Let $A = [a_{ij}]_{n×n}$ be a matrix, then match List-I with List-II
|
List-I |
List-II |
|
(A) $|A|= 0$ |
(I) A is a symmetric matrix |
|
(B) $|A|≠0$ |
(II) A is a skew-symmetric matrix |
|
(C) $A^T = A$ |
(III) A is a singular matrix |
|
(D) $A^T=-A$ |
(IV) A is a non-singular matrix |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
|
List-I |
List-II |
|
(A) $|A|= 0$ |
(III) A is a singular matrix |
|
(B) $|A|≠0$ |
(IV) A is a non-singular matrix |
|
(C) $A^T = A$ |
(I) A is a symmetric matrix |
|
(D) $A^T=-A$ |
(II) A is a skew-symmetric matrix |
Given: Let A = [aij]n×n be a matrix.
(A) |A| = 0: If the determinant of a matrix is 0, then the matrix is called singular.
So, (A) ⟶ (III)
(B) |A| ≠ 0: If the determinant of a matrix is non-zero, then the matrix is called non-singular.
So, (B) ⟶ (IV)
(C) AT = A: This is the definition of a symmetric matrix.
So, (C) ⟶ (I)
(D) AT = –A: This is the definition of a skew-symmetric matrix.
So, (D) ⟶ (II)