Calculator User Manual
500      Appendix  A:  Functions  and  Instructions
8992APPA.DOC TI-89 / TI-92 Plus: Appendix A (US English) Susan Gullord Revised: 02/23/01 1:48 PM Printed: 02/23/01 2:21 PM Page 500 of 132
ShowStat
CATALOG
ShowStat
Displays a dialog box containing the last
computed statistics results if they are still
valid. Statistics results are cleared
automatically if the data to compute them
has changed.
Use this instruction after a statistics
calculation, such as 
LinReg
.
{1,2,3,4,5}
!
L1
¸
{1 2 3 4 5}
{
0,2,6,
1
0,25
}
!
L2
¸
{
0 2 6 
1
0 25
}
TwoVar L
1
,L2
¸
S
h
owStat
¸
sign()
MATH/Number menu
sign(
expression1
) 
⇒
expression
sign(
list1
) 
⇒
list
sign(
matrix1
) 
⇒
matrix
For real and complex 
expression1
, returns
expression1
/
abs(
expression1
)
 when
expression1
ƒ
 0.
Returns 1 if 
expression1 
is positive.
Returns 
ë
1 if 
expression1 
is negative.
sign(
0
)
 returns 
„
1 if the complex format
mode is 
REAL
; otherwise, it returns itself.
sign(
0
)
 represents the unit circle in the
complex domain.
For a list or matrix, returns the signs of all
the elements.
sign(
ë
3.2)
¸ ë
1.
sign({2,3,4,
ë
5})
¸
{1
1
1
ë
1}
sign(1+abs(x))
¸
1
If complex format mode is 
REAL
:
sign([
ë
3,0,3])
¸
[
ë
1 „1 1]
simult()
MATH/Matrix menu
simult(
coeffMatrix
, 
constVector
[
, 
tol
]
) 
⇒
matrix
Returns a column vector that contains the
solutions to a system of linear equations.
coeffMatrix
 must be a square matrix that
contains the coefficients of the equations.
constVector
 must have the same number of
rows (same dimension) as 
coeffMatrix
 and
contain the constants.
Optionally, any matrix element is treated as
zero if its absolute value is less than 
tol
. This
tolerance is used only if the matrix has
floating-point entries and does not contain
any symbolic variables that have not been
assigned a value. Otherwise, 
tol
 is ignored.
• 
If you use 
¥¸
 or set the mode to
Exact/Approx=APPROXIMATE
, computations
are done using floating-point arithmetic.
• 
If 
tol
 is omitted or not used, the default
tolerance is calculated as:
5
E
ë
14 
ù
max(dim(
coeffMatrix
))
ù
rowNorm(
coeffMatrix
)
Solve for x and y:  x + 2y = 1 
3x + 4y = 
ë
1
simult([1,2;3,4],[1;
ë
1])
¸
[
ë
3
2
]
The solution is x=
ë
3 and y=2.
Solve: ax + by = 1 
cx + dy = 2
[a,b;c,d]
!
matx1
¸
[
a b
c d
]
simu
l
t
(
matx
1
,
[1
;2
])
¸
ë
(2
ø
b
ì
d)
a
ø
d
ì
b
ø
c
2
ø
a
ì
c
a
ø
d
ì
b
ø
c










