Question

If \(A = \begin{bmatrix} 4 & 2 \\[0.3em] -3 & 3 \end{bmatrix}\), then \(A^{-1} =\)

• A. \( \frac{1}{6}(7I - A) \)
• B. \( \frac{1}{4}(5I - A) \)
• C. \( \frac{1}{3}(7I - A) \)
• D. \( \frac{1}{18}(7I - A) \)

Hint:

Use the identity \( A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) \), and remember that expressions like \( (kI - A) \) can simplify inverse computations.

Answer 1

The Correct Option is D

Let’s verify by using the identity \( A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \).

Given \(A = \begin{bmatrix} 4 & 2 \\[0.3em] -3 & 3 \end{bmatrix}\)

Determinant of A: \[ \text{det}(A) = (4)(3) - (-3)(2) = 12 + 6 = 18 \]

Adjoint of A: \(\text{adj}(A) = \begin{bmatrix} 3 & -2 \\[0.3em] 3 & 4 \end{bmatrix}\)

Thus,\(A^{-1} = \frac{1}{18} \begin{bmatrix} 3 & -2 \\[0.3em] 3 & 4 \end{bmatrix}\)

Now checking:

 

So: \(A^{-1} = \frac{1}{18}(7I - A)\)