Solving Square Systems of Linear Equations, Inverse Matrices

Matrices 2. Solving Square Systems of Linear Equations; Inverse Matrices Solving square systems of linear equations; inverse matrices. Linear algebra is essentially about solving systems of linear equations, an important application of mathematics to real-world problems in engineering, business, and science, especially the social sciences. Here we will just stick to the most important case, where the system is square, i.e., there are as many variables as there are equations. In low dimensions such systems look as follows (we give a 2 × 2 system and a 3 × 3 system): (7) a11 x1 + a12 x2 = b1 a11 x1 + a12 x2 + a13 x3 = b1 a21 x1 + a22 x2 = b2 a21 x1 + a22 x2 + a23 x3 = b2 a31 x1 + a32 x2 + a33 x3 = b3 In these systems, the aij and bi are given, and we want to solve for the xi . As a simple mathematical example, consider the linear change of coordinates given by the equations x1 = a11 y1 + a12 y2 + a13 y3 x2 = a21 y1 + a22 y2 + a23 y3 x3 = a31 y1 + a32 y2 + a33 y3 If we know the y-coordinates of a point, then these equations tell us its x-coordinates immediately. But if instead we are given the x-coordinates, to find the y-coordinates we must solve a system of equations like (7) above, with the yi as the unknowns. Using matrix multiplication, we can abbreviate the system on the right in (7) by     b1 x1 (8) Ax = b, x =  x2  , b =  b2  , x3 b3 where A is the square matrix of coefficients ( aij ). (The 2 × 2 system and the n × n system would be written analogously; all of them are abbreviated by the same equation Ax = b, notice.) You have had experience with solving small systems like (7) by elimination: multiplying the equations by constants and subtracting them from each other, the purpose being to eliminate all the variables but one. When elimination is done systematically, it is an efficient method. Here however we want to talk about another method more compatible with hand­ held calculators and MatLab, and which leads more rapidly to certain key ideas and results in linear algebra. Inverse matrices. Referring to the system (8), suppose we can find a square matrix M , the same size as A, such that (9) MA = I (the identity matrix). 1 2 MATRIX INVERSES We can then solve (8) by matrix multiplication, using the successive steps, Ax = b M (A x) = M b (10) x = M b; where the step M (Ax) = x is justified by M (A x) = (M A)x, by associative law; = I x, by (9); = x, because I is the identity matrix . Moreover, the solution is unique, since (10) gives an explicit formula for it. The same procedure solves the problem of determining the inverse to the linear change of coordinates x = Ay, as the next example illustrates.     1 2 −3 2 Example 2.1 Let A = and M = . Verify that M satisfies (9) 2 3 2 −1 above, and use it to solve the first system below for xi and the second for the yi in terms of the xi : x1 + 2x2 = −1 x1 = y1 + 2y2 2x1 + 3x2 = 4   x2 = 2y1 + 3y2    2 1 0 = , by matrix multiplication. To −1 0 1        −1 11 x1 −3 2 = , so the solve the first system, we have by (10), = 4 −6 x2 2 −1 solution is x1 = 11, x2 = −6. By reasoning similar to that used above in going from Ax = b to x = M b, the solution to x = Ay is y = M x, so that we get Solution. We have 1 2 2 3 −3 2 y1 = −3x1 + 2x2 y2 = 2x1 − x2 as the expression for the yi in terms of the xi . Our problem now is: how do we get the matrix M ? In practice, you mostly press a key on the calculator, or type a Matlab command. But we need to be able to work abstractly with the matrix — i.e., with symbols, not just numbers, and for this some theoretical ideas are important. The first is that M doesn’t always exist. M exists ⇔ |A| 6= 0. The implication ⇒ follows immediately from the law M-5 in section M.1 (det(AB) = det(A)det(B)) , since MA = I ⇒ |M ||A| = |I| = 1 ⇒ |A| = 6 0. MATRIX INVERSES 3 The implication in the other direction requires more; for the low-dimensional cases, we will produce a formula for M . Let’s go to the formal definition first, and give M its proper name, A−1 : Definition. Let A be an n × n matrix, with |A| 6= 0. Then the inverse of A is an n × n matrix, written A−1 , such that (11) A−1 A = In , AA−1 = In (It is actually enough to verify either equation; the other follows automatically — see the exercises.) Using the above notation, our previous reasoning (9) - (10) shows that (12) |A| = 6 0 ⇒ the unique solution of A x = b is x = A−1 b; (12) |A| = 6 0 ⇒ the solution of x = A y for the yi is y = A−1 x. Calculating the inverse of a 3 × 3 matrix Let A be the matrix. The formulas for its inverse A−1 and for an auxiliary matrix adj A called the adjoint of A (or in some books the adjugate of A) are (13) A−1  A 1 1  11 = adj A = A21 |A| |A| A31 A12 A22 A32 T A13 A23  . A33 In the formula, Aij is the cofactor of the element aij in the matrix, i.e., its minor with its sign changed by the checkerboard rule (see section 1 on determinants). Formula (13) shows that the steps in calculating the inverse matrix are: 1. Calculate the matrix of minors. 2. Change the signs of the entries according to the checkerboard rule. 3. Transpose the resulting matrix; this gives adj A. 4. Divide every entry by |A|. (If inconvenient, for example if it would produce a matrix having fractions for every entry, you can just leave the 1/|A| factor outside, as in the formula. Note that step 4 can only be taken if |A| = 6 0, so if you haven’t checked this before, you’ll be reminded of it now.) The notation Aij for a cofactor makes it look like a matrix, rather than a signed determinant; this isn’t good, but we can live with it. 4 MATRIX INVERSES  0 −1 1 1. 0 1  1 Example 2.2 Find the inverse to A =  0 1 Solution. We calculate that |A| = 2. Then the steps are (T means transpose):    1 1 −1 1 0 −1 0 1 1 → 0 2 0 1 −1 1 1 0 1 matrix A cofactor matrix   1  1  0 1 0 1 2 2  1 2 −1  →  1 1 − 1  2 2 1 −1 0 1 − 21 0 2 adj A inverse of A  → T To get practice in matrix multiplication, check that A · A−1 = I, or to avoid the fractions, check that A · adj (A) = 2I. The same procedure works for calculating the inverse of a 2 × 2 matrix A. We do it for a general matrix, since it will save you time in differential equations if you can learn the resulting formula.   a b → c d matrix A  d −c −b a cofactors   → T Example 2.3 Find the inverses to: a)  d −b −c a adj A 1 0 3 2 Solution. a) Use the formula: |A| = 2, so A−1    1 d −b a |A| −c inverse of A →   1 2 2 b)  2 −1 1  1 3 2     1 2 0 1 0 . = = − 23 21 2 −3 1  b) Follow the previous scheme:     −5 2 −5 −3 7 1 2 2  2 −1 1  →  2 0 0 −1  →  −3 7 −1 4 3 −5 1 3 2   4 3 → −5  −5 2 4 1  −3 0 3  = A−1 . 3 7 −1 −5  Both solutions should be checked by multiplying the answer by the respective A. Proof of formula (13) for the inverse matrix. We want to show A · A−1 = I, or equivalently, A · adj A = |A|I; when this last is written out using (13) (remembering to transpose the matrix on the right there), it becomes      a11 a12 a13 A11 A21 A31 |A| 0 0  a21 a22 a23   A12 A22 A32  =  0 |A| 0  . (14) a31 a32 a33 A13 A23 A33 0 0 |A| To prove (14), it will be enough to look at two typical entries in the matrix on the right — say the first two in the top row. According to the rule for multiplying the two matrices on the left, what we have to show is that (15) a11 A11 + a12 A12 + a13 A13 = |A|; (16) a11 A21 + a12 A22 + a13 A23 = 0 MATRIX INVERSES 5 These two equations are both evaluating determinants by Laplace expansions: the first equation (15) evaluates the determinant on the left below by the cofactors of the first row; the second equation (16) evaluates the determinant on the right below by the cofactors of the second row (notice that the cofactors of the second row don’t care what’s actually in the second row, since to calculate them you only need to know the other two rows). a11 a21 a31 a12 a22 a32 a13 a23 a33 a11 a11 a31 a12 a12 a32 a13 a13 a33 The two equations (15) and (16) now follow, since the determinant on the left is just |A|, while the determinant on the right is 0, since two of its rows are the same. � The procedure we have given for calculating an inverse works for n × n matrices, but gets to be too cumbersome if n > 3, and other methods are used. The calculation of A−1 for reasonable-sized A is a standard package in computer algebra programs and MatLab. Unfor­ tunately, social scientists often want the inverses of very large matrices, and for this special techniques have had to be devised, which produce approximate but acceptable results.