Question

If A=\([\begin{smallmatrix} 2 & -1 \\ 3 & -2 \end{smallmatrix}]\) ,then the inverse of the matrix A³ is

• A. A
• B. 1
• C. -1
• D. -A

Answer 1

The Correct Option is A

We are given: \[ A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix} \] We are asked to find the inverse of \( A^3 \), i.e.: \[ (A^3)^{-1} \] Using the identity: \[ (AB)^{-1} = B^{-1} A^{-1} \] we get: \[ (A^3)^{-1} = (A \cdot A \cdot A)^{-1} = A^{-1} \cdot A^{-1} \cdot A^{-1} = (A^{-1})^3 \] Now suppose \( A^{-1} = -A \). Then: \[ (A^{-1})^3 = (-A)^3 = -A^3 \] So: \[ (A^3)^{-1} = -A^3 \Rightarrow \boxed{(A^3)^{-1} = -A^3} \Rightarrow A^3 \cdot (A^3)^{-1} = A^3 \cdot (-A^3) = -I \] But this contradicts the identity property. So let's compute the inverse of \( A \) directly. Let: \[ A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix} \] The inverse of a 2x2 matrix is: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] Here, \[ \text{det}(A) = (2)(-2) - (-1)(3) = -4 + 3 = -1 \] So: \[ A^{-1} = \frac{1}{-1} \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix} = A \] Thus, \[ A^{-1} = A \Rightarrow (A^3)^{-1} = (A^{-1})^3 = A^3 \] Correct answer: A 

Answer 2

We are given the matrix A = \(\begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}\) and we need to find the inverse of A³, i.e., (A³)^-1.

First, let's find the inverse of A (A^-1).

For a 2x2 matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\), the inverse is given by \(\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\)

The determinant of A is (2 * -2) - (-1 * 3) = -4 + 3 = -1.

Therefore, A^-1 = \(\frac{1}{-1}\begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}\)

Notice that A^-1 = A. This means A is an involutory matrix. Therefore, A² = I, where I is the identity matrix.

Now, we need to find (A³)^-1.

(A³)^-1 = (A * A * A)^-1 = A^-1 * A^-1 * A^-1 = (A^-1)³

Since A^-1 = A,

(A³)^-1 = A³ = A² * A = I * A = A

So (A³)^-1 = A

Answer: A