You can check that 3 x 4 4 t t t t 3 3 4 (d) so there is no solution since the last row says that x 3 7 3 7
so there is a unique solution namely x 6 Consider the matrix equation [ 3 x [ 6 (a) Show that the equation has a unique solution and find that solution We reduce the augmented system [ 3 6 [ to see that the unique solution is x [ Write the corresponding system of equations graph the two lines and verify that your solution from (a) is the intersection point of the two lines The corresponding system of equation is 3x y x + y 6 We graph these two lines and see that ( ) is the unique intersection point 3x y x + y 6 (c) [ Write the corresponding linear combination problem and then use the graph of and the column 6 vectors of the matrix to verify that your solution from (a) gives the correct linear combination The corresponding linear combination problem and unique solution are [ [ [ [ [ [ 3 3 x + x + 6 6 We graph this linear combination below
[ [ 7 Consider the matrix equation x 6 (a) Show that the system has no solution We reduce the augmented system [ 6 [ 7 to see that there is no solution since the last row says that 7 Graph the lines of the corresponding system of equations How does this graph relate to the fact that there is no solution? The corresponding system is x + y 6x 3y which are the two lines shown below There is no solution exactly because the lines are parallel x + y 6x 3y (c) [ Graph the vector along with the column vectors of the matrix How can you interpret the fact that there is no solution in terms of linear combinations? The[ column vectors lie in the same line so every linear combination of them will lie on the line Since is not on the line it is impossible to get to it using the column vectors
8 Consider following vectors in R 3 a b 3 c d e 5 For each of the sets below determine if the set is linearly dependent or linearly independent If the set is linearly dependent give one explicit dependence relation between the vectors (a) a b} They are linearly independent since b is clearly not a multiple of a a d} They are linearly dependent since d is a multiple of a One dependence relation is a + d (c) a b c} To check for independence we reduce the matrix with a b c as columns: 3 R R 7 7 R 3 + R 3 R 3 + R 4 Since there are 3 pivots and 3 vectors a b c are linearly independent (d) a b e} To check for independence we reduce the matrix with a b e as columns: 3 5 R R R R R 3 + R R 3 + R Since there are only pivots they are linearly dependent To find a dependence relation we note that the system augmented by the zero vector gives the equations x + x 3 and x + x 3 If we choose x 3 we get x and x Therefore one dependence relation is a + b e ie + 3 5 In other words we can think for example of e as redundant because it can be expressed as e a+b