Chapter 4Systems of Linear Equations; MatricesSection 6Matrix Equations and Systems of Linear Equationsqy qMatrix EquationsLet’s review one property of solving equations involvingLet s review one property of solving equations involving real numbers. Recall If ax = b then x = , or  A similar property of matrices will be used to solve systems of linear equations. 1baba2 Many of the basic properties of matrices are similar to the properties of real numbers, with the exception that matrix multiplication is not commutative. Basic Properties of MatricesAssuming that all products and sums are defined for theAssuming that all products and sums are defined for the indicated matrices A, B, C, I, and 0, we have Addition Properties• Associative: (A + B) + C = A + (B+ C)• Commutative: A + B = B + A• Additive Identity: A + 0 = 0 + A = A3y• Additive Inverse: A + (–A) = (–A) + A = 0Basic Properties of Matrices(continued)Multiplication PropertiesMultiplication Properties• Associative Property: A(BC) = (AB)C• Multiplicative identity: AI = IA = A• Multiplicative inverse: If A is a square matrix and A–1exists, then AA–1= A–1A = ICombined Properties4Co b ed ope es• Left distributive: A(B + C) = AB + AC• Right distributive: (B + C)A = BA + CABasic Properties of Matrices(continued)EqualityEquality• Addition: If A = B, then A + C = B + C• Left multiplication: If A = B, then CA = CB• Right multiplication: If A = B, then AC = BCThe use of these properties is best illustrated by an example of solving a matrix equation.5solving a matrix equation.Example: Given an n × n matrix A and an n × p matrix B and a third matrix denoted by X, we will solve the matrix equation AX = B for X.Solving a Matrix EquationGi iAiXbAX B=Given; since Ais n×n, Xmust by n×p.Multiply on the left by A-1.Associative property of matrices.()()1111AX BAAX A BAAX AB−−−−===6Property of matrix inverses.Property of the identity matrix.()11nIXABXAB−−==Example Example:Use matrix inverses21xy z++=Example:Use matrix inverses to solve the system 22223xyxyz+=++=7Example  Example: Use matrix inverses to 21xy z++=solve the system  Solution:• Write out the matrix of coefficients A, the matrix Xcontaining the variables x, y, ddth l ti22223xyxyz+=++=112210A⎡⎤⎢⎥=⎢⎥8and z, and the column matrix B containing the numbers on the right hand side of the equal sign. 122⎢⎥⎢⎥⎣⎦123B⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦xXyz⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦Example(continued)• Form the matrix equation AX = B. Multiply the 3 × 3 matrixAby the 3×1matrixXto verify that this112210122⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦matrix Aby the 3 ×1 matrix Xto verify that this multiplication produces the 3 × 3 system at the bottom: xyz⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦123⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦9122⎢⎥⎣⎦2122223xy zxyxyz++=+=++=z⎢⎥⎣⎦⎣⎦Example(continued)If the matrix A–1exists, then the solution is determined bysolution is determined by multiplying A–1by the matrix B. Since A–1is 3 × 3 and B is 3 × 1, the resulting product will have dimensions 3 × 1 and will store the values of x, y and z.A-1can be determined by the1XAB−=11 11222−⎡⎤⎢⎥⎡⎤10Acan be determined by the methods of a previous section or by using a computer or calculator. The resulting equation is shown at the right: 122210 1 23113444X⎢⎥⎡⎤⎢⎥⎢⎥=−⎢⎥⎢⎥⎢⎥⎢⎥−−⎣⎦⎢⎥⎣⎦ExampleSolutionThe product ofA–1andBisTh l i b d ffThe product of Aand BisThe solution can be read off from the X matrix: x = 0, y = 2,z = -1/2Written as an ordered triple fbthltii1XAB−=X =02−1⎡⎢⎢⎢⎢⎢⎤⎥⎥⎥⎥⎥11 1122210 1 23113X−⎡⎤⎢⎥⎡⎤⎢⎥⎢⎥=−⎢⎥⎢⎥⎢⎥⎢⎥−−⎣⎦⎢⎥⎣⎦11of numbers, the solution is (0, 2, –1/2). −2⎣⎢⎦⎥444⎢⎥⎣⎦Another ExampleExample: Solve the system on the right using the inverse matrix method21222xyz++=right using the inverse matrix method.222334xyzxy z−+=++=12Another Example SolutionExample: Solve the system on the right using the inverse matrix method21222xyz++=right using the inverse matrix method.Solution:The coefficient matrix A is displayed at the right. The inverse of A does not exist. (We can determine this by using 222334xyzxy z−+=++=121212⎡⎤⎢⎥−⎢⎥13a calculator.) We cannot use the inverse matrix method. Whenever the inverse of a matrix does not exist, we say that the matrix is singular.313⎢⎥⎢⎥⎣⎦Cases When Matrix Techniques Do Not Work There are two cases when inverse methods will not work: 1. If the coefficient matrix is singular 2. If the number of variables is not the same as the number of equations. 14Application Production scheduling:Labor and material costs forProduction scheduling:Labor and material costs for manufacturing two guitar models are given in the table below: Suppose that in a given week $1800 is used for labor and $1200 used for materials. How many of each model should be produced to use exactly each of these allocations? Guitar modelLabor costMaterial15Guitar modelLabor costMaterial costA $30 $20B $40 $30Application SolutionLet x be the number of model A guitars to produce andyThis gives us the system ofA guitars to produce and yrepresent the number of model B guitars. Then, multiplying the labor costs for each guitar by the number of guitars produced, we have 30x+40y= 1800This gives us the system of linear equations: 30x + 40y = 1800 20x + 30y = 1200 We can write this as a matrix equation: 1630x+ 40y= 1800 Since the material costs are $20 and $30 for models A and B respectively, we have 20x + 30y = 1200. ⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡1200180030204030yxApplication Solution(continued)Solution:1XAB−=The inverse of matrix A is Solution:Produce 60 model A guitars and no model B guitars. 30 4020 30A⎡⎤=⎢⎥⎣⎦0.3 0.40.2 0.3−⎡⎤⎢⎥−⎣⎦170.2