## Matrix Equations on a TI-83 Plus

Consider the following system of linear equations:

4x_{1} + 3x_{2} = 7
x_{1} + x_{2} = -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:

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

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

2
2

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

4
3
1
1

5. When you are done, press
.

### Enter the Constant Matrix

1.

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.

### Perform the Matrix Arithmetic A^{-1}B

1.
and select row 1 (matrix A).

2.

3.
and select row 2 (matrix B).

The display should look like this:

4.

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:

x_{1} = 10
x_{2} = -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.
**[C]
**

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

**[A]**
**[C]
**

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