Column 3 is fine, so it remains to add Row 2 multiplied by 2 to Row 1. We obtain

Transcription

1 Section Exercise : We are given the following augumented matrix We have to bring it to the diagonal form The entries below the diagonal are already zero, so we work from bottom to top Adding the fourth row with coefficients 3 and to the first and second rows, we obtain Column 3 is fine, so it remains to add Row multiplied by to Row We obtain Thus ( 3, 5, 6, 3) is the unique solution Exercise 6: The obvious augmented matrix with the unique solution (,, ) is Possible matrices can be obtained by applying one or a few row operation to the above matrix Here are two more examples: 3 6 and

2 Section Exercise : We have to reduce the matrix to reduced echelon form We take the first non-zero column which is Column and choose the entry for pilot Using row replacement, we create zeros below : Now (Step ) we ignore Row and Column The entry is the pivot now: The matrix is in echelon form now Its pivots are the entries,, and, located in Columns,, and respectively Now we start Step 5 First we multiply Row 3 by / and eliminate non-zero entries above this pivot: Next, we multiply the second row by / and reduce the entry above the pivot to zero: The matrix is in reduced echelon form now Exercise 8: We have to find the general solution, given the following augmented matrix: [ ] 7 7 We bring it first to echelon form by adding Row multiplied by to Row [ ] 7 Next, we bring the matrix to reduced echelon form We have to make leading entries to be equal to by multiplying the second row by and then eliminate the above it: [ ] 9

3 We see that the system has infinitely many solutions: x = 9, x =, x 3 is free Section 3 Exercise 8: The question, if rephrased, asks for which h the system of equations corresponding to the augmented matrix 3 h is consistent We convert the matrix to echelon form pivoting on the first column and then on the second column: 3 h 5 h 3 Hence, y is in the span of v and v if and only if h = 7/ 3 h 5 h + 7 Section Exercise 6: Let , b = We have find for which b the system Ax = b has a solution We start with augmented matrix and bring to echelon form: 3 b 3 6 b 5 8 b 3 3 b 7 6 3b + b 5b + b 3 b b b 3 3 b 7 6 3b + b (3b + b ) 5b + b 3 We can choose b, b, b 3 so that (3b +b ) 5b +b 3 = b +b +b 3 is non-zero (For example, b = b = and b 3 = ) So, the columns of A do not span R 3 The set of b for which the system Ax = b has a solution consists of those triples with b + b + b 3 = Exercise 3: It is intuitively clear that 3 vectors cannot span R but let argue using theorems from the course Let the vectors be a,a,a 3 R Let A be the 3-matrix [a,a,a 3 ] To test if these vectors span R we have to test whether the system Ax = b has a solution for every b R The latter happens if and only if A has pivots in every row But the number of pivots in A is at most 3 (the number of 3

4 columns), so we cannot have a pivot in every row Section 5 Exercise : We have to describe all solutions of Ax = in vector parametric form, where We bring A to reduced echelon form: A Thus x, x, x 5 are the free variables The general solution of Ax = has the form x = 5x 8x x 5, x 3 = 7x x 5 x 6 = We can rewrite it as x = x 5 + x x 5 = x u + x v + x 5 w Thus x = ru + sv + tw (where we replaced the free variables by parameters r, s, t) is the required parametric vector form Section 6 Exercise 8: We have to find coefficents in the following chemical reaction: (x ) KMnO + (x ) MnSO + (x 3 ) H O (x ) MnO + (x 5 ) K SO + (x 6 )H SO

5 Since there are 5 types of atoms, namely (K, Mn, O, S, H), we use vectors in R 5 We get x + x + x 3 = x + x 5 + x 6 This corresponds to a matrix equation Ax = with / / 3/ 5/ / Here x 6 is a free variable Since the last column involves division by, let us take x 6 = This gives us x = (, 3,, 5,, ) Let us make a check For example, of oxygen atoms we have = in the left-hand side and = on the right-hand side, which is reassuring Section 7 Exercise 8: We have to check whether the given 3 -matrix has linearly independent columns Actually, we can tell that the columns must be linearly independent for any matrix A with these dimensions Indeed, after we bring A to echelon form, there will be at most 3 pivots (at most one per row) So, the corresponding linear homogeneous system x = has free variables and, hence, non-trivial solutions Exercise : Suppose an m n matrix A has n pivot columns We have to explain why for each b the equation Ax = b has at most one solution 5

6 As no two pivots can be in one column, we have exactly one pivot in each column If we had two different solutions to Ax = b, say y and z, then we would have A(z y) = But the homogeneous system Ax = has no free variables and, therefore, cannot have any non-trivial solutions Section 8 Exercise : We have 5 a T(v) = Av = 5 b 5 c = a 5 + b 5 + c 5 = 5a 5b 5c For u = (,, ) we obtain T(u) = (5,, ) (Recall that (x,,x p ) is a shorhand notation for the vector whose components are x,,x p ) Exercise : Suppose vectors v,,v p span R n, and let T : R n R n be a linear transformation Suppose that T(v i ) = for i =,,p Show that T is the zero transfomation Solution: Take any x R n As v,,v p span R n, we have x = c v + +c p v p for some real numbers c,,c p (This representation may be not unique, but this does not affect our argument) By the properties of linear transformations, we have T(x) = T(c v + + c p v p ) = c T(v ) + + c p T(v p ) =, the last equality being true as T(v i ) = for every i Section 9 Exercise 8 & : Find the standard matrix of the tranformation T : R R that first reflects points through the horizontal x -axis and then reflects points through the line x = x Show that this tranformation is a rotation about the origin We have to find the images of the vectors e and e The above two steps act on e as follows: [ ] [ ] [ ] For e we have [ ] [ ] [ ] 6

7 Hence, the standard matrix is [ ] We see that T rotates each of e and e by π/ anticlockwise Since T and the π/-rotation are both linear transformations that coincide on basis vectors, they are the same (Alternatively, note that the matrix A is equal to the matrix of Example 3 in Section 9 for φ = π/) 7