HP 38g Graphing Calculator - Manual (Page 138)

Solving Systems of Linear Equations
Remember that a system of equations can be represented by
a matrix equation :
Equation Form Matrix Form
ax + by + cz =
dx + ey + fz = kj^
gx + hy + iz =
a b c X
K
d e
f y
= k.
g
h i z
Using the matrix form, the solution is the vector of variables,
as shown below.
Constants
Vector
[^1» ^ 2 *^ 3 ]
Coefficients Matrix Variables
Vector (result)
[ [ a , b . c ] { d . e , f \ [ S , h . i ] ] [x, y. z]
The coefficient matrix must be square (the number of
coefficients per equation equals the number of equations).
To solve linear 1. tn Home, enter the constants vector or the name of a
equations
Example
stored constants vector (Ml ...M9, MO). (This must be a
vector, which has a single set of brackets, not a matrix,
which has multiple sets of brackets.)
2. Press [7].
3. Enter the coefficients matrix or the name of a stored
square coefficients matrix (Ml ...M9, MO).
4. Press
[ e n t e r ]
. The resulting variables vector is displayed.
Find all [x, y, z] satisfying
2x + 3y + 4z = 6
X
+ y - z = 0
4x - y + 2z = 6
The constants vector is [6,0,6]. The coefficients matrix is
[[2,3,4],[1,1,-1],[4, -1,2]]. The solution for this system of
linear equations (that is, constants vector divided by
coefficients matrix) is [1,0,1].
6-8 Using Matrices