MATH Exam -Solutions pts Write your answers on separate paper. You do not need to copy the questions. Show your work!!!. ( pts) Find the reduced row echelon form of the matrix Solution : 4 4 6 4 4 R R R R R 4R R R R R R R +4R R R +4R R R R R R R R R R The reduced row echelon form of 4 4 6 4 4 4 4 6 4 4 4 4 8 4 8 4 4 4 6 4 6 4 4 4 is. 4. [. ( pts) Find all the values of h for which the matrix matrix of a consistent linear system. 4 h is the augmented Solution : We reduce the augmented matrix into an echelon form (we do not need the RREF).
[ 4 h R R +R h [ 4 The system is consistent if we do not have a row [ with. So the system is consistent if and only if h. The system is consistent if and only if h.. ( pts) Let a span{ a, a, a }?, a, a 6 and b 9. Is b in Solution : We have that b in span{ a, a, a } if and only if there exist x, x, x R with x a +x a +x a b. This is equivalent to the vector equation A x b having a solution where A [ a a a. So we reduce the augmented matrix to echelon form (again, we do not need to reduce to RREF). 6 9 R R R R R 4R 6 4 6 4 We see that this is the augmented matrix of a consistent system. So b in span{ a, a, a }. Note that the augmented matrix shows that the system has a unique solution. b in span{ a, a, a }. 4. ( pts) Describe the solutions of the following system in parametric vector form: { x 4x x + x 4 x x x + 4x 4 Solution : We reduce the augmented matrix to RREF.
[ 4 4 R R 4 [ 4 R R R [ 4 R R +R [ This is the augmented matrix of the system Solving for x and x, we get In parametric vector form, we get x x x 4 x x 4 x, x 4 : free x + x + x 4 x + x 4 x, x 4 : free x x x x 4 + x + x 4 x + x 4 x 4 + x + x 4 where x, x 4 R x + s + t where s, t R.. ( pts) Given the vectors v 4, v and v h h. (a) ( pts) Find all values of h for which v, v and v are linearly dependent. (b) ( pts) What is a linear dependence relation between v, v and v for those values of h? Solution : (a) We know that v, v and v are linearly dependent if and only if the vector equation A x has a non-trivial solution where A [ v v v. So we reduce the coefficient matrix to echelon form (again, we do not need to reduce to RREF at this
point). We can also work with the augmented matrix (but the last column is always the zero-column). h 4 h R R +R R R +4R R R +R h + 4 h + 8 h + 4 6 h In order to have a non-trivial solution, we must have that 6 h. So h 6. The vectors are linearly dependent if and only if h 6. (b) Let h 6. To find a linear dependence relation, we need to reduce the coefficient matrix to RREF. We get (remember that h 6) R R +R This corresponds to the homogeneous system x 4x x x x : free 4 Choosing a certain non-zero value for x, say x, we get that x 4 and x. This means that A linear dependence relation is 4 v + v + v. 6. ( pts) Let T : R R be the linear transformation with T ( e ) [ [ and T ( e ). [, T ( e ) (a) ( pts) Find the standard matrix for T. (b) ( pts) Find T ( v) where v. (c) ( pts) Is T onto? Justify your answer! (d) ( pts) Is T one-to-one? Justify your answer! 4
Solution : (a) Recall that the standard matrix for T is A [ T ( e ) T ( e ) T ( e ). So The standard matrix for T is A [. (b) There are basically two ways to find T ( v). T ( v) A v A [ [ + ( ) ( ) + + ( ) + [ Another method is to use linear combinations. Note that v e e + e. Since T is linear, we get [ [ [ [ T ( v) T ( e e + e ) T ( e ) T ( e )+T ( e ) + T ( v) [ (c) T is onto if the columns of A span R. So every row in A must have a pivot. We easily get that [ So A has two pivots. Hence T is onto. [ R R R T is onto. (d) T is one-to-one if the equation A x has no non-trivial solutions. Since A has two pivots but three columns, there will be a column that does not have a pivot. This corresponds to a free variable. So the equation A x has non-trivial solutions. Hence T is not one-to-one. T is not one-to-one.
[ 4. ( pts) Let A and B [ 4. Calculate A T B. Solution : We easily get that A T B [ 4 [ 4 [ 4 ( ) + 4 + 4 4 + ( ) ( ) + + 4 + ( ) [ 6 4 8 9 A T B [ 6 4 8 9 8. ( pts) Find the inverse (if it exists) of the matrix 8 4 Solution : To find the inverse of A, we find the RREF of the augmented matrix [ A I. 8 4 R R R R R R R R R R R R +R R R 8R R R R R R R Since the RREF of A is I, we know that A is invertible and. 8 8 8 8 8 6 6
A 9. ( pts) Let A be a matrix, B an m n matrix and C a 4 matrix. What are m and n if the product ABC exists? Solution : Since AB exists, we have that m. Since BC exists, we have that n.. ( pts)let A be a matrix and b R. (a) ( pts) Can the equation A x b have a unique solution? (b) ( pts) Give an example for A (in RREF) and b such that the equation A x b has no solutions. (c) ( pts) Give an example for A (in RREF) and b such that the equation A x b has infinitely many solutions. Solution : (a) The equation A x b can not have a unique solution. When we reduce the augmented matrix [ A b to RREF, there will be at least one column of A that does not have a pivot (since A has two rows and three columns). So either the vector equation A x b does not have a solution or there is a free variable, in which case there are infinitely many solutions. [ [ (b) A and b b where b. (c) A [ and b b [ b b.. ( pts) Let A be a 4 4 matrix such that the equation A x has only the trivial solution. (a) ( pts) Explain why the matrix A A T A is invertible. (b) ( pts) Find an expression for the inverse of A A T A in terms of the inverse of A. Solution : (a) Since the equation equation A x has only the trivial solution, we know that the columns of A are linearly independent. Since A is a 4 4 matrix, this implies that A is invertible (another way: since A x has only the trivial solution, there are no free variables. So every column in A has a pivot. Then every row in A has a pivot as well. So the RREF of A is I 4. Thus A is invertible). Since A is invertible, we have that A T is invertible. Since the product of invertible matrices is invertible, we get that AAA T A is invertible. So A A T A is invertible. (b) Recall that the inverse of a product is the product of the inverses in reverse order. So ( AAA T A ) A ( A T ) A A A ( A T ) ( A )
We know that ( A T ) (A ) T. Hence ( A A T A ) A ( A ) T ( A ) Extra Credit ( pts) Let T : R R be a linear transformation with T [. Find the standard matrix for T. ([ ) [ and T ([ ) Solution : To find the standard matrix for T, we need to know T [( e ) and T ( e ). [ We can find these values by writing e and e as linear combinations of u : and u :. For e : e x u + x u [ [ [ x + x So we have to solve a system of equations! [ [ R R R [ R R [ R R R [ [ x x So x and x. Hence e u + u. Since T is linear, we get [ [ [ 4 T ( e ) T ( u + u ) T ( u ) + T ( u ) + Similarly for e : e y u + y u [ [ [ y + y [ [ y y Again, we have to solve a system of equations (with the same coefficient matrix). [ [ R R R [ R R [ R R R [ [ 8
So y and y. Hence e u u. Since T is linear, we get [ [ [ T ( e ) T ( u u ) T ( u ) T ( u ) Hence the standard matrix for T is [ T ( e ) T ( e ) [ 4 Note that we could have combined the calculations for e and e. The result would have been the algorithm to find the inverse of a matrix! 9