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7)$a11, a12 ,a13, a21,a22,a23, a31,a32,a33$

if all diagonal elements are 1,1,1 then no of remaining elements for the matrix (a12,a13,a23)=(a21,a31,a32)= (0,0,1),(0,1,0),(1,0,0)

so 3 symmetric matrices.

if all diagonal elements are 1,0,0 then no of remaining elements for the matrix (a12,a13,a23)=(a21,a31,a32)=(0,1,1),(1,1,0),(1,0,1)

so 3 symmetric matrices.

if all diagonal elements are 0,0,1 then no of remaining elements for the matrix (a12,a13,a23)=(a21,a31,a32)=(0,1,1),(1,1,0),(1,0,1)

so 3 symmetric matrices.

if all diagonal elements are 0,1,0 then no of remaining elements for the matrix (a12,a13,a23)=(a21,a31,a32)=(0,1,1),(1,1,0),(1,0,1)

so 3 symmetric matrices.

so total 12 symmetric matrices.

8)for diagonal elements (1,1,1) only system is having infinite no of solutions.

i.e) x=1,y+z=0--------if z is any real number then y(=-z) also any real number.so infinite solutions.

so atleast 4 and less than 7

9)all 12 matrices are consistent(9 having unique solutions and 3 having infinite no of solutions)
• edited by
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Case1) The diagonal elements need to be (0,0,1) and any triangulat matrix needs to be (1,1,0).

OR

Case 2) The diagonal elements need to be (1,1,1) and any triangulat matrix needs to be (1,0,0).

 

Lets note down all matrices:

Case 1)

[011,100,101]   - inconsistent

[011,110,100]   - unique

[111,100,100]   - multiple solns

[010,101,011]   - unique

[010,111,010]   - inconsistent

[110,101,010]   - unique

[001,001,111]   - inconsistent

[001,011,110]   - unique

[101,001,110]   - unique

Case 2)

[100,011,011]   - multiple solns

[101,010,101]   - inconsistent

[110,110,001]   - inconsistent
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