how can we perform 6x6 matrix without values to find the inverse and determination

to create a 6x6 matrix and find the inverse and determination of the above matrix without any values

x=[a(1,1) a(1,2) a(1,3) a(1,4) a(1,5) a(1,6);

a(2,1) a(2,2) a(2,3) a(2,4) a(2,5) a(2,6);

a(3,1) a(3,2) a(3,3) a(3,4) a(3,5) a(3,6);

a(4,1) a(4,2) a(4,3) a(4,4) a(4,5) a(4,6);

a(5,1) a(5,2) a(5,3) a(5,4) a(5,5) a(5,6);

a(6,1) a(6,2) a(6,3) a(6,4) a(6,5) a(6,6)]

to find

| X | = ?

X-inv = 1/| X | (adj X)

Answer 1

[Mohammad Alhashash]

You can do that using the Symbolic Math Toolbox. And you can check wether you have this toolbox or not using the simple command :

ver

If you do have it, then you have to define the arguements a(i,j) in slightly different manner using the underscore characters instead of comma, i.e. :

syms a_1_1 a_1_2 a_1_3 a_1_4 a_1_5 a_1_6
syms a_2_1 a_2_2 a_2_3 a_2_4 a_2_5 a_2_6
syms a_3_1 a_3_2 a_3_3 a_3_4 a_3_5 a_3_6
syms a_4_1 a_4_2 a_4_3 a_4_4 a_4_5 a_4_6
syms a_5_1 a_5_2 a_5_3 a_5_4 a_5_5 a_5_6
syms a_6_1 a_6_2 a_6_3 a_6_4 a_6_5 a_6_6

then redefine our matrix X using these arguements:

X = [a_1_1 a_1_2 a_1_3 a_1_4 a_1_5 a_1_6;...
a_2_1 a_2_2 a_2_3 a_2_4 a_2_5 a_2_6;...
a_3_1 a_3_2 a_3_3 a_3_4 a_3_5 a_3_6;...
a_4_1 a_4_2 a_4_3 a_4_4 a_4_5 a_4_6;...
a_5_1 a_5_2 a_5_3 a_5_4 a_5_5 a_5_6;...
a_6_1 a_6_2 a_6_3 a_6_4 a_6_5 a_6_6]

Then simply use the inv and det commands to determine what you seek for:

X_det = det(X);
X_inv = inv(X);

Keep in mind that the inverse answer will be very long and may extened beyond the matlab command window allowable space.

[prabakaran Rajendiran]

Mohammad,

asper ur instruction i check the syms toolbox. but i dont have that one how can i get the tool box and how to add the tool box in my matlab.

[John D'Errico]

You will need to buy it from The MathWorks.

[prabakaran Rajendiran]

my college have full suite with 50 license

Answer 2

[Manvi Goel]

You can calculate the inverse and determinant of a any n*n matrix using these inbuilt functions:

det(x)
inv(x)

Answer 3

[Alex Mcaulley]

If you have the symbolic toolbox, you can do something like this:

syms a11 a12 a21 a22

x = [a11,a12;a21,a22];

det(x)

ans =
 
a11*a22 - a12*a21

adjoint(x)

ans =
 
[  a22, -a12]
[ -a21,  a11]

inv(x)

ans =
 
[  a22/(a11*a22 - a12*a21), -a12/(a11*a22 - a12*a21)]
[ -a21/(a11*a22 - a12*a21),  a11/(a11*a22 - a12*a21)]