Question

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

• A. \( \begin{bmatrix}2 & -1 -3 & 2 \end{bmatrix} \)
• B. \( \begin{bmatrix}2 & 1 3 & 2 \end{bmatrix} \)
• C. \( \begin{bmatrix}2 & -1 3 & -2 \end{bmatrix} \)
• D. \( \begin{bmatrix}-2 & 1 3 & -2 \end{bmatrix} \)
• E. \( \begin{bmatrix}1 & 0 0 & 1 \end{bmatrix} \)

Hint:

For \(2 \times 2\) matrices, always remember: swap diagonal, change signs of off-diagonal, then divide by determinant.

Answer 1

The Correct Option is A

Concept: To find the inverse of a \(2 \times 2\) matrix, we use the formula: \[ A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b -c & a \end{bmatrix} \] where \( |A| = ad - bc \).

Step 1: First, identify the elements of the matrix.
The given matrix is: \[ A = \begin{bmatrix}2 & 1 3 & 2 \end{bmatrix} \] So we compare with general form \( \begin{bmatrix}a & b c & d\end{bmatrix} \), giving: \[ a=2,\; b=1,\; c=3,\; d=2 \]

Step 2: Now compute the determinant.
\[ |A| = (2)(2) - (1)(3) = 4 - 3 = 1 \] This step is very important because the inverse exists only if determinant is non-zero.

Step 3: Since determinant is \(1 \neq 0\), inverse exists.
This confirms we can proceed safely.

Step 4: Apply inverse formula carefully.
We swap the diagonal elements \(a\) and \(d\), and change the sign of off-diagonal elements: \[ \begin{bmatrix} d & -b -c & a \end{bmatrix} = \begin{bmatrix} 2 & -1 -3 & 2 \end{bmatrix} \]

Step 5: Multiply by \( \frac{1}{|A|} \).
Since determinant is 1: \[ A^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -1 -3 & 2 \end{bmatrix} \]

Step 6: Final answer: \[ \boxed{\begin{bmatrix} 2 & -1 -3 & 2 \end{bmatrix}} \]