Question:
If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, then $A^{-1} =$
If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, then $A^{-1} =$
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For $2 \times 2$ matrices, swap the main diagonal elements, change signs of off-diagonal elements, and divide by the determinant.
Updated On: May 31, 2026
- $-\frac{1}{2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$
- $\frac{1}{2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$
- $-\frac{1}{2} \begin{pmatrix} 1 & -2 \\ -3 & 4 \end{pmatrix}$
- $\frac{1}{2} \begin{pmatrix} 1 & -2 \\ -3 & 4 \end{pmatrix}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The inverse of a \(2 \times 2\) matrix \[ A=\begin{pmatrix} a & b\\ c & d \end{pmatrix} \] is given by:
Step 2: Meaning
To find the inverse of a \(2 \times 2\) matrix:
- Calculate the determinant \(\det(A)=ad-bc\).
- Swap the diagonal elements.
- Change the signs of the off-diagonal elements.
- Divide the resulting matrix by the determinant.
Step 3: Analysis
Given:
\[ A=\begin{pmatrix} 1 & 2\\ 3 & 4 \end{pmatrix} \]
First, calculate the determinant:
\[ \det(A)=(1)(4)-(2)(3)=4-6=-2 \]
Now find the adjoint matrix:
\[ \operatorname{adj}(A)= \begin{pmatrix} 4 & -2\\ -3 & 1 \end{pmatrix} \]
Therefore,
\[ A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2\\ -3 & 1 \end{pmatrix} = -\frac{1}{2} \begin{pmatrix} 4 & -2\\ -3 & 1 \end{pmatrix} \]
Step 4: Conclusion
The inverse of the matrix \(A\) is:
\[ A^{-1} = -\frac{1}{2} \begin{pmatrix} 4 & -2\\ -3 & 1 \end{pmatrix} \]
Final Answer: (A)
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