Question

Let \(A = \begin{bmatrix} 3 & 5\\ -2 & -3 \end{bmatrix}\). If \(BA^2 = A\), where \(B\) is a \(2 \times 2\) matrix, then \(B = \)

• A. \(\begin{bmatrix} -3 & -5\\ 2 & 3 \end{bmatrix}\)
• B. \(\begin{bmatrix} 3 & -5 \\ 2 & -3 \end{bmatrix}\)
• C. \(\begin{bmatrix} 3 & -2 \\ 5 & -3 \end{bmatrix}\)
• D. \(\begin{bmatrix} 3 & 2 \\ -5 & -3 \end{bmatrix}\)
• E. \(\begin{bmatrix} 3 & 5 \\ -2 & -3 \end{bmatrix}\)

Hint:

To find the inverse of a \(2 \times 2\) matrix quickly: swap the diagonal elements, change the signs of the off-diagonal elements, and divide by the determinant.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given a matrix equation \(BA^2 = A\). If matrix \(A\) is invertible, we can simplify this equation by multiplying both sides by the inverse of \(A\).

Step 2: Key Formula or Approach:
1. Inverse of a \(2 \times 2\) matrix: If \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), then \(A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b -c & a \end{bmatrix}\).
2. If \(BA^2 = A\), then post-multiplying by \(A^{-1}\): \(BA = I \implies B = A^{-1}\).

Step 3: Detailed Explanation:
First, check if \(A\) is invertible by finding its determinant:
\[ |A| = (3)(-3) - (5)(-2) = -9 + 10 = 1 \]
Since \(|A| \neq 0\), \(A\) is invertible.
From \(BA^2 = A\), multiply both sides by \(A^{-1}\) on the right:
\[ (BA^2)A^{-1} = AA^{-1} \]
\[ BA(AA^{-1}) = I \]
\[ BA(I) = I \implies BA = I \]
This means \(B\) is the inverse of \(A\):
\[ B = A^{-1} = \frac{1}{1} \begin{bmatrix} -3 & -5 \\ -(-2) & 3 \end{bmatrix} \]
\[ B = \begin{bmatrix} -3 & -5 \\ 2 & 3 \end{bmatrix} \]

Step 4: Final Answer:
The matrix \(B\) is \(\begin{bmatrix} -3 & -5\\ 2 & 3 \end{bmatrix}\).