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Let A = [aij]n×n[a_{ij}]_{n \times n} be a matrix. Then

Match List-I with List-II

List-IList-II
(A) AT=AA^T = A(I) A is a singular matrix
(B) AT=−AA^T = -A(II) A is a non-singular matrix
(C) ∣A∣=0\vert A\vert = 0(III) A is a skew symmetric matrix
(D) ∣A∣≠0\vert A\vert \neq 0(IV) A is a symmetric matrix

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

Let's match the matrix properties with their correct definitions.


First, let's understand what each property in List-I means:

(A)(A) AT=AA^T = A means when you transpose the matrix (flip across diagonal), it stays the same

(B)(B) AT=−AA^T = -A means when you transpose the matrix, every element changes sign

(C)(C) ∣A∣=0|A| = 0 means the determinant of the matrix equals zero

(D)(D) ∣A∣≠0|A| \neq 0 means the determinant of the matrix is not zero


Now let's recall the definitions in List-II:

(I)(I) Singular matrix: A matrix with determinant =0= 0

(II)(II) Non-singular matrix: A matrix with determinant ≠0\neq 0

(III)(III) Skew symmetric matrix: A matrix where AT=−AA^T = -A

(IV)(IV) Symmetric matrix: A matrix where AT=AA^T = A


Matching them directly:

(A)(A) AT=AA^T = A matches with (IV)(IV) Symmetric matrix

(B)(B) AT=−AA^T = -A matches with (III)(III) Skew symmetric matrix

(C)(C) ∣A∣=0|A| = 0 matches with (I)(I) Singular matrix

(D)(D) ∣A∣≠0|A| \neq 0 matches with (II)(II) Non-singular matrix


Answer: A→IVA \rightarrow IV, B→IIIB \rightarrow III, C→IC \rightarrow I, D→IID \rightarrow II

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