Question 2: How do you solve a matrix equation using the matrix inverse?
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1 Question : How do you solve a matrix equation using the matrix inverse? In the previous question, we wrote systems of equations as a matrix equation AX B. In this format, the matrix A contains the coefficients on the variables, matrix X contains the variables, and matrix B contains the constants. Solving the system of equations means that we need to solve for the variable matrix X. This is accomplished by multiplying both sides of the matrix equation by the inverse of the coefficient matrix, A. A AX A B The product of a matrix A and its inverse is the identity matrix I. This means we can simplify the matrix equation to IX A B The product of an identity matrix with X is simply X, so the solution to the matrix equation is X A B How To Solve a System of Linear Equations with Inverses. Make sure the system is in proper format with variable terms on the left side of the equation and constants on the right side. The variable terms should be listed in the same order in each equation. Identify the coefficient matrix A, the variable matrix X, and the constant matrix B. 3. Compute the inverse A. If the matrix is not invertible, it is not possible to solve the system with inverses. 4. Compute the product A B. 8
2 5. The solution to the matrix equation AX B is The corresponding solution for the system of linear X A B. equations is found by matching the variables in X with the corresponding entries in the product A B. Example 3 Solve a Linear System with the Inverse Solve the system of linear equations using the inverse of the coefficient matrix. x y 5 3x4y Solution In 0, we wrote this system as the matrix equation AX where B, x 5 A, X, B 3 4 y and In section 3.3, we found the inverse of A, A 4 3 We can use the inverse to compute the solution to the system. The solution is found by multiplying the inverse of A times the constant matrix B, 9
3 X A B Insert the matrices. The product of a x matrix and a x matrix is a x matrix Multiply the row entries in the inverse times the corresponding column entries in B 6 Since the individual entries in X correspond to the variables x and y, this tells us that x and y 6. We can check these values in the original system to make sure they solve the system: TRUE TRUE Example 4 Solve a Linear System with the Inverse Solve the system of linear equations with the inverse of the coefficient matrix. x x x 3 x x 3x 3 x x 7 3 Solution This system of equations is equivalent to the matrix equation AX B where x A 3, X y, B and 0 z 7 0
4 The variable matrix may be solved for using the inverse of the coefficient matrix, X A B. We found the inverse of this coefficient matrix in Example 5 of section 3.3. X A B A was computed in Example 5 of section 3.3 Multiply the entries in the rows of the entries in the column of B A by Equating the entries in the variable matrix with the entries in this product, we observe that x 3, y 4, and z 5. For systems of linear equations with unique solutions, we can use inverses to solve for the variables. Many of the applications in Chapter may be solved using this strategy. Example 5 Mixing Ethanol Blends In Example of section. we created a system of equations to describe a mix of E0 and E85 ethanol, E0 E E E85 where E0 is the amount of 0% ethanol pumped in gallons and E85 is the amount of 85% ethanol pumped in gallons. The first equation describes the total amount in the mixture, 0 gallons. The second
5 equation describes the total amount of ethanol in the mixture, 0% of 0 gallons or gallons. a. Solve this system of equations by finding the inverse of the coefficient matrix. Solution This system of equations was solved in Chapter using the Substitution Method. In this section we ll solve the same system by writing the system as a matrix equation. We can write this system of equations as the matrix equation AX B by identifying the matrices, E0 0 A, X, B E85 and The solution to the matrix equation is of the matrix A. X A B and requires the inverse We begin the process of finding the inverse of A by placing the coefficient matrix in a new matrix alongside a x identity matrix Since the entry in the first row, first column is already a, we ll make the rest of the column into zeros using row operations. To do this, multiply the first row by -0.0 and add it to the second row. Place the sum in the second row: 0.0 R : R : R R becomes R
6 With the first column transformed, proceed to the second column and use row operations to create a in the second row, second column. We put a in the second row, second column by multiplying the second row by Since , this is the same as multiplying by : R becomes R Now multiply the second row by - and add it to the first row. Place the result in the first row: R : 5 3 R : R R becomes R The inverse matrix is the right hand side of this matrix, A The solution to the original system of equations is 3
7 X A B Multiply the entries in the rows of the entries in the column of B A Combine the terms in each entry and simplify all fractions by Since the variable matrix X represents E0 E85, this means that the amount of E0 needed is gallons. gallons and the amount of E85 needed is b. If the number of gallons in the mixture should be gallons, how much 0% ethanol and 85% ethanol must be pumped? Solution If the total number of gallons in the mixture is increased to gallons, the total amount of ethanol in the mixture is.0 or.4 gallons. The system of equations becomes E0 E E E85.4 The only change to the system of equations is in the constants, not the coefficients. 4
8 This means that we may use the existing coefficient matrix A and the corresponding inverse A to find the solution of this system, E0 E E E85.4 The constant matrix for this system is B.4 and the solution is X A B Use the same inverse as in part a Multiply the entries in the rows of by the entries in the column of B A Combine the terms in each entry and simplify all fractions Since the coefficients in the system have not changed, there is no need to recompute the inverse. We may use the inverse from part a to calculate the product. We must mix 5 5 gallons of E0 and 8 5 gallons of E85. 5
9 This example illustrated the power of using inverses to solve a system of linear equations. As long as the coefficient matrix does not change, the same inverse may be used to solve several different problems with different constant matrices. The solutions are simply the product of the inverse of the coefficient matrices and the different 0 constants. For part a, the constant was B and for part b the constant was B.4. There was no need to recomputed the inverse for the different parts. 6
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