Question

If $A^{-1} = \frac{-1}{2} \begin{bmatrix} 1 & -4 \\ -1 & 2 \end{bmatrix}$, then $2A + I_2 = \dots$ where $I_2$ is a unit matrix of order 2.

• A. $\begin{bmatrix} 5 & 8 \\ 1 & 2 \end{bmatrix}$
• B. $\begin{bmatrix} 5 & 8 \\ 2 & 2 \end{bmatrix}$
• C. $\begin{bmatrix} 2 & 4 \\ 1 & 1 \end{bmatrix}$
• D. $\begin{bmatrix} 5 & 8 \\ 2 & 3 \end{bmatrix}$

Hint:

To find the inverse of a $2 \times 2$ matrix quickly: swap the main diagonal elements, change the signs of the off-diagonal elements, and divide everything by the determinant!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the inverse of a $2 \times 2$ matrix $A$, denoted as $A^{-1}$, and need to compute the matrix $2A + I_2$.

Step 2: Key Formula or Approach:
For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is given by $A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
We can reverse this process to find $A$ from its inverse.

Step 3: Detailed Explanation:
Given $A^{-1} = -\frac{1}{2} \begin{bmatrix} 1 & -4 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -1/2 & 2 \\ 1/2 & -1 \end{bmatrix}$.
To find $A$, we can use the property $(A^{-1})^{-1} = A$.
The determinant of $A^{-1}$ is: $\det(A^{-1}) = (-1/2)(-1) - (2)(1/2) = 1/2 - 1 = -1/2$.
$A = \frac{1}{\det(A^{-1})} \begin{bmatrix} -1 & -2 \\ -1/2 & -1/2 \end{bmatrix} = \frac{1}{-1/2} \begin{bmatrix} -1 & -2 \\ -1/2 & -1/2 \end{bmatrix} = -2 \begin{bmatrix} -1 & -2 \\ -1/2 & -1/2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 1 & 1 \end{bmatrix}$.
Now compute $2A + I_2$:
$$2 \begin{bmatrix} 2 & 4 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 2 & 3 \end{bmatrix}$$

Step 4: Final Answer:
The resulting matrix is $\begin{bmatrix} 5 & 8 \\ 2 & 3 \end{bmatrix}$, which corresponds to option (D).