MATH 152 Exam 1-Solutions 135 pts. Write your answers on separate paper. You do not need to copy the questions. Show your work!!!

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1 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 : 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 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).

2 [ 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 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.

3 [ 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

4 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 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

5 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.

6 [ 4. ( pts) Let A and B [ 4. Calculate A T B. Solution : We easily get that A T B [ 4 [ 4 [ 4 ( ) ( ) ( ) ( ) [ A T B [ ( 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

7 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 )

8 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

9 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