Question

If \( A = \begin{bmatrix} 2 & 1 5 & 3 \end{bmatrix} \), then \( A^{-1} = \) ____.

• A. \( \begin{bmatrix} 3 & -1 -5 & 2 \end{bmatrix} \)
• B. \( \begin{bmatrix} 3 & 1 -5 & 2 \end{bmatrix} \)
• C. \( \begin{bmatrix} 3 & 1 5 & 2 \end{bmatrix} \)
• D. None

Hint:

When the determinant $|A| = 1$, the inverse of the matrix is simply its adjoint. Always check the determinant first; if it's 1, you can write the answer by inspection!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The inverse of a $2 \times 2$ matrix exists only if its determinant is non-zero. The inverse is calculated by swapping the main diagonal elements, changing the signs of the off-diagonal elements, and dividing by the determinant.

Step 2: Key Formula or Approach:
For a matrix \( A = \begin{bmatrix} a & b c & d \end{bmatrix} \), the inverse is: \[ A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{ad - bc} \begin{bmatrix} d & -b -c & a \end{bmatrix} \]

Step 3: Detailed Explanation:
Given \( A = \begin{bmatrix} 2 & 1 5 & 3 \end{bmatrix} \): 1. Find the determinant \( |A| \): \[ |A| = (2 \times 3) - (1 \times 5) = 6 - 5 = 1 \] 2. Find the Adjoint of A: Swap diagonal elements ($2$ and $3$) and change signs of $1$ and $5$: \[ \text{adj}(A) = \begin{bmatrix} 3 & -1 -5 & 2 \end{bmatrix} \] 3. Calculate \( A^{-1} \): \[ A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -1 -5 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -1 -5 & 2 \end{bmatrix} \]

Step 4: Final Answer:
The inverse matrix is \( \begin{bmatrix} 3 & -1 -5 & 2 \end{bmatrix} \).

Answer 2

Instead of computing the inverse from scratch, we can verify which of the given matrices is genuinely the inverse of \( A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} \) by directly multiplying \( A \) with each candidate and checking whether the product is the identity matrix \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).

1. \( \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix} \): Multiplying, the top-left entry is \( (2)(3) + (1)(-5) = 6 - 5 = 1 \), top-right is \( (2)(-1) + (1)(2) = -2 + 2 = 0 \), bottom-left is \( (5)(3) + (3)(-5) = 15 - 15 = 0 \), and bottom-right is \( (5)(-1) + (3)(2) = -5 + 6 = 1 \). The product is exactly \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), so this is indeed \( A^{-1} \).
2. \( \begin{bmatrix} 3 & 1 \\ -5 & 2 \end{bmatrix} \): The top-right entry of the product is \( (2)(1) + (1)(2) = 2 + 2 = 4 \), which is not \( 0 \). Since a true inverse must give exactly the identity matrix, this candidate fails.
3. \( \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} \): The top-left entry of the product is \( (2)(3) + (1)(5) = 6 + 5 = 11 \), nowhere near \( 1 \). This candidate is not the inverse either.
4. None: Since the first candidate multiplies with \( A \) to give exactly the identity matrix, an inverse does exist among the given options, so this choice is incorrect.

Only the first matrix satisfies \( A \cdot A^{-1} = I \) exactly.

Therefore, the correct answer is \( \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix} \).