LESSON 28 (7.4 - 7.6) MATRIX OPERATIONS WITH CALCULATORS
AND SOLVING SYSTEMS OF EQUATIONS USING MATRICES
You should learn to:
1. Enter matrices on your calculator.
3. Solve systems of linear equations using matrices.
2. Use your calculator to perform matrix operations.
Terms to know: inverse (of a matrix), singular matrix, solution matrix, ordered triple
Most matrix operations can be done by hand faster than they can be done on a calculator. However, matrix
multiplication is more difficult, so it might be to your advantage to do complicated matrix multiplications using your
calculator.
To enter a matrix on your calculator, press the matrix key (MATRIX) and then arrow left to access EDIT
( then ENTER). Type in the number of rows × columns, and then type each element (entry). You can repeat this
process to enter other matrices with different names. Matrix operations can be performed on your calculator’s home
screen by entering the matrix names separated by the operation(s) that you wish to perform. (Be careful not to use
double brackets when writing your answers.)
 2

Example 1: Let A  1

 2
a.
AB
1
0
1
3
 1 0  2

4  and B   4 0 0  . Use your calculator to find:
 2 0 3 
1 
b. BA
 6 1 1 
 8 4 12 


10 1 3
 4 0 13 
 7 0 10 


 4 0 1
c. 3B  5 A
 13 5 21
 7
0 20 

 16 5 4 
Calculators are beneficial when finding inverses of matrices. Matrix inverses are used to solve systems of linear
equations. To find the inverse of a square matrix (same number of row and columns) using a calculator, use the x 1
key after entering the matrix onto home screen.
Example 2: Find the inverses of the matrices in Example 1. (Express elements as fractions.) Hint: To change
decimals to fractions, use MATH 1 ( Frac ) on your calculator.
 92

a. A1  187

 181
1
9
4
9
2
9



1 
18 
2
9
11
18
b. B 1  Singular Matrix
Only square matrices can have inverses. However, not every square matrix has an inverse. Square matrices without
inverses are said to be singular, and the calculator will denote that (as was the case in Example 2b).
Solving Systems of Equations Using Inverses
If a system of linear equations is consistent and independent (has a unique solution), then the solutions for
the system can be found using matrix multiplication involving a matrix inverse. The procedure can best be illustrated
using an actual system of equations.
Procedure for Solving:
1.
2.
3.
4.
5.
Make sure that the variables are lined up correctly (in the same order for each equation).
Write a square matrix for the coefficients of the variables and call it A.
Write a single column matrix for the constant on the right side of each equation and call it B .
Use your calculator to multiply A inverse times B. A1 B
Write the solutions to the system in whatever form is specified.
Example 3: Use an inverse matrix to solve the system of equations below:
If we have the system:
A
x yz 2
2x  y  2 z  1
 x  3 y  2 z  2
1 1 1
 2 1 2 


 1 3 2
We can write it into matrices:
2
B = 1 
 
 2 
We now have a system of matrices in the form AX = B
1 1 1
1
A B   2 1 2 


 1 3 2 
1
2
 x
 3
 1    y   1 
 
 
 
 2
 z 
 4 
1   x  2 
1 1
 2  1 2    y   1 

    
 1 3  2  z   2
x  3,
y  1, z  4
This result is the matrix of the solutions to the system of equations (the solution matrix). The solutions
from top to bottom in the column stand for the variables as they appear from left to right in the equations for
the system.
Write the solutions as an ordered triple. (3,1, 4)
Remember:
A1B   solution matrix 
Example 4: Solve the system:
 1 2 3
 1 3 0 


 2 5 5 
1
Example 5: Solve the system:
 3 0 1
0 2 1 


5 1 1
1
x  2 y  3z  9
 x  3y
 4
2 x  5 y  5 z  17
9
 x
1
 4    y    1
 
 
 
17 
 z 
 2 
x  1,
y  1, z  2
3 x  w  4
w  2y  4
y  5 x  w  3
3 x  0 y  w  4
(1, 1, 2)
0x  2 y  w  4
5 x  y  w  3
 4 
 x
2
 4    y    3
 
 
 
 3
 z 
10 
x  2,
y  3, w  10
Given enough points on polynomial functions, equations for those functions can be found by solving a system of
equations that uses the given points.
2
Example 6: Find an equation of the quadratic function y  ax +bx +c .which passes through the points
 1,3 , 1,1 , and  2,6 .
using (1,3)  3  a(1) 2  b(1)  c
 3  a bc
using (1,1)
 1  a(1)  b(1)  c
 1 a bc
using (2, 6)
 6  a(2)  b(2)  c
 6  4 a  2b  c
1 1 1
1 1 1


 4 2 1
2
2
1
3
a 
2
1    b    1
 
 
 
6 
 c 
 0 
so,
y  ax 2  bx  c 
a  2, b  1, c  0
y  2 x 2  (1) x  0
y  2 x2  x
ASSIGNMENT 28 (Calculator) Pages 547-550 (12-20 even, 26, 50, 51, 54, 56, 60) + Page 523 (69, 72, 79)
(NonCalculator) Page 537 (21, 26, 37)