Section Exercise : We are given the following augumented matrix 3 7 6 3 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 7 5 6 3 Column 3 is fine, so it remains to add Row multiplied by to Row We obtain 3 5 6 3 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 3 6 3 6
Section Exercise : We have to reduce the matrix 3 5 7 3 5 7 9 5 7 9 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 : 3 5 7 8 8 6 3 Now (Step ) we ignore Row and Column The entry is the pivot now: 3 5 7 8 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: 3 5 8 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
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 5 8 3 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 3 3 6 5 8, 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
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 5 6 9 7 8 We bring A to reduced echelon form: 5 6 9 A 7 5 8 7 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 8 7 + 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
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 3 8 6 3 6 5/ / 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 + 3 + = in the left-hand side and 5 + + = 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
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
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