Hello Sara:

In the command line:
-glt s gltname
the "s" refers to the number of rows in the GLT matrix.

However, in the command line:
-glt_label k glabel
the "k" refers to the index number of the GLT.

Using your GLT matrix glt.Full_F.mat, you could execute the following:

3dDeconvolve \
-input myData+orig \
-nfirst 3 \
-polort 1 \
-num_stimts 9 \
-stim_file 1 myStim1.1D -stim_label 1 myStim1 \
-stim_file 2 myStim2.1D -stim_label 2 myStim2 \
-stim_file 3 myStim3.1D -stim_label 3 myStim3 \
-stim_file 4 'myMotion.1D[0]' -stim_base 4 -stim_label 4 Roll \
-stim_file 5 'myMotion.1D[1]' -stim_base 5 -stim_label 5 Pitch \
-stim_file 6 'myMotion.1D[2]' -stim_base 6 -stim_label 6 Yaw \
-stim_file 7 'myMotion.1D[3]' -stim_base 7 -stim_label 7 dS \
-stim_file 8 'myMotion.1D[4]' -stim_base 8 -stim_label 8 dL \
-stim_file 9 'myMotion.1D[5]' -stim_base 9 -stim_label 9 dP \
-progress 1000 \
-glt 3 "glt.Full_F.mat" -glt_label 1 "Full_F"
-fout -bucket myResults

The output bucket dataset myResults+orig will contain 25 sub-bricks.
Note that the Full F-stat sub-brick is identical to the GLT F-stat sub-brick,
the 1st stim function Coef sub-brick is identical to the GLT LC[0] coef.
sub-brick, the 2nd stim function Coef sub-brick is identical to the GLT LC[1]
coef. sub-brick, and the 3rd stim function Coef sub-brick is identical to
the GLT LC[2] coef. sub-brick. As they should be.

Since the motion parameters have been identified as being part of the Baseline
Model, they do not contribute to the reduction in error sum of squares for the
Full Model (relative to the Baseline Model). Hence, the statistical test
of the Full Model is identical to your GLT test.

So, in this case, by using the -stim_base option, the -glt option is
unnecessarily redundant, or redundantly unnecessary.

Doug Ward