Question:

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

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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!
Updated On: May 19, 2026
  • \( \begin{bmatrix} 3 & -1 -5 & 2 \end{bmatrix} \)
  • \( \begin{bmatrix} 3 & 1 -5 & 2 \end{bmatrix} \)
  • \( \begin{bmatrix} 3 & 1 5 & 2 \end{bmatrix} \)
  • None
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The Correct Option is A

Solution and Explanation

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 \( \begin{bmatrix} 3 & -1 -5 & 2 \end{bmatrix} \).
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