Question:
Let \( A, B \) be \( 3^{\text{rd}} \) order non-singular square matrices and \( K \) is a real number. Which of the following is true?
Let \( A, B \) be \( 3^{\text{rd}} \) order non-singular square matrices and \( K \) is a real number. Which of the following is true?
Show Hint
Keep these crucial adjugate shortcut rules handy for an \( n \times n \) matrix:
\[ \text{adj}(KA) = K^{n-1} \text{adj}(A) \]
\[ |\text{adj}(A)| = |A|^{n-1} \]
\[ \text{adj}(\text{adj}(A)) = |A|^{n-2}A \]
Updated On: Jun 8, 2026
- \( \text{Adj}(AB)=(\text{Adj } B)(\text{Adj } A) \) and \( \text{adj}(A^{-1}) \neq (\text{adj } A)^{-1} \)
- \( \text{Adj}(KA)=K \text{ Adj}(A) \) and \( |KA|=K^{3}|A| \)
- \( |B^{-1}AB|=|A| \) and \( (A+B)^{2}=A^{2}+2AB+B^{2} \)
- \( (\text{adj } A)^{-1}=\frac{A}{|A|} \) and \( (AB)^{-1}=B^{-1}A^{-1} \)
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Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
Let us recall the standard algebraic and adjugate properties for a non-singular matrix \( A \) of order \( n \):
• \( A \cdot \text{adj}(A) = |A|I \implies \text{adj}(A) = |A|A^{-1} \)
• Taking the inverse: \( (\text{adj } A)^{-1} = (|A|A^{-1})^{-1} = \frac{1}{|A|}(A^{-1})^{-1} = \frac{A}{|A|} \)
• Reversal law of inverses: \( (AB)^{-1} = B^{-1}A^{-1} \)
Step 1: Verifying option (D).
From the basic property defined above, \( (\text{adj } A)^{-1} = \frac{A}{|A|} \) is perfectly correct. Additionally, the standard reversal rule for matrix inverses states that \( (AB)^{-1} = B^{-1}A^{-1} \). Therefore, both statements in option (D) are true.
Step 2: Analyzing why other options are incorrect.
• In option (A): The property \( \text{adj}(A^{-1}) = (\text{adj } A)^{-1} \) is always true, which makes the statement \( \text{adj}(A^{-1}) \neq (\text{adj } A)^{-1} \) false.
• In option (B): For a matrix of order \( n = 3 \), the scalar property is \( \text{adj}(KA) = K^{n-1}\text{adj}(A) = K^2\text{adj}(A) \). The given equation states it equals \( K\text{adj}(A) \), which is false.
• In option (C): Matrix multiplication is non-commutative (\( AB \neq BA \)), so the expansion of \( (A+B)^2 \) is \( A^2 + AB + BA + B^2 \), making the given expression false.
• \( A \cdot \text{adj}(A) = |A|I \implies \text{adj}(A) = |A|A^{-1} \)
• Taking the inverse: \( (\text{adj } A)^{-1} = (|A|A^{-1})^{-1} = \frac{1}{|A|}(A^{-1})^{-1} = \frac{A}{|A|} \)
• Reversal law of inverses: \( (AB)^{-1} = B^{-1}A^{-1} \)
Step 1: Verifying option (D).
From the basic property defined above, \( (\text{adj } A)^{-1} = \frac{A}{|A|} \) is perfectly correct. Additionally, the standard reversal rule for matrix inverses states that \( (AB)^{-1} = B^{-1}A^{-1} \). Therefore, both statements in option (D) are true.
Step 2: Analyzing why other options are incorrect.
• In option (A): The property \( \text{adj}(A^{-1}) = (\text{adj } A)^{-1} \) is always true, which makes the statement \( \text{adj}(A^{-1}) \neq (\text{adj } A)^{-1} \) false.
• In option (B): For a matrix of order \( n = 3 \), the scalar property is \( \text{adj}(KA) = K^{n-1}\text{adj}(A) = K^2\text{adj}(A) \). The given equation states it equals \( K\text{adj}(A) \), which is false.
• In option (C): Matrix multiplication is non-commutative (\( AB \neq BA \)), so the expansion of \( (A+B)^2 \) is \( A^2 + AB + BA + B^2 \), making the given expression false.
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