11 a 12 a 13 a 21 a 22 a b 12 b 13 b 21 b 22 b b 11 a 12 + b 12 a 13 + b 13 a 21 + b 21 a 22 + b 22 a 23 + b 23

Transcription

1 Chapter 2 (3 3) Matrices The methods used described in the previous chapter for solving sets of linear equations are equally applicable to 3 3 matrices. The algebra becomes more drawn out for larger matrices, so the usefulness compact matrix notation is obvious, and the visualisation of the geometrical properties of matrices becomes increasingly important matrix operations Matrix addition is exactly analogous to that in 2 2 matrices in the last chapter. We simply add the matrices element by element A = a 11 a 12 a 13 a 21 a 22 a 23 (2.1) a 31 a 32 a 33 B = b 11 b 12 b 13 b 21 b 22 b 23 (2.2) b 31 b 32 b 33 A + B = a 11 + b 11 a 12 + b 12 a 13 + b 13 a 21 + b 21 a 22 + b 22 a 23 + b 23. (2.3) a 31 + b 31 a 32 + b 32 a 33 + b 33 Matrix multiplication is also analogous, for example we obtain the first element by multiplying the elements in the row of the first matrix by the corresponding elements in the column of the second matrix. If we form the product AB from the matrices given above then we obtain AB = a 11b 11 + a 12 b 21 + a 13 b 31 a 11 b 12 + a 12 b 22 + a 13 b 32 a 11 b 13 + a 12 b 23 + a 13 b 33 a 21 b 11 + a 22 b 21 + a 23 b 31 a 21 b 12 + a 22 b 22 + a 23 b 32 a 21 b 13 + a 22 b 23 + a 23 b 33. (2.4) a 31 b 11 + a 32 b 21 + a 33 b 31 a 31 b 12 + a 32 b 22 + a 33 b 32 a 31 b 13 + a 32 b 23 + a 33 b Networks We can use the 3 3 matrix to describe a linear system with 3 inputs and 3 outputs, just as we discussed in the previous chapter for 2 2 matrices. The inputs and outputs here may again be the voltages, currents, displacements etc. This is illustrated in the figure below. 12

2 CHAPTER 2. (3 3) MATRICES 13 Inputs x 1 Outputs y 1 x 2 A y 2 x 3 y 3 The connection between the inputs and outputs can be described by a system of linear equations These equations can be rewritten in matrix form Y = y 1 y 2 = a 11 a 12 a 13 a 21 a 22 a 23 y 3 a 31 a 32 a 33 y 1 = a 11 x 1 + a 12 x 2 + a 13 x 3 (2.5) y 2 = a 21 x 1 + a 22 x 2 + a 23 x 3 (2.6) y 3 = a 31 x 1 + a 32 x 2 + a 33 x 3. (2.7) x 1 x 2 x 3 = AX (2.8) We can then put these systems in parallel or in series in the same way as discussed in the previous chapter. 2.3 Simultaneous Equations Consider now a system of 3 linear equations such as x + y + z = 1 (2.9) 3x 2y + z = 3 (2.10) x + 4y z = 17. (2.11) We would like to solve this system of equations for x, y and z. Is this always possible? Graphically each one of these equations is represented by a plane. We wish to know if these planes intersect at a point (x, y, z) which is the solution for which we are looking. Compared with the 2 dimensional case there are many more circumstances; all three planes might be parallel,

3 CHAPTER 2. (3 3) MATRICES 14 two planes might be parallel and the third intersect them in two lines that never cross, the planes may intersect in three lines that never cross, all three planes may intersect in a single line, finally the planes intersect at a point, the solution! It is clear from these many scenarios that we must be more careful when looking into whether we have a solution. Notice that although we are looking at the same problem of solving a set of linear

4 CHAPTER 2. (3 3) MATRICES 15 equations, the problem is much richer in 3 dimensions than in 2. This is usually the case in physics: low dimensional problems are generally easier to solve and less complicated than the same problem in higher dimensions. We can now use the same algorithms as we did in 2 dimensions to solve this problem. Consider the general set of 3 linear equations a 11 a 12 a 13 x c 1 a 21 a 22 a 23 y = c 2 (2.12) a 31 a 32 a 33 z c 3 Gaussian Elimination The algorithm for Gaussian elimination proceeds by using the first equation to eliminate x from the remaining two equations: xa 11 + ya 12 + za 13 = c 1 (2.13) xa 21 + ya 22 + za 23 = c 2 (2.14) xa 31 + ya 32 + za 33 = c 3 (2.15) y(a 22 a 11 a 12 a 21 ) + z(a 23 a 11 a 13 a 21 ) = a 11 c 2 a 21 c 1 (2.16) y(a 32 a 11 a 12 a 31 ) + z(a 33 a 11 a 13 a 31 ) = a 11 c 3 a 31 c 1. (2.17) We can then eliminate y from the previous one. z [(a 33 a 11 a 13 a 31 )(a 22 a 11 a 12 a 21 ) (a 23 a 11 a 13 a 21 )(a 32 a 11 a 12 a 31 )] = (2.18) [(a 11 c 3 a 31 c 1 )(a 22 a 11 a 12 a 21 ) (a 11 c 2 a 21 c 1 )(a 32 a 11 a 12 a 31 )] (2.19) This rather complicated equation can be simplified to give z = c 1(a 21 a 32 a 22 a 31 ) c 2 (a 11 a 32 a 31 a 12 ) + c 3 (a 11 a 22 a 21 a 12 ) a 11 (a 22 a 33 a 23 a 32 ) a 12 (a 21 a 33 a 31 a 23 ) + a 13 (a 21 a 32 a 31 a 22 ). (2.20) There are similar long expressions for the solutions y and z. Cramer s Rule The long expressions for the solution in the previous section can be written more concisely if we define the determinant of a 3 3 matrix. a 11 a 12 a 13 deta = a 21 a 22 a 23 a 31 a 32 a 33 = a 11(a 22 a 33 a 23 a 32 ) a 12 (a 21 a 33 a 31 a 23 ) + a 13 (a 21 a 32 a 31 a 22 ). (2.21) This may seem like a complicated expression, but it can be expressed in terms of 2 2 determinants, which makes it easier to remember. a deta = a 22 a a 32 a 33 a 12 a 21 a 23 a 31 a 33 + a 13 a 21 a 22 a 31 a 32 (2.22) We perform this expansion by first choosing a row and writing down the first element. By covering up the rest of the row and column of that element we get the 2 2 determinant for this element. We

5 CHAPTER 2. (3 3) MATRICES 16 then repeat this for the rest of the elements in the row. However, the sign infront of the element must alternate according to the following pattern (2.23) This expansion method can be performed using any row or column as a basis for the expansion. Determinants can be manipulated in a similar way to matrices. We can add or subtract a multiple one row from another without changing the result. We can also add or subtract a multiple of one column from another without changing the result. If two rows are swapped around then the determinant changes sign. Similarly if two columns are swapped then the determinant changes sign. With this compact notation of the determinant we can use Cramer s rule to write down the solution to the 3 3 matrix problem 1 c 1 a 12 a 13 x = det A c 2 a 22 a 23 (2.24) c 3 a 32 a 33 1 a 11 c 1 a 13 y = det A a 21 c 2 a 23 (2.25) a 31 c 3 a 33 1 a 11 a 12 c 1 z = det A a 11 a 22 c 2 (2.26) a 31 a 32 c 3 You should check, by expanding the determinant, that this give the same solution as in the previous section. Note that no solution exists for the case deta = 0. This corresponds one of the cases described above where the planes either do not intersect at all, or do not intersect at a unique point. Matrix Inverse We can also calculate a solution by calculating the inverse of the matrix A and then writing X = A 1 C. Forming the inverse is a little more complicated for a 3 3 matrix. We must first calculate what is known as the matrix of cofactors cofa. To calculate this we cover up the associated row and column of an element, then compute the determinant of the remaining 2 2 matrix that is left. We then write down the result multiplied by the sign according to , (2.27) or alternatively ( 1) i+j where i and j are the indexes of the row and column respectively. Once we have computed this matrix, we can calculate the matrix inverse from A 1 = 1 deta (cofa)t (2.28) where the superscript T denotes the transpose of the matrix i.e. swapping the rows and columns around. Note that this algorithm works for the 2 2 matrix case, but is much simpler there.

6 CHAPTER 2. (3 3) MATRICES 17 Example of 3 3 matrix inverse Consider the matrix A = (2.29) To calculate the matrix of cofactors for this we start with the row 1 column 1 element, and cover up the remaining elements in its row and column: (2.30) The determinant we calculate from the remaining 4 elements is = 2 (2.31) so the row 1 column 1 element of our cofactor matrix is 2. For the next element we perform the same trick: (2.32) We are left with the determinant = 1 (2.33) so the row 1 column 2 element for our cofactor matrix is 1 (note the extra 1 factor here as ( 1) 1+2 = 1). We continue this process for the remaining elements in the matrix to produce cofa = (2.34) Now, the determinant of our matrix A is 6. To form the inverse we need the transpose of cofa divided by det A i.e. A 1 = (2.35) You should check that if we multiply this by A then the result is the identity matrix Geometric interpretation The 3 3 matrix can be interpreted as a linear transformation of 3 dimensional space, taking a point (x, y, z) to its image point point (x, y, z ) according to the rule x = a 11 x + a 12 y + a 13 z (2.36) y = a 21 x + a 22 y + a 23 z (2.37) z = a 31 x + a 32 y + a 33 z (2.38)

7 CHAPTER 2. (3 3) MATRICES 18 There are various deformations just as in the 2 dimensional case, but now we have more freedom. For example, we can still define a rotation matrix in the same way, but we have the freedom to choose the axis about which this rotation occurs, for example a rotation about the z axis is R z = whereas a rotation about the x axis is R x = cosθ sinθ 0 sin θ cosθ cosθ sin θ 0 sin θ cosθ, (2.39). (2.40) Examples of matrices in physics: Liquid Crystalline order parameter In soft condensed matter, the ordering of molecules is important to calculate the macroscopic properties of a material. The average orientation of rod-like molecules is encoded in the order parameter Q. This is a matrix that is defined so that Q = 0 for a disordered system, and Q 0 for an ordered case. Order parameters are important for understanding a phase transition e.g. the liquid-gas transition. It is carefully defined so that it has zero trace. The trace of a matrix is simply the sum of the diagonal components e.g. tra = a 11 + a 22 + a 33. (2.41) The energy of a system of rods can be defined in terms of Q, and in equilibrium the system adopts the value of Q corresponding to the minimum energy. In the problems you can calculate the energy of a liquid crystalline system using this matrix Q. 2.5 Problems 1A. If A = A. If P = illustrate? and B =, and Q = find A + B, A B, B A, AB and BA., then find QP, and PQ. What does this 3A. If C is the 3 by 3 square matrix C = Find the square of this martrix C 2 = CC, i.e., multiply C by itself. What do you notice about the matrix C 2? 4A. Find the value of the determinant of the following matrices, and state which of them can be inverted.

8 CHAPTER 2. (3 3) MATRICES 19 (a) (b) (c) B. Form the co-factor matrix of Q = Calculate the determinant of Q. Form Q by transposing the cofactor matrix and dividing it by the determinant. Verify that QQ 1 = Q 1 Q = 1. 6B. Find the reciprocal matrix of and verify that AA 1 = A 1 A = 1. A A. Consider the 2 sets of 3 simultaneous linear equations and x + 2y 2z = 5 2x 2y 2z = 6 x + 2y = 7 2x 2y 4z = 3 x + 3y + 4z = 4 x 2y 3z = 10 One of these two sets of equations has a solution, the other is not soluble. Which one? You do not need to try to solve both sets of equations to find the answer. Just consider the matrix method of solving sets of simultaneous linear equations and look for a necessary condition for this method to work. 8B. Consider the set of 3 simultaneous linear equations 2x + y + 3z = 4 x + y z = 6 y + 2z = 1 Write down the 3 by 3 matrix of coefficients O, and the column matrix of right-hand sides R. Then solve for x, y, and z using matrices.

9 CHAPTER 2. (3 3) MATRICES 20 9C. Solve the following equations by matrix methods (a) (b) 4x 3y + z = 11 2x + y 4z = 1 x + 2y 2z = 1 x + 5y + 3z = 1 5x + y z = 2 x + 2y + z = 3 10A. Find the rotation matrix R such that y = Rx represents an anticlockwise rotation about the z-axis through an angle of π/2 radians. 11A. The free energy of a liquid crystal is expressed as a power series expansion in terms of a matrix Q as follows F = atr[q 2 ] + btr[q 3 ] + c ( tr[q 2 ] ) Calculate the free energy for following matrix Q Q = 2S 0 0 S S Make a rough sketch of F(S) you have calculated for the two cases (i) a = 1, b = 1, c = 1, and 0.5 < S < 1 (ii) a = 1, b = 1, c = 1, and 0.5 < S < 1.