## Output from imlasn4.sas

Source
0 Graphs

```NOTE: Capture of log output started.
```
```NOTE: %INCLUDE (level 1) file n:\psy6140\examples\iml\imlasn4.sas is file
n:\psy6140\examples\iml\imlasn4.sas.```
```227 +title 'Green & Carroll problems';
228 +    *  ------------------------ ;
229 +
230 +proc iml;```
`IML Ready`
```231 +   reset print log fuzz fw=6;
232 +
233 +*-- 4.7b,c ;
234 +A = { 3 2 4, 2 0 2, 4 2 3};```
```                A             3 rows      3 cols    (numeric)

3      2      4
2      0      2
4      2      3

```
`235 +r = echelon(A);`
```                R             3 rows      3 cols    (numeric)

1      0      0
0      1      0
0      0      1

```
`236 +rank = (((echelon(A)^=0)[,+]) ^=0)[+,];`
```                RANK          1 row       1 col     (numeric)

3

```
```237 +
238 +A = { 4 3 2 18, 2 1 3 10, 5 7 2 29};```
```                A             3 rows      4 cols    (numeric)

4      3      2     18
2      1      3     10
5      7      2     29

```
`239 +r = echelon(A);`
```                R             3 rows      4 cols    (numeric)

1      0      0   2.36
0      1      0   2.16
0      0      1   1.04

```
`240 +rank = (((echelon(A)^=0)[,+]) ^=0)[+,];`
```                RANK          1 row       1 col     (numeric)

3

```
```241 +   *-- Only 3 columns are linearly independent;
242 +   *   so, we can solve for one as linear comb of other three;
243 +   *   e.g., solve A[,{1 2 3}] * x = A[,4];
244 +x = solve(A[,1:3], A[,4]);```
```                X             3 rows      1 col     (numeric)

2.36
2.16
1.04

```
`245 +print (A[,4]) '=' (A[,1:3] * x);`
```                             #TEM1001   #TEM1004
18 =       18
10         10
29         29
```
```246 +
247 +*-- 4.9a;
248 +A = {2 1 4, 1 -1 1, 1 2 3};```
```                A             3 rows      3 cols    (numeric)

2      1      4
1     -1      1
1      2      3

```
`249 +b = {13, 2, 10};`
```                B             3 rows      1 col     (numeric)

13
2
10

```
`250 +r = echelon(A);`
```                R             3 rows      3 cols    (numeric)

1      0 1.6667
0      1 0.6667
0      0      0

```
`251 +r = echelon(A||b);`
```                R             3 rows      4 cols    (numeric)

1      0 1.6667      0
0      1 0.6667      0
0      0      0      1

```
```252 +   *-- Inconsistent: No exact solution;
253 +
254 +*-- 4.9b;
255 +A = {1 2 6, 1 3 -1, 1 2 0};```
```                A             3 rows      3 cols    (numeric)

1      2      6
1      3     -1
1      2      0

```
`256 +b = {16, 12, 10};`
```                B             3 rows      1 col     (numeric)

16
12
10

```
`257 +r = echelon(A);`
```                R             3 rows      3 cols    (numeric)

1      0      0
0      1      0
0      0      1

```
`258 +r = echelon(A||b);`
```                R             3 rows      4 cols    (numeric)

1      0      0      4
0      1      0      3
0      0      1      1

```
```259 +   *-- Unique solution;
260 +x = solve(A, b);```
```                X             3 rows      1 col     (numeric)

4
3
1

```
`261 +print (A * x) '=' b;`
```                              #TEM1001        B
16 =     16
12       12
10       10
```
`262 +quit;`
`Exiting IML.`
```NOTE: The PROCEDURE IML used 1.2 seconds.

```
`263 +`
`NOTE: %INCLUDE (level 1) ending.`