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: Write the inverse clearly.
We are given $A^{-1}=-\tfrac{1}{2}\begin{bmatrix}1&-4\\-1&2\end{bmatrix}=\begin{bmatrix}-\tfrac{1}{2}&2\\\tfrac{1}{2}&-1\end{bmatrix}$.
Step 2: Plan to recover A.
Since $A$ is the inverse of $A^{-1}$, we can invert $A^{-1}$ using the quick $2\times 2$ rule: swap the diagonal, negate the off-diagonal, divide by the determinant.
Step 3: Find the determinant of the inverse.
$\det(A^{-1})=(-\tfrac{1}{2})(-1)-(2)(\tfrac{1}{2})=\tfrac{1}{2}-1=-\tfrac{1}{2}$.
Step 4: Apply the inverse rule.
Swap diagonal and negate off-diagonal: $\begin{bmatrix}-1&-2\\-\tfrac{1}{2}&-\tfrac{1}{2}\end{bmatrix}$. Divide by $-\tfrac{1}{2}$, which means multiply by $-2$: $A=\begin{bmatrix}2&4\\1&1\end{bmatrix}$.
Step 5: Build 2A.
Doubling each entry gives $2A=\begin{bmatrix}4&8\\2&2\end{bmatrix}$.
Step 6: Add the identity.
Adding $I_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ gives $\begin{bmatrix}5&8\\2&3\end{bmatrix}$, which is option (D).
\[ \boxed{\begin{bmatrix}5&8\\2&3\end{bmatrix}} \]