Q100 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics
Inverse of $\begin{bmatrix} 3 & -2 \\ 1 & 4 \end{bmatrix}$ is
- Correct answerA$\begin{bmatrix} \dfrac{2}{7} & \dfrac{1}{7} \\ -\dfrac{1}{14} & \dfrac{3}{14} \end{bmatrix}$
- B$\begin{bmatrix} \dfrac{3}{14} & -\dfrac{1}{7} \\ \dfrac{1}{14} & \dfrac{2}{7} \end{bmatrix}$
- C$\begin{bmatrix} \dfrac{2}{7} & -\dfrac{1}{7} \\ \dfrac{1}{14} & \dfrac{3}{14} \end{bmatrix}$
- D$\begin{bmatrix} -\dfrac{3}{14} & \dfrac{1}{7} \\ \dfrac{1}{14} & \dfrac{2}{7} \end{bmatrix}$
Solution
Let $A = \begin{bmatrix} 3 & -2 \\ 1 & 4 \end{bmatrix}$
The determinant of $A$ is $|A| = (3)(4) - (-2)(1) = 12 + 2 = 14$
The adjugate of $A$ is $\text{adj}(A) = \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix}$
The inverse of $A$ is given by $A^{-1} = \dfrac{1}{|A|} \text{adj}(A)$
$A^{-1} = \dfrac{1}{14} \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} \dfrac{4}{14} & \dfrac{2}{14} \\ -\dfrac{1}{14} & \dfrac{3}{14} \end{bmatrix} = \begin{bmatrix} \dfrac{2}{7} & \dfrac{1}{7} \\ -\dfrac{1}{14} & \dfrac{3}{14} \end{bmatrix}$
Answer: $\begin{bmatrix} \dfrac{2}{7} & \dfrac{1}{7} \\ -\dfrac{1}{14} & \dfrac{3}{14} \end{bmatrix}$
The determinant of $A$ is $|A| = (3)(4) - (-2)(1) = 12 + 2 = 14$
The adjugate of $A$ is $\text{adj}(A) = \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix}$
The inverse of $A$ is given by $A^{-1} = \dfrac{1}{|A|} \text{adj}(A)$
$A^{-1} = \dfrac{1}{14} \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} \dfrac{4}{14} & \dfrac{2}{14} \\ -\dfrac{1}{14} & \dfrac{3}{14} \end{bmatrix} = \begin{bmatrix} \dfrac{2}{7} & \dfrac{1}{7} \\ -\dfrac{1}{14} & \dfrac{3}{14} \end{bmatrix}$
Answer: $\begin{bmatrix} \dfrac{2}{7} & \dfrac{1}{7} \\ -\dfrac{1}{14} & \dfrac{3}{14} \end{bmatrix}$
Hence, the correct answer is option: 'A'.