1 00:00:01,839 --> 00:00:06,630 For the non-homogeneous system of equations, which means that for your Right Hand Side, 2 00:00:06,630 --> 00:00:15,040 at least one of b1, b2, … bn is non-zero, let’s see how to solve this: 3 00:00:15,040 --> 00:00:20,920 The general matrix representation is AX equals B 4 00:00:20,920 --> 00:00:29,539 Pre-multiply it by A inverse: so A inverse times A times X equals A inverse times B 5 00:00:29,539 --> 00:00:35,280 (You have to do the operation on both sides of the equation, to keep the relation the 6 00:00:35,280 --> 00:00:40,530 same) And provided A inverse exists, then you will 7 00:00:40,530 --> 00:00:46,540 recognize this part: A inverse times A as the Identity matrix of Order n [times X] equals 8 00:00:46,540 --> 00:00:53,940 A inverse times B. Any matrix multiplied by Identity is just 9 00:00:53,940 --> 00:01:03,760 itself, so that gives us the solution: X equals A inverse times B, provided A inverse 10 00:01:03,760 --> 00:01:06,200 exists, i.e. 11 00:01:06,200 --> 00:01:16,860 [X equals] adjoint(A) over det(A) times B So this is the method to find the solution 12 00:01:16,860 --> 00:01:21,999 to a set of linear equations in n unknowns. 13 00:01:21,999 --> 00:01:28,060 Now there are some special cases: This thing [X = adj(A) over det(A) times B] 14 00:01:28,060 --> 00:01:37,140 works fine if the det(A) is non-zero What happens if the determinant of A is zero? 15 00:01:37,140 --> 00:01:47,729 Special case: if det(A) is zero, look at the numerator of this expression [X = adj(A) over 16 00:01:47,729 --> 00:01:54,229 det(A) times B]: If [adj(A) times B], i.e. if this product 17 00:01:54,229 --> 00:02:10,380 is not zero, the system of equations [AX = B] is Inconsistent, and it has NO SOLUTIONS. 18 00:02:10,380 --> 00:02:22,300 But if the numerator [adj(A) times B] is also zero (in addition to the denominator being 19 00:02:22,300 --> 00:02:35,129 zero), then the system is Consistent, and it has an INFINITE NUMBER OF SOLUTIONS. 20 00:02:35,129 --> 00:02:50,720 Ok… so this is one way, using Matrix Operations, to solve a system of simultaneous linear equations. 21 00:02:50,720 --> 00:03:10,730 A second way, is using Determinants, or what is known as “Cramer’s Rule”: 22 00:03:10,730 --> 00:03:18,450 For the same system of equations [AX = B], the solution for each unknown variable is 23 00:03:18,450 --> 00:03:31,780 given as: xi equals 1 over det(A) times this determinant: 24 00:03:31,780 --> 00:03:38,110 so a11, a21, … an1 25 00:03:38,110 --> 00:03:45,069 a21, a22, … an2 and then observe this carefully: 26 00:03:45,069 --> 00:03:56,049 find the ith column of the determinant: so a1i: replace it by b1 27 00:03:56,049 --> 00:04:06,889 a2i: replace it by b2 so this is a substitution operation: 28 00:04:06,889 --> 00:04:11,370 you don’t keep the original term a1i, a2i, … 29 00:04:11,370 --> 00:04:18,250 And so on… keep replacing this [column] by the elements of the Right Hand Side … bn 30 00:04:18,250 --> 00:04:31,419 And then the rest of the determinant stays the same: all the way up to a1n, a2n, ann 31 00:04:31,419 --> 00:04:42,400 So this whole operation: we can write it in short-hand as: [xi equals] det(Ai) over det(A), 32 00:04:42,400 --> 00:04:49,220 where the det(Ai) basically means that you are substituting the ith column of the determinant 33 00:04:49,220 --> 00:04:52,880 det(A) by the Right Hand Side, i.e. the elements of matrix B. 34 00:04:52,880 --> 00:05:02,180 This is known as Cramer’s Rule. 35 00:05:02,180 --> 00:05:09,090 In reality: these two (methods to solve a system of linear equations) are identical. 36 00:05:09,090 --> 00:05:13,099 If you care to write this [X = adj(A) over det(A) times B] whole thing down symbolically, 37 00:05:13,099 --> 00:05:16,770 you can show that this works out to be [xi equals] det(Ai) over det(A) exactly. 38 00:05:16,770 --> 00:05:25,979 i.e. each element of X [xi] will work out to be adj(A) over det(A) times matrix B, which 39 00:05:25,979 --> 00:05:33,659 will work out to be this [det(Ai) over det(A)] Now again, what happens if the denominator 40 00:05:33,659 --> 00:05:35,889 is zero? 41 00:05:35,889 --> 00:05:44,419 Special case: if det(A) equals zero, well again – look at the numerator: 42 00:05:44,419 --> 00:05:55,460 If any of the numerators det(Ai) is also zero: Inconsistent set of equations (i.e. there 43 00:05:55,460 --> 00:05:59,120 is NO SOLUTION). 44 00:05:59,120 --> 00:06:05,750 I’m sorry – if any of the determinants in the numerator is non-zero: it is Inconsistent 45 00:06:05,750 --> 00:06:07,720 (NO SOLUTION). 46 00:06:07,720 --> 00:06:19,580 If all det(Ai) determinants are zero for all i: the system is Consistent, and it has an 47 00:06:19,580 --> 00:06:27,199 INFINITE NUMBER OF SOLUTIONS. 48 00:06:27,199 --> 00:06:34,159 So we’ve looked at 2 methods of how to solve a system of simultaneous linear equations 49 00:06:34,159 --> 00:06:39,930 in n unknowns, for the “Non-homogeneous case”, which means at least one of the elements 50 00:06:39,930 --> 00:06:44,600 of your Right Hand Side (the B matrix) is non-zero. 51 00:06:44,600 --> 00:07:01,530 Now the only thing to look at that is remaining, is the “Homogeneous” system: which means 52 00:07:01,530 --> 00:07:12,680 that every single one of the Right Hand Side (b1, b2, … bn) is zero. 53 00:07:12,680 --> 00:07:17,159 Again, we can look at it using this method [i.e. 54 00:07:17,159 --> 00:07:26,460 Cramer’s Rule]: If the determinant det(A) is non-zero: there 55 00:07:26,460 --> 00:07:36,530 is only a TRIVIAL SOLUTION, which means every single variable x1, x2, … all the way to 56 00:07:36,530 --> 00:07:39,370 xn is zero. 57 00:07:39,370 --> 00:07:49,380 But if the determinant det(A) is zero: then there is an INFINITE NUMBER OF SOLUTIONS. 58 00:07:49,380 --> 00:08:02,439 So these are the 2 general methods to find out the values of the unknown variables x1, 59 00:08:02,439 --> 00:08:04,689 x2, … all the way through xn. 60 00:08:04,689 --> 00:08:11,490 Let’s take a Solved Example to better understand how to work these methods. 61 00:08:11,490 --> 00:08:30,140 Example: Let’s say Solve this system of equations: 62 00:08:32,320 --> 00:08:48,700 x + 7y – 3z = 11 25y + z = –3 63 00:08:48,700 --> 00:09:01,060 And 3x – 6y + 2z = 0 Now as soon as you have a set of equations 64 00:09:01,060 --> 00:09:07,090 given to you, there are a few things to check: So first of all, how many unknowns? 65 00:09:07,090 --> 00:09:14,340 x, y and z: 3 unknowns, and you have 3 equations (in 3 unknowns): so the number of equations 66 00:09:14,340 --> 00:09:16,420 is equal to the number of unknowns. 67 00:09:16,420 --> 00:09:24,350 Now the Right Hand part of this set of equations: 11, –3, 0: so clearly it is NOT a Homogeneous 68 00:09:24,350 --> 00:09:30,350 system, because you have these non-zero elements {11, –3} on the Right Hand Side. 69 00:09:30,350 --> 00:09:34,610 The next thing to check: would be this determinant 70 00:09:34,610 --> 00:09:41,170 det(A): this part: AX equals B: let us first write it in that form: 71 00:09:42,540 --> 00:09:52,060 So A would be 1, 7, –3 it’s 0 times x so 0, 25, 1 72 00:09:52,060 --> 00:09:57,640 And 3, –6, 2 This is your A matrix 73 00:09:57,640 --> 00:10:03,560 Times the variable matrix: in this case x y z 74 00:10:03,560 --> 00:10:13,100 Equals B, which is 11, – 3, 0 So after writing it in this form AX equals 75 00:10:13,120 --> 00:10:19,640 B, the next thing to check is whether determinant, det(A) is zero or not. 76 00:10:19,640 --> 00:10:24,100 Check this calculation: so [determinant of] 1, 7, –3 77 00:10:24,100 --> 00:10:29,340 0, 25, 1 3, –6, 2 78 00:10:29,340 --> 00:10:36,440 Works out to be: 1 times 56 79 00:10:36,440 --> 00:10:42,400 I’m expanding by this [first] column NOTE that: if any column or row of a determinant 80 00:10:42,400 --> 00:10:47,680 has one or more zero elements, it is always easier to expand by that row or column 81 00:10:47,690 --> 00:10:51,710 So that’s why I’m choosing to expand by this [first] column (instead of by any other 82 00:10:51,710 --> 00:10:58,820 row or column) 1 times (50 plus 6), so 56, minus 0 times 83 00:10:58,820 --> 00:11:06,820 (something) (the cofactor of this element, which we don’t care about), plus 3 times 84 00:11:06,820 --> 00:11:16,190 (7 minus –75), so 82 So the value of the determinant, det(A) works 85 00:11:16,190 --> 00:11:29,240 out to be 302, which is not zero. Which means: we can find out these unknown 86 00:11:29,240 --> 00:11:35,700 values [x, y and z] Because we’re taking this as a Worked Out 87 00:11:35,700 --> 00:11:39,360 example, I want to demonstrate this by both methods. 88 00:11:39,360 --> 00:11:46,750 Let’s first try to find out the values of x, y and z by the Matrix Method. 89 00:11:46,750 --> 00:11:56,070 Using the Matrix Method: well, x, y and z: you need to find out as A inverse times B 90 00:11:56,070 --> 00:12:00,610 using this definition [X equals A inverse times B] 91 00:12:00,610 --> 00:12:12,890 And to do that: you first need to find out A inverse, so that is adj(A) times B over 92 00:12:12,890 --> 00:12:20,860 det(A) For calculating the adjoint: you need to find 93 00:12:20,860 --> 00:12:29,890 out the Co-Factors of each element of A. Let’s take an example. To recall: C11 is 94 00:12:29,890 --> 00:12:40,290 (–1) to the power of (i + j), so (1 + 1), and then , 95 00:12:40,290 --> 00:12:48,230 i.e. mentally block off the ith row and jth column and calculate the determinant of the 96 00:12:48,230 --> 00:12:54,340 Minor of aij: so 50 minus –6, which equals (–1)2(56) equals 56 97 00:12:54,340 --> 00:13:00,510 And so on… I’ll leave it as an exercise for the reader to calculate each Cofactor 98 00:13:00,510 --> 00:13:09,000 And what you can show, is: Adj(A) is basically the transpose of the Co-factors 99 00:13:09,000 --> 00:13:16,830 of all the elements of A, which works out to be this: 100 00:13:16,830 --> 00:13:28,830 56, 4, 82 3, –7, –1 101 00:13:28,830 --> 00:13:42,030 – 75, 27, 25 So A inverse is simply the adjoint adj(A) 102 00:13:42,030 --> 00:13:54,400 divided by the determinant det(A), so 1 over 302 times 00:14:00,590 gonna write it again> And then, once you have A inverse: you get 104 00:14:00,590 --> 00:14:09,440 your unknown variables x, y, z is [equal to] A inverse times B 105 00:14:09,440 --> 00:14:26,571 Well… I’ll write it here: 1 over 302 times 106 00:14:26,571 --> 00:14:32,490 56, 4, 82 3, –7, –1 107 00:14:32,490 --> 00:14:37,450 – 75, 27, 25 times 108 00:14:37,450 --> 00:14:41,460 your Right Hand Side matrix, which is 11, –3, 0 109 00:14:41,460 --> 00:14:48,250 And you can do this calculation; it works out to be: 110 00:14:48,250 --> 00:14:55,940 Let’s still keep this {302} outside: 1 over 302 times (first do the multiplication here 111 00:14:55,940 --> 00:15:02,180 and you can show that this works out to be) 112 00:15:02,180 --> 00:15:05,660 –906 113 00:15:05,660 --> 00:15:10,170 Which finally gives you x is 2 114 00:15:10,170 --> 00:15:16,390 y is 0 and z is –3 115 00:15:16,390 --> 00:15:20,850 So this is one way to find out the values of x, y and z