Question

If $ A = \begin{bmatrix} 2 & 2 \\3 & 4 \end{bmatrix}, \quad \text{then} \quad A^{-1} \text{ equals to} $

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

Hint:

For any 2x2 matrix, the inverse can be found using the formula \(\frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\-c & a \end{bmatrix}\), provided the determinant is not zero.

Answer 1

The Correct Option is B

To find the inverse of matrix \( A = \begin{bmatrix} 2 & 2 \\ 3 & 4 \end{bmatrix} \), we follow these steps:

1. Calculate the Determinant:
\[ |A| = (2)(4) - (2)(3) = 8 - 6 = \boxed{2} \]

2. Find the Matrix of Cofactors:
\[ \begin{align*} A_{11} &= +1 \cdot 4 = \boxed{4} \\ A_{12} &= -1 \cdot 3 = \boxed{-3} \\ A_{21} &= -1 \cdot 2 = \boxed{-2} \\ A_{22} &= +1 \cdot 2 = \boxed{2} \end{align*} \]

3. Form the Adjugate Matrix:
Transpose the cofactor matrix:
\[ \text{adj } A = \begin{bmatrix} 4 & -2 \\ -3 & 2 \end{bmatrix} \]

4. Compute the Inverse:
Using the formula \( A^{-1} = \frac{1}{|A|} \text{adj } A \):
\[ A^{-1} = \frac{1}{2} \begin{bmatrix} 4 & -2 \\ -3 & 2 \end{bmatrix} = \boxed{\begin{bmatrix} 2 & -1 \\ -\frac{3}{2} & 1 \end{bmatrix}} \]

Key Verification:
We can verify by checking \( AA^{-1} = I \):
\[ \begin{bmatrix} 2 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -\frac{3}{2} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Answer 2

To find \(A^{-1}\), we use the formula for the inverse of a 2x2 matrix: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b -c & a \end{bmatrix} \] where \(a = 2, b = 2, c = 3, d = 4\). The determinant is: \[ \text{det}(A) = ad - bc = 2(4) - 2(3) = 8 - 6 = 2 \] Now, using the formula for the inverse: \[ A^{-1} = \frac{1}{2} \begin{bmatrix} 4 & -2 \\-3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\-3/2 & 1 \end{bmatrix} \]