Question

If \(A = \left[ \begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array} \right]\) then which of the following is true ?

• A. \(A^2 = A^{-1}\)
• B. \(A = -(\text{adj}A)\)
• C. \(A^{-1} = A\)
• D. \(A^{-1} = -A\)

Hint:

Square the matrix. If A times A gives the identity, A is its own inverse.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a 3 x 3 matrix \(A\) with -1 on the anti-diagonal and 0 elsewhere. We must find which relation among \(A^2\), \(A^{-1}\), \(\text{adj}A\) and \(A\) is true.

Step 2: Key Formula or Approach:
If \(A^2 = I\) then \(A\cdot A = I\), so \(A\) is its own inverse. Also \(\text{adj}A = |A|\,A^{-1}\).

Step 3: Compute \(A^2\):
Row 1 of \(A\) is \((0,0,-1)\). Multiplying it with the columns of \(A\) gives \((-1)(-1)=1\) in column 1 and 0 in the others. Row 2 \((0,-1,0)\) gives 1 only in column 2. Row 3 \((-1,0,0)\) gives 1 only in column 3. So \[ A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \]

Step 4: Conclude about the inverse:
Since \(A\cdot A = I\), the inverse of \(A\) is \(A\) itself, so \(A^{-1} = A\). This is option (C).

Step 5: Check the other options:
Option (A): \(A^2 = I\), but \(A^{-1} = A \neq I\), so \(A^2 \neq A^{-1}\). Option (D): \(A^{-1} = -A\) would mean \(A = -A\), which is false because \(A \neq 0\). Option (B): \(|A| = (-1)\cdot(-1)^3 = 1\), since the anti-diagonal arrangement is an odd permutation. So \(\text{adj}A = |A|A^{-1} = A\), not \(-A\), and (B) is false.

Final Answer:
Only \(A^{-1} = A\) is true. \[ \boxed{A^{-1}=A} \]