Question

If $A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}, B = \begin{bmatrix} 1 & 1 \\ 4 & -1 \end{bmatrix}$, then $(A+B)^{-1}$ is

• A. $\begin{bmatrix} -\frac{1}{2} & 0 \\ -\frac{3}{2} & \frac{1}{2} \end{bmatrix}$
• B. $\begin{bmatrix} \frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$
• C. $\begin{bmatrix} \frac{1}{2} & 0 \\ -\frac{3}{2} & \frac{1}{2} \end{bmatrix}$
• D. $\begin{bmatrix} \frac{1}{2} & 0 \\ \frac{3}{2} & \frac{1}{2} \end{bmatrix}$

Hint:

To find the inverse of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, first find its determinant $ad-bc$. If the determinant is non-zero, the inverse is $\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

Answer 1

The Correct Option is B

Step 1: First, calculate the sum of matrices $A$ and $B$, i.e., $A+B$.
\[ A+B = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 1 \\ 4 & -1 \end{bmatrix} = \begin{bmatrix} 1+1 & -1+1 \\ 2+4 & -1-1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 6 & -2 \end{bmatrix} \]
Step 2: Next, calculate the determinant of the matrix $A+B$.
For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is $ad - bc$. \[ \det(A+B) = (2)(-2) - (0)(6) = -4 - 0 = -4 \] Since the determinant is non-zero ($-4 \neq 0$), the inverse exists. 
Step 3: Use the formula for the inverse of a $2 \times 2$ matrix.
If $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then \[ M^{-1} = \frac{1}{\det(M)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] 
Step 4: Apply the formula to find $(A+B)^{-1}$.
\[ (A+B)^{-1} = \frac{1}{-4} \begin{bmatrix} -2 & 0 \\ -6 & 2 \end{bmatrix} \] Multiply each element by $-\frac{1}{4}$: \[ (A+B)^{-1} = \begin{bmatrix} \frac{-2}{-4} & \frac{0}{-4} \\ \frac{-6}{-4} & \frac{2}{-4} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} \]