(i) Case I : All three diagonal elements are 1

No. of matrices = 3 C 1 = 3

Case II : Two diagonal elements are zero & one element is one

No. of matrices = 3 C 1 . 3 C 1 = 9

Total matrices = 3 + 9 = 12

(ii) A =

For unique solution det ≠ 0

Case I

det(a) = = 1 – a 2 – b 2 – c 2 + 2abc ≠ 0

Here a, b, c is selected from 1, 0, 0. (No case is possible)

Case II

(i) det(a) = = 2abc – c 2 ≠ 0

Here a, b, c are selected from 1, 1, 0. (2 cases are possible)

(ii) det(a) = = 2abc – b 2 ≠ 0

Here a, b, c are selected from 1, 1, 0. (2 cases are possible)

(iii) det(a) = = 2abc – a 2 ≠ 0

Here a, b, c are selected from 1, 1, 0. (2 cases are possible)

Hence there are exactly 6 matrices for unique solution . Hence option B is correct

(iii) A =

Case I :

= a, b, c are selected from 1, 0, 0

⇒ x + ay + bz = 1

ax + y + cz = 0

bx + cy + z = 0

(i) If a = 1 , b = c = 0

then x + y = 1 Inconsitent system of equation

x + y = 0

(ii) If a = 0 = c, b = 1

then x + z = 1

y = 0 Inconsitent system of equation

x + z = 0

(iii) If c = 1, a = b = 0

then x = 1, z = 0, y = 0

Case II :

(i) = a, b, c are selected from 1, 1, 0

⇒ x + ay + bz = 1

ax + cz = 0

bx + cy = 0 Clearly, In all three cases, solutions are possible so system is consistent.

(ii) =

⇒ ay + bz = 1

ax + y + cz = 0

bx + cy = 0

Clearly , b = 0, a= c= 1 gives y = 1

x + y + z = 0 Inconsistent system y = 0

More than 2 matrices are possible. Hence option B is correct

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