S09 MTH 37 Linear Algebra NEW PRACTICE QUIZ 4, SOLUTIONS Prof. G.Todorov February, 009 Please, justify your answers. 3 0. Let A = 0 3. 7 Determine whether the column vectors of A are dependent or independent. If they are independent, say why. If they are dependent, exhibit a linear dependence relation among them. 3 0 7 7 7 7 A = 0 3 0 3 0 3 0 3 0 /3 7 3 0 3 0 0 This is NOT rref A, however it is clear that there are only leading, so we can conclude that the columns of A are linearly dependent! In order to write a linear dependence relation, should find KerA, and therefore rrefa. 0 /3 0 /3 Leading variables: x, x. Non-leading variables: x 3 0 x + 0x + /3x 3 = 0 x + /3x 3 = 0 x = /3x 3 x = +4/3x 3 General solution for KerA: x x x 3 /3t /3 = 4/3t = 4/3 t. t 3 0 0 A linear dependence: ( /3) 0 + (4/3) 3 + () = 0 = 0. 7 0. For which value(s) of the constant k do the vectors below form a basis of R 4? 3 v = 0 0, v = 0, v 3 = 0, v 4 =. k 4 3 k These vectors form a basis for R 4 if rank rrefa = 4, where A = v v v 3 v 4. 3 3 A = 0 0 0 0 { v, v, v 3, v 4 } form a basis for R 4 for all k (/), k R. k 4 3 k 0 k +
3. Consider the liner transformation given by the multiplication by matrix A as A x. 3 4 0 The matrix A = 6 7 8 9 0 3 4 has row-reduced echelon form: rrefa = 3 4 0. 6 7 8 8 0 0 (a) Fill in R A R 4. (b) Find the image of A, ImA. ImA = span 6 6 (c) Find a basis for ImA., 7 7 (A basis for ImA) = 6 6 (d) Find the dimension of ImA. dimima =, 7 7 (e) Find the kernel of A, KerA. (!!!! You need to show your work here!!!!!) x s + t + 3u 3 x KerA = x 3 x 4 = s 3t 4u s t = 0 s + t + u, for all s, t, u R 0 x u 3 KerA = span 0,, 0 (f) Find a basis for KerA. 3 (A basis for KerA) = 0,, 0 (g) Find the dimension of KerA. dimkera = 3
4. Consider the x4 matrix A = v v v 3 v 4. We are told the vector 4, is in the kernel of A. Write v 4 as a linear combination of v, v, v 3. v + 4 v 3 v 3 + v 4 = 0 and therefore v 4 = (/) v (4/) v + (3/) v 3.. Let V be the subspace of R 3 defined by the equation 3x + x x 3 = 0. (a) Express V as the kernel of a matrix A. A = [ 3 ] /3 /3 V = KerA = span, 0 (!!!!!! Show work!!!!!!) 0 (b) Express V as the image of a matrix B. /3 /3 /3 /3 B = 0. V = ImB = span, 0 0 0 6. Let V be the subspace of R 3 defined by the equations x + x = 0 3x + x x 3 = 0. (a) Express V as the kernel of a matrix A. 0 A = V = KerA = span (!!!!!! Show your work!!!!!!) 3 (b) Express V as the image of a matrix B. B =. V = ImB = span 7. Find the set of all vectors x R 3 such that x = 0. x + x + x 3 = 0 ker [ ] = span, 0 (!!! And again, show your work!!!) 0 8. Find the set of all vectors x R 3 such that x = 0 and x = 0. 0 ker [ ] 0 = span (!!! And again, show your work!!!) 3
9. Examples (a) Give an example of a matrix A with dimkera = and dimima =. 0 0 0 0 0, or 0 30 Or make your own 8 0 interesting examples. 0 0 3 Change the size of the matrix. 3 4 (b) Give an example of a square matrix A with dimkera = and dimima =. 0 0 Make your own 0 0 interesting examples. Think how you can make many such matrices. 0 0 (c) Give an example of a matrix A with dimkera = 0 and dimima =. 0 0 0 0 0 0 0 0, 0 0, 0 0, 0 0, 0 0 (d) Give an example of a matrix A with dimkera = and dimima = 0.,,... What else can you make? (e) Give an example of a square matrix A with KerA = R. (f) Give an example of a square matrix A with ImA = R. 0 a b,,, ad bc 0. Anything else? 0 0 3 c d (g) Give an example of a square matrix A with ImA R 3 and dimima =. 0 0 0, 0, 0, 0 0, 3 4,, 4 6 (h) Give an example of an invertible matrix A.,.3,,,, ad bc 0, 0 3 0 0 0 0 0 a b 0 c d (i) Give an example of a non-invertible matrix A., 0,, All examples from (g). Your examples? (j) Give an example of a linear transformation R R 3., 4 Make your own interesting examples. Think how you can make many such.
0. True-False T - F If a 4 4 matrix A has ranka = 3 then dimkera = 3. F T - F If a vector v is in KerA, then the vector v is in KerA. T T - F T - F Consider a system of equations in variables. let A be the matrix of coefficients. If the system has many solutions, then dimima = F Consider a system of equations in variables. let A be the matrix of coefficients. If the system has exactly one solution, then dimkera =. F T - F If a 3 3 matrix A has dimima = then dimkera =. T T - F If a 3 matrix A has dimima = then dimkera =. T. Always-Sometimes-Never (A - S - N) If A is 3 3 matrix, then dimima 3. Always (A - S - N) If A is 4 4 matrix, then dimima 3. Sometimes (A - S - N) If A is matrix, then dimima = 3. Sometimes (A - S - N) If A is matrix, then dimima = 6. Never (A - S - N) If A is matrix, then dimima = 6. Never (A - S - N) If A is matrix, then dimima + dimkera =. Never (A - S - N) If a system A x = 0 has a non-leading variable, then KerA has dimension at least. Always (A - S - N) Let A x = 0. Then #{leading variables} + #{non-leading variables}=#{columns of A}. Always (A - S - N) If multiplication by a matrix A, as A x, defines a linear transformation R R 3 then dimima = #{leading } in the rrefa. Always (A - S - N) If multiplication by a matrix A, as A x, defines a linear transformation R R 3 then dimima = 3. Never (A - S - N) If multiplication by a matrix A, as A x, defines a linear transformation R R 3 then dimima =. Sometimes (A - S - N) If multiplication by a matrix A, as A x, defines a linear transformation R R 3 then dimima. Always