Matrix Equations on a TI-83 Plus

Consider the following system of linear equations:

4x1 + 3x2 =  7
 x1 +  x2 = -1

Here is the same system written as a matrix equation:

In this equation, [A] is the matrix of coefficients, [B] is the matrix of constants, and [C] is the solution matrix. The solution matrix C would normally be called X but the TI-83 plus only has 10 matrices named A through J. Solving this matrix equation for [C] gives us:

Enter the Coefficient Matrix

1. To enter a matrix, begin by entering this keystroke combination: 2nd Matrix

2. Use the cursor keys to select the Edit option and then select row 1 (matrix A). Press Enter.

3. Enter the dimensions of the coefficient matrix (2 x 2):

2 Enter 2 Enter

4. Enter the values of the coefficient matrix (row by row):

4 Enter 3 Enter 1 Enter 1 Enter

5. When you are done, press 2nd Quit.

Enter the Constant Matrix

1. 2nd Matrix

2. Select the Edit option and then row 2 (matrix B).

3. Enter the dimensions of the constant matrix (2 x 1).

4. Enter the values of the constant matrix (7 and -1).

5. 2nd Quit

Perform the Matrix Arithmetic A-1B

1. 2nd Matrix and select row 1 (matrix A).

2. Inverse Mult

3. 2nd Matrix and select row 2 (matrix B).

The display should look like this:

4. Enter

The result of the matrix arithmetic is a 2 x 1 matrix containing the values 10 and -11. The solution to the system of equations is:

x1 =  10
x2 = -11

Verifying the Solution

Store the solution matrix as matrix C. (These instructions assume that the solution matrix is still being displayed and you have not pressed any keys since pressing the Enter key in the previous step.)

1. Store 2nd Matrix [C] Enter

2. Now verify the solution by performing the matrix multiplication [A][C]:

2nd Matrix [A] Mult 2nd Matrix [C] Enter

3. The result should be the matrix of constants (the same as matrix B containing 7 and -1).