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Home Questions HD4V3Y93EX
Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Question:

Match List-I with List-II

List-I

List-II

(A) $\text{A(adj A)}$

(I) $\frac{1}{|A|}$

(B) $\text{|adj A|}$

(II) $|A|^n$

(C) $|A^{-1}|$

(III) $|A|I$

(D) $\text{|A(adj A)|}$

(IV) $|A|^{n-1}$

Choose the correct answer from the options given below:

Options:

(A)-(I), (B)-(IV), (C)-(II), (D)-(III)

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Correct Answer:

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Explanation:

The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Given:

(A) A(adj A)

Property: A(adj A) = |A| I → determinant is $|A(adj A)| = |A|^n$, but the expression itself equals |A|I. So, (A) matches (III)

(B) |adj A|

Property: Determinant of adjoint: $|adj A| = |A|^{n-1}$, where n is the order of matrix. So, (B) matches (IV)

(C) $|A^{-1}|$

Property: Determinant of inverse: $|A^{-1}| = 1/|A|$. So, (C) matches (I)

(D) |A(adj A)|

Property: $|A(adj A)| = |A|^n$. So, (D) matches (II)

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