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?

• A. \( \text{Adj}(AB)=(\text{Adj } B)(\text{Adj } A) \) and \( \text{adj}(A^{-1}) \neq (\text{adj } A)^{-1} \)
• B. \( \text{Adj}(KA)=K \text{ Adj}(A) \) and \( |KA|=K^{3}|A| \)
• C. \( |B^{-1}AB|=|A| \) and \( (A+B)^{2}=A^{2}+2AB+B^{2} \)
• D. \( (\text{adj } A)^{-1}=\frac{A}{|A|} \) and \( (AB)^{-1}=B^{-1}A^{-1} \)

Hint:

Keep these crucial adjugate shortcut rules handy for an \( n \times n \) matrix: \[ \text{adj}(KA) = K^{n-1} \text{adj}(A) \] \[ |\text{adj}(A)| = |A|^{n-1} \] \[ \text{adj}(\text{adj}(A)) = |A|^{n-2}A \]

Answer 1

The Correct Option is D

Step 1: List the key matrix rules.
For a non-singular matrix we know $A\cdot \text{adj}(A) = |A|I$, so $\text{adj}(A) = |A|\,A^{-1}$. Also the inverse of a product flips the order.
Step 2: Build the first part of option D.
Take the inverse of $\text{adj}(A)=|A|A^{-1}$: \[ (\text{adj }A)^{-1} = \frac{1}{|A|}(A^{-1})^{-1} = \frac{A}{|A|} \] So this statement is true.
Step 3: Check the second part of option D.
The reversal law for inverses says \[ (AB)^{-1} = B^{-1}A^{-1} \] This is a standard true rule. So both parts of D are correct.
Step 4: Test option A.
The rule $\text{adj}(A^{-1}) = (\text{adj }A)^{-1}$ is always true, so saying they are not equal is false. Option A fails.
Step 5: Test option B.
For order 3, the correct rule is $\text{adj}(KA) = K^{2}\text{adj}(A)$, not $K\,\text{adj}(A)$. So option B is wrong.
Step 6: Test option C and conclude.
Matrices do not always commute, so $(A+B)^2 = A^2 + AB + BA + B^2$, not $A^2+2AB+B^2$. Option C is wrong. Only D survives. \[ \boxed{(\text{adj }A)^{-1}=\tfrac{A}{|A|},\ (AB)^{-1}=B^{-1}A^{-1}} \]