# Section 3.4 Question 2

### 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-1.

A-1 AX = A-1 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-1 B

The product of an identity matrix I with X is simply X, so the solution to the matrix equation is

X = A-1 B

How To Solve a System of Linear Equations with Inverses

1. 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.

2. Compute the inverse A-1. If the matrix is not invertible, it is not possible to solve the system with inverses.

3. Compute the product A-1 B.

4. The solution to the matrix equation AX = B is X = A-1 B. The corresponding solution for the system of linear equations is found by matching the variables in X with the corresponding entries in the product .

### Example 3      Solve a Linear System with the Inverse

Solve the system of linear equations using the inverse of the coefficient matrix.

Solution In Example 1, we wrote this system as the matrix equation , where

In section 3.3, we found the inverse of A,

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,

Since the individual entries in X correspond to the variables x and y, this tells us that x = -21 and y = 16. We can check these values in the original system to make sure they solve the system:

### Example 4      Solve a Linear System with the Inverse

Solve the system of linear equations with the inverse of the coefficient matrix.

Solution This system of equations is equivalent to the matrix equation AX = B where

The variable matrix may be solved for using the inverse of the coefficient matrix, X = A-1 B. We found the inverse of this coefficient matrix in Example 5 of section 3.3.

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 2 may be solved using this strategy.

### Example 5      Mixing Ethanol Blends

In Example 12 of section 2.2 we created a system of equations to describe a mix of E10 and E85 ethanol,

where E10 is the amount of 10% 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, 10 gallons. The second equation describes the total amount of ethanol in the mixture, 20% of 10 gallons or 2 gallons.

a. Solve this system of equations by finding the inverse of the coefficient matrix.

Solution This system of equations was solved in Chapter 2 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,

The solution to the matrix equation is X = A-1 B and requires the inverse of the matrix A.

We begin the process of finding the inverse of A by placing the coefficient matrix in a new matrix alongside a 2 x 2 identity matrix.

Since the entry in the first row, first column is already a 1, we’ll make the rest of the column into zeros using row operations. To do this, multiply the first row by -0.10 and add it to the second row. Place the sum in the second row:

With the first column transformed, proceed to the second column and use row operations to create a 1 in the second row, second column.

We put a 1 in the second row, second column by multiplying the second row by 1/0.75. Since 0.75 = 3/4, this is the same as multiplying by 4/3:

Now multiply the second row by -1 and add it to the first row. Place the result in the first row:

The inverse matrix is the right hand side of this matrix,

The solution to the original system of equations is

Since the variable matrix X represents , this means that the amount of E10 needed is 26/3 gallons and the amount of E85 needed is 4/3 gallons.

b. If the number of gallons in the mixture should be 12 gallons, how much 10% ethanol and 85% ethanol must be pumped?

Solution If the total number of gallons in the mixture is increased to 12 gallons, the total amount of ethanol in the mixture is or 2.4 gallons. The system of equations becomes

The only change to the system of equations is in the constants, not the coefficients.

This means that we may use the existing coefficient matrix A and the corresponding inverse A-1 to find the solution of this system,

The constant matrix for this system is

and the solution is

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 52/5 gallons of E10 and 8/5 gallons of E85.

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 constants. For part a, the constant was  and for part b the constant was . There was no need to recompute the inverse for the different parts.