Question

If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, then $A^{-1} =$

• A. $-\frac{1}{2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$
• B. $\frac{1}{2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$
• C. $-\frac{1}{2} \begin{pmatrix} 1 & -2 \\ -3 & 4 \end{pmatrix}$
• D. $\frac{1}{2} \begin{pmatrix} 1 & -2 \\ -3 & 4 \end{pmatrix}$

Hint:

For $2 \times 2$ matrices, swap the main diagonal elements, change signs of off-diagonal elements, and divide by the determinant.

Answer 1

The Correct Option is A

Step 1: Recall the 2x2 inverse rule.
For a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the inverse is one over the determinant times a swapped matrix.
\[ M^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

Step 2: Find the determinant.
For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, compute $ad - bc$.
\[ \det(A) = (1)(4) - (2)(3) = 4 - 6 = -2 \]

Step 3: Build the swapped matrix.
Swap the two diagonal entries (the $1$ and $4$) and change the signs of the other two (the $2$ and $3$).
\[ \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} \]

Step 4: Divide by the determinant.
Multiply the swapped matrix by $\frac{1}{-2}$.
\[ A^{-1} = \frac{1}{-2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} \]

Step 5: Write it neatly.
Since $\frac{1}{-2} = -\frac{1}{2}$, pull the minus out front.
\[ A^{-1} = -\frac{1}{2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} \]

Step 6: State the answer.
That is the required inverse.
\[ \boxed{-\tfrac{1}{2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}} \]