TI-84 MANUAL 3: Solving Systems Using Matrices
[ADVANCED ALGEBRA 2] Mr. Bermel
Objectives: Students will be able to…
(1) Define appropriate matrices from a system of equations
(2) Write matrix equations that can be used to solve for missing variables
(3) Enter matrices into the TI84 calculator
(4) Solve a system of equations using the matrix functions of the TI84 calculator
Step 1: To solve a system of equation using matrices, one must first define three matrices: (1) a
 x
 x
matrix of coefficients [A], (2) a solution matrix   or  y  and (3) a matrix of constants [B].
 
 y
 z 
For example consider the systems
2 x  5 y  4
S1 = 
6 x  10 y  3
 x  3 y  z  21

S2= 2 x  y  z  9
2 x  y  5 z  31

In system S1, the required matrices are as follows: In system S2, the required matrices are as follows:
2  5
(1) [A]  
 (this is a 2  2 matrix)
6 10 
 x
(2)  
 y
  4
(3) [B]    (this is a 2  1 matrix)
3
2  5  x   4
The matrix equation is 
    
6 10   y   3 
 x
Which is equivalent to [ A]     [ B]
 y
[ A]1 means multiplication inverse of matrix A
 x
Which means our solution is    [ A]1  [ B]
 y
1  3 1 
(1) [A]  2  1 1 (this is a 3 3 matrix)
2 1 5
 x
(2)  y 
 
 z 
 21
(3) [B]    9  (this is a 3 1 matrix)
  31
1  3 1  x   21
The matrix equation is 2  1 1   y     9 
2 1 5  z    31
 x
Which is equivalent to [ A]   y   [ B] .
 z 
 x
Which means our solution is  y   [ A]1  [ B]
 z 
TI-84 MANUAL 3: Solving Systems Using Matrices
[ADVANCED ALGEBRA 2] Mr. Bermel
Step 2: We must now enter our (1) coefficient matrix and (2) constant matrix into our TI84.
For system S1…
(a) Enter the coefficients into matrix [A]
Directions:
2nd Matrix > Edit 1:[A]
<enter 2  2 as the dimensions>
<enter the coefficients as described in step 1>
(b) Enter the constants into matrix [B]
Directions:
2nd Matrix > Edit 1:[B]
<enter 2  1 as the dimensions>
<enter the constants as described in step 1>
Step 3: Find the solution matrix
Directions:
<Start at the home screen>
2nd Matrix Names 1:[A]
<press the inverse key x
1
>
nd
2 MATRIX NAMES 2:[B]
<enter>
(matrix [A] will appear on the home screen)
1
( [ A] will appear on the home screen)
1
( [ A] [B] will appear on the home screen)
(this will display the solution)
The solution is
( x  0.5, y  0.6)
For system S2…
(a) Enter the coefficients into matrix [A]
Directions:
2nd Matrix > Edit 1:[A]
<enter 3 3 as the dimensions>
<enter the coefficients as described in step 1>
(b) Enter the constants into matrix [B]
Directions:
2nd Matrix > Edit 1:[B]
<enter 3 1 as the dimensions>
<enter the constants as described in step 1>
Step 3: Find the solution matrix
Directions:
<Start at the home screen>
2nd Matrix Names 1:[A]
<press the inverse key x
1
>
2nd MATRIX NAMES 2:[B]
<enter>
(matrix [A] will appear on the home screen)
1
( [ A] will appear on the home screen)
1
( [ A] [B] will appear on the home screen)
(this will display the solution)
The solution is
( x  2, y  5, z  8 )