CHAPTER 5 SYSTEMS OF EQUATIONS. x y

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1 page 1 of Section 5.1 CHAPTER 5 SYSTEMS OF EQUATIONS SECTION 5.1 GAUSSIAN ELIMINATION matrix form of a system of equations The system 2x + 3y + 4z 1 5x + y + 7z 2 can be written as Ax where b A [ ] 5 7 The system is abbreviated by writing, x x y z, b [ 1 ] 2 (1) The matrix A is called the coefficient matrix. The 2 4 matrix in (1) is called the augmented matrix and is denoted A b. Gaussian elimination Row ops on A b amount to interchanging two equations or multiplying an equation by a nonzero constant or adding a multiple of one equation to another. They do not change the solution so they may be used to simplify the system. In particular, performing row ops on A b until A is in echelon form is called Gaussian elimination. There are two possibilities (Fig 1). 1. The row ops produce a row of the form (2) nonzero Then the system has no solution and is called inconsistent. For example, if a system row ops to then it has no solutions because the third row is the equation x 4 which is impossible to satisfy. 2. The row ops do not produce a row of the form nonzero. Then the system has a solution (at least one) and is called consistent. There are two subcases (see the summary in the box below). 2a. All the echelon cols of A have pivots. Then there is exactly one solution. For example if the row ops produce and the variables are named x,y,z then the solution is x2, y5, z.

2 page 2 of Section 5.1 2b. Not all the echelon cols of A have pivots. Then there are infinitely many solutions. It is always possible to solve for some of the variables in terms of the others. Those others are called free variables or parameters. One way to do this is to choose the variables corresponding to the cols without pivots to be free. If the system Ax b has n variables (so that A has n cols) and rank A r (so that there are r cols with pivots in the echelon form of A) then the solution has n r free variables. For example, if the row ops produce and the variables are named x,y,z,w then one way to write the solution is x 5-3w - 2y z - 4w (w, y free) Another way to write the sol is x 5-3s - 2t y t z - 4s w s A x b Has an echelon row of the form... nonzero INCONSISTENT NO SOLUTIONS No echelon row of the form... nonzero CONSISTENT All echelon cols have pivots ONE SOLUTION Has an echelon col without a pivot INFINITELY MANY SOLS # parameters # cols w/o pivots n-r where n # variables r rank A example 1 Solve x 4 + 4x x 4-2x x x 5 solution Begin with and do row ops

3 page 3 of Section 5.1 to get R1 1 2 R1 R2-3R 1 + R2 R3-9R1 + R3 R2 1 4 R2 R1 R2 + R1 R3-1R2 + R3 R3 1 4 R3 R2 2R3 + R / /4 Choose and x 4, the variables corresponding to the cols without pivots, to be the free variables (other choices are possible but I like these) and solve for x, x, x in terms of them. A final sol is x x 4-2 (x, 2 x 4 free) Another version of the sol is x x 4 t s 1-4t - 2s example 2 Suppose A is 4 3 and Ax b has infinitely many solutions. Consider the new system Ax c. Will it have infinitely many solutions also; i.e., what happens if you change the righthand side of the system. solution Since Ax b has infinitely many sols, the echelon form of A must have at least one col without a pivot and A b must row op into something like this: or or warning I'm not saying that the original system looks like this. But A b must row operate into something like this. Now look at Ax c, same A but different righthand side. Can't have just one sol because the echelon form of A has at least one col without a pivot. Can have infinitely many sols or no solutions: If A b row opped to

4 page 4 of Section 5.1 it's possible for A c to row op to say and have no solutions, and it's also possible for A c to row op to π e and have infinitely many sols. So if Ax b has infinitely many sols then Ax c has either infinitely many sols or no solutions (but can't have just one solution). solving with unreduced echelon form and back substitution (much more efficient) Row operate on the system so that the coeff matrix is in unreduced echelon form (upper triangular form). Then starting with the last row, solve for the first variable in each row and back substitute as you go along. For example, if the row operations produce then x 5 3 from the third row - 8x 5 - x 4 4 from second row - 8(3) - x 4 4 back substitute x x 5-5x from first row 2-7(3) - 5x 4-3( -18-x 4 4 ) - 2 back substitute x 4 The solution can also be written as x 5 3 x 4 t t 4 s s t

5 page 5 of Section 5.1 PROBLEMS FOR SECTION Here are some systems of equations, already in echelon form. Let the variables be named x,x, Solve and identify the free variables. Then express the solution using parameters r,s,t,... (a) (b) (c) (d) 1 (e) Let u (1,,1,1) v (,1,1,) w (1,1,2,1) y (2,-1,1,2) Decide if y is in the subspace spanned by u,v,w using (a) ideas from Section 3.1 (b) ideas from this section If it is, express y as a combination of u,v,w. 3. Look at the system x - 2y + 3z 1 2x + ky + z -x + 3y + (k-3)z For what values of k will it have (a) no solutions (b) one solution (c) infinitely many solutions 4. True or False. If False, what would the correct conclusion be. (a) If there are fewer equations than unknowns then the unknowns are "underdetermined" and there will be infinitely many sols. (b) If there are more equations than unknowns then the unknowns are "overdetermined" and there are no solutions. (c) If there are the same number of unknowns as equations then everything is hunky dory and there will be exactly one solution. 5. Suppose A is 5 7 and the 2 2 subdet in the northeast corner is nonzero but all the 3 3 subdets are. What can you conclude about the number of solutions and number of free variables for the system Ax b.. Solve using row ops and back substitution (a) (b) 2x + 8y + 5z + 2u- v 8 4y + z + 3u + 3v 9 2x + 12y + 11z + 5u - 3v 17 4y + z + 3u + v 9 (c) 2x + y + z 1 4x + y -2-2x + 2y + z 7

6 page of Section A system of equations in unknowns x,y,z,w row ops to Solve and, if possible, make the free variables (a) w and y (b) x and w (c) y and z (d) x and y 8. Suppose A is 4 3. If Ax solutions to Ax c. b has one solution, what can you conclude about 9. What can you conclude about sols to Ax (a) A is 3 5 with rank 3 (b) A is 3 5 with rank 2 (c) A is 5 3 with rank 3 b if 1. Let M For what vectors b is Mx b consistent. In that consistent case, solve. 11. You have a fixed 3 4 matrix A (3 rows, 4 columns) with rank 3. You're looking for a right inverse for A; i.e., you're looking for a matrix B y 1 z 1 y 2 z 2 y 3 z 3 x 4 y 4 z 4 so that AB I. Can you find such a B. If so, how many. Suggestion: Think about systems of equations and remember that A is fixed and B is filled with unknowns.

7 page 1 of Section 5.2 SECTION 5.2 HOMOGENEOUS SYSTEMS homog systems and the null space of a matrix A homog system is one of the form Mx. It is always consistent since it at least has the (trivial) solution,,..., x n. A homog system will always have either one solution (the trivial one) or infinitely many sols (the trivial one plus infinitely many others). Furthermore, the set of solutions to Mx of M. is a subspace called the null space If M has n cols (so that the system has n variables) and rank M r then the null space is an (n r)-dim subspace of R n ; the dimension of the null space is the number of free variables. The dimension of the null space of M is called the nullity of M. At worst, the null space is a dim subspace containing only. Here's an example to illustrate why the sols to a homog system are a subspace and to show how to find a basis for the null space. Suppose M Then row ops to x (1) x 5 + 3x 4-5x 5-2x The solution can also be written as -5r - 2s + 7t t r + 3s x 4 s x 5 r x In vector notation, the solution is (2) x 4 x 5 x r s t 7 1 u v w

8 page 2 of Section 5.2 The set of sols is the set of all combinations of u,v,w so it is a subspace of R. Furthermore u,v,w are ind (look at their 2nd, 4th and 5th components to see that no one can be a combination of the other). So the sols are a 3 dim subspace with basis u,v,w. The dimension of the subspace matches the number of free variables. Now that you know the sols are a subspace, here's another way to extract a basis from the solution in (1) without rewriting it as (2). Assign values to the three free variables so as to get 3 ind solutions. The easiest way to do this is to let 1, x 4, x 5 to get solution (7,1,,,,); then let, x 4 1, x 5 to get solution (-2,,3,1,,); then let, x 4, x 5 1 to get solution (-5,,,,1,). homogeneous versus non-homogeneous The set of solutions to a homogeneous system of equations is a subspace but the set of solutions to a non homogeneous system of equations can never be a subspace. For one thing it never contains because a non homogeneous system never has the solution x, x,..., x. 1 2 n example 1 Suppose the system Mx row ops to Then x 4 (3) -3x 4-2 The sols are a 2 dim subspace of R 4 since there are 2 free variables. To pick a basis let x 1, x in (3) to get solution 4 2 u (-3,,-4,1); and set x 4, 1 in (3) to get solution v (-2,1,,). Then u and v are two ind solutions and are a basis for the subspace of solutions. summary Suppose A is 8 1 (8 rows and 1 columns) with rank r. The row space of A is an r dim subspace of R 1. The col space of A is an r dim subspace of R 8. The null space of A (the set of sols to Ax ) is a (1-r) dim subspace of R 1. how to show that a set of vectors is a subspace Section 2.5 gave two ways to show that a set of vectors is a subspace: method 1 Show that it's closed under addition and scalar mult. method 2 Show that the set is composed of all combinations of a bunch of vectors. Now you have a third way. method 3 If the set of vectors is the solution of a homogeneous system of equations then it's a subspace. example 2

9 page 3 of Section 5.2 Look at the set of vectors of the form (a,b,c,-a). In example 3 of Section 2.5, I used two methods to show that the set is a subspace. And I found a basis for the subspace. Here's a third way. The set consists of all vectors (x,x,x,x ) where x -x. So the set is the solution of the homogeneous system x 4 (one equation, four unknowns) So the set is a subspace. The solution to the system is x 4 - ;,, free. There are 3 free variables so the subspace of sols is 3 dim. Here's how to get a basis for the subspace: Let 1,, to get sol (1,,,-1). Let, 1, to get sol (,1,,). Let,, 1 to get sol (,,1,). A basis is (1,,,-1), (,1,,), (,,1,). PROBLEMS FOR SECTION Write the solutions in parametric form and find a basis for the subspace of solutions. (a) (b) x + 2y + z x + 3y + 2z (c) 2x + y + z 4x + y -2x + 2y + z 2. Find a basis for the null space of M (a) M (b) M (c) M [1 1 2] 3. Find an orthogonal basis for the subspace of solutions to x + 2y + z - w 2x + 5y + 3z - w 4. Describe the set of solutions geometrically. (d) (a) Solve Ax b (b) (e) (c) where A is the 3 5 zero matrix and

10 page 4 of Section 5.2 (a) b (b) b. Suppose A has six cols, the cols are dep but the last five cols are ind. What can you conclude about the number of solutions (and the number of free variables) to Ax b where (a) b (b) b 7. Let u (1,,2), v (1,1,4). Find all the vectors orthog to both u and v (a) using geometry and calculus (b) using the ideas of this section 8. Let u (1,1,4,-2). Find as many independent vectors as possible orthogonal to u. 9. Use ideas from this section to show that the set of points is a subspace and find a basis. (a) The set of points of the form (x,..., x ) where x -x + x and x 2x (b) The set of points of the form (a,a,2a,2a,b) 1. Let A and B be fixed 1 7 matrices. Look at the set of column vectors x in R 7 such that Ax Bx, i.e., the set of vectors x such that A and B "do the same thing to them". Use ideas of this section to show that this set is a subspace. 11. Solve the system of equations 2 3 basis for the subspace of solutions. (call the unknowns x and y) and find a

11 page 1 of Section 5.3 SECTION 5.3 ORTHOGONAL COMPLEMENTS definition of the orthogonal complement of a subspace Suppose V is a subspace of R n. The set of all vectors orthog to every vector in V is called the orthogonal complement of V and denoted by V perp or V Ú. In other words, u is in V Ú iff u is orthog to every vector in V. For example, in R 3, a plane through the origin is a 2 dim subspace; its orthogonal complement is the line through the origin perpendicular to the plane (Fig 1). The line consists precisely of those points u such that the u is perp to every vector in the plane. point on the line V perp point in the plane V point in the plane FIG 1 all about the orthog complement Let V be a subspace of R n. (1) V Ú is also a subspace of R n. (2) The sum of the dimensions of V and V Ú is n (e.g., if V is a dim sub space of R 3 then V Ú is a 24 dim subspace of R 3 ). (3) Not only does V Ú contain everything orthog to all the vectors in V but it can be shown that V in turn contains everything orthog to all the vectors in V Ú so that the spaces V and V Ú are orthog complements of each other. For example in Fig 1, the line is the orthog complement of the plane and the plane is the orthogonal complement of the line. (4) Suppose V is spanned by vectors u,v,w,p. To find V Ú, let A be the matrix with rows u,v,w,p. Then the set of solutions to Ax (the null space of A) is V Ú. (5) The null space and the row space of a matrix are orthog complements. proof of (4) Suppose V is spanned by u,v,w,p. To find V Ú, find all x orthog to the spanning vectors by solving ( ) u x, v x, w x, p x. Once x is orthog to u,v,w,p then x is orthog to every vector in V (see problem # in Section 2.5). So the solution to ( ) is the orthog complement of V. The system in ( ) can be written as Ax where the rows of A are u,v,w,p, and x

12 page 2 of Section 5.3 is written as a column So to find V Ú, solve Ax, i.e., find the null space of A. proof of (5) This follows from (4) which shows that if a matrix A has rows u,v,w,p then the null space of the matrix is the orthog complement of the space spanned by u,v,w,p. proof of (1) and (2) Suppose V is a -dim subspace of R 3. Let A be a marix whose rows are a basis for V (so that V is the row space of A). A is 3. By (4), V Ú is the null space of A which makes it a subspace, proving (1). Furthermore, rank A number of cols in A 3 dim of null space n-r 24 So dim V Ú is 24, proving (2). proof of (3) omitted example 1 Let (subtle) u (,1,,1,) v (,,1,,2) Let V be the space spanned by u,v. Find a basis for V Ú solution Let A be the matrix with rows u,v. Solve the system Ax : It's already in echelon form. The solution is -2x 5-4x 4 To find a basis for V Ú, set 1, x 4, x 5 to get p (1,,,,); set, x 4 1, x 5 to get q (,-1,,1,); set, x 4, x 5 1 to get r (,,-2,,1) A basis for V Ú is p,q,r. Note that V and V Ú are subspaces of R 5 so, by (2), their dimensions should add up to 5. And they do: dim V 2 (since u and v are ind) and dim V Ú 3. how many perps Suppose u,v,w are independent vectors in R 3. You can get 27 independent vectors orthogonal to u,v,w, but no more than that. You can get 28 independent vectors orthogonal to u and v, but no more than that. You can get 29 independent vectors orthogonal to u, but no more than that.

13 page 3 of Section 5.3 proof The number of independent vectors orthog to u,v,w is the same as the dimension of the orthog complement of the subspace spanned by u,v,w. The subspace spanned by u,v,w has dimension 3 (since u,v,w are ind). By (2), the orthog complement has dimension 27. So that's the most independent vectors you can get which are orthog to u,v and w. Similarly, the number of independent vectors orthog to u is the same as the dimension of the orthog complement of the 1 dim subspace spanned by u. By (2), the orthog complement has dimension 29. mathematical catechism question What does it mean to say that U is the orthogonal complement of a subspace W of R 8. answer It means that U is the set of all vectors in R 8 that are orthogonal to every vector in W. PROBLEMS FOR SECTION Find a basis for the orthog complement of the subspace spanned by p (1,1,,), q (,1,,1). 2. Find all the vectors perp to p (1,,,,,1), q (1,1,,,,), r (,,1,1,,). 3. Let A Find the dimension of and a basis for (a) the row space of A (b) the col space of A (c) the null space of A (d) the orthogonal complement of the null space of A 4. Let A (a) Find a basis for the orthogonal complement of the column space. (b) Continue from part (a) and find an orthogonal basis. 5. Do you remember from calculus that the graph in 3 space of 2x + 3y + 4z is a plane through the origin with normal vector n 2i + 3j + 4k. Furthermore, the only vectors perp to the plane are n and multiples of n. Here's the 5 dimensional version of this idea. Look at the set of points (a "hyperplane") satisfying the equation ( ) x 4 + x 5. A vector is called orthog to the hyperplane if it is orthog to every vector in the hyperplane. Let n (2,3,5,-7,1) Use the suff from this section to show that n is a orthog to the hyperplane and furthermore the only vectors orthog to the hyperplane are n and multiples of n.

14 page 4 of Section 5.3. Let L 1 and L 2 be perpendicular lines through the origin in space. They are subspaces since a line through the origin in R 3 is a subspace. Are they orthogonal complements. 7. Let W be a subspace of R n. Let w be in W. Can you decide if w and u are orthogonal if (a) u is the orthogonal complement of W (b) u is not in the orthogonal complement of W 8. Suppose (3,1,2) is in the row space of M. Is it possible for (2,1,1) to be in the null space. 8. No. the row space of a matrix and its null space are orthogonal complements. so a vector in the null space has to be orthog to everything in the row space. But (2,1,1) is not orthog to (3,1,2).

15 page 1 of Section 5.4 SECTION 5.4 SQUARE SYSTEMS Throughout this section, matrices are square, and all systems of equations have the same number of equations as unknowns. number of solutions to Mx b (a) If M is invertible then Mx b (b) If M is not invertible then Mx solutions. proof of (b) has one solution, namely x b M -1 b. has either no solutions or infinitely many Suppose M is 3 3 and is not invertible. Then M row ops into something like or 1 3 etc. And M b row ops to something like in which case there are no solutions or it row ops to something like in which case there are infinitely many solutions (because there is a free variable corresponding to the column without a pivot). You can't get exactly one solution because when the system is consistent there will always be at least one free variable. number of solutions to Mx This is a special case of (a) and (b) above. Remember that a homogeneous system always has at least the trivial solution x (aa) If M is invertible then Mx (bb) If M is not invertible then Mx the trivial solution. has only the trivial solution x I'm going to add this to the invertible rule.. has more (infinitely many more) than just. invertible rule. Let M be n n. The following are equivalent; i.e., either all are true or all are false. (1) M is invertible (nonsingular). (2) M. (3) Echelon form of M is I. (4) Rows of M are independent. (5) Cols of M are independent. () Rank of M is n. (7) Mx has only the trivial solution x. In other words, the null space of M contains only. In other words, if x then Mx

16 page 2 of Section 5.4 Cramer's rule Look at Mx b where M is square. (1) If M then either the system is inconsistent or it has infinitely many solutions (this is just a restatement of (b) from above). (2) If M then the system has one solution (this is a restatement of (a)). Furthermore, in case (2), the solution is x ( ) etc. proof of the formula in ( ) replace col 1 of M with b det replace col 2 of M with b det M M M -1 b and then take det and then take det which boils down to Here's where the formulas in ( ) come from for a 3 3 matrix M. If M, the sol is x M -1 b. Use the adjoint method for finding M -1 to get 1 M cof of m 11 cof of m 12 cof of m 13 cof of m 21 cof of m 22 cof of m 23 cof of m 31 cof of m 32 cof of m 33 T b 1 b 2 b 3 So 1 M (b 1 cof of m 12 + b 2 cos of m 22 + b 3 cof of m 32 ) 1 m 21 m 23 m 11 m 13 m 11 m 13 M -b 1 + b m m b m m m m m 11 b 1 m 13 m 21 b 2 m 23 m 31 b 3 m 33 det M Similarly for and. example 1 The system 2x + y + z x - y + 5z y - z 4 has one solution because The solution is

17 page 3 of Section x y z - 2 PROBLEMS FOR SECTION Suppose M is 3 3 and M. What important fact(s) can you conclude about M. 2. Try Cramer's rule and if that doesn't help, solve some other way. 2x - 5y + 4z 5 2y + 4z 1 2y + 4z (a) x - 3y + 2z 1 (b) x + y + 4z 4 (c) x + y + 4z 4 y + z 7 x + 2z 1 x + 2z 1 3. Look at the system 3x + y 3 2x - y 5 Solve it three times. (a) Use Cramer's rule. (b) Use an inverse matrix (c) Use Gaussian elimination. 4. Use Cramer's rule to solve ax + by c dx + ey f for x and y. What assumption is necessary for your solution to be valid. 5. Let A be 3 3 and let b 5 9 8, c 1 2 Suppose Ax b has no solutions. What can you conclude about (a) the number of sols (and the number of free variables) to Ax (b) the number of sols (and the number of free variables) to Ax c where c

18 page 4 of Section 5.4 (c) A (d) rank A. Let A 1/ 2/ 1/ -2/ 5 1/ 5 1/ 3 2/ 3-5/ 3 x y z Without any agony, solve A