Q69 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
If $A=\left[\begin{array}{ll}2 & 2 \\ 3 & 4\end{array}\right]$, then $A^{-1}$ equals to
- A$\left[\begin{array}{cc}2 & 1 \\ -3 / 2 & -1\end{array}\right]$
- B$\left[\begin{array}{cc}-2 & -1 \\ 3 / 2 & 1\end{array}\right]$
- C$\left[\begin{array}{cc}-2 & 1 \\ 3 / 2 & -1\end{array}\right]$
- Correct answerD$\left[\begin{array}{cc}2 & -1 \\ -3 / 2 & 1\end{array}\right]$
Solution
Given the matrix $A = \begin{bmatrix} 2 & 2 \\ 3 & 4 \end{bmatrix}$.
The determinant of $A$ is calculated as $|A| = (2 \times 4) - (2 \times 3) = 8 - 6 = 2$.
The inverse of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $A^{-1} = \dfrac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Substituting the values $a=2, b=2, c=3, d=4$ and $|A|=2$, we get:
$A^{-1} = \dfrac{1}{2} \begin{bmatrix} 4 & -2 \\ -3 & 2 \end{bmatrix}$
$A^{-1} = \begin{bmatrix} \dfrac{4}{2} & \dfrac{-2}{2} \\ \dfrac{-3}{2} & \dfrac{2}{2} \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3/2 & 1 \end{bmatrix}$.
Answer: $\left[\begin{array}{cc}2 & -1 \\ -3 / 2 & 1\end{array}\right]$
The determinant of $A$ is calculated as $|A| = (2 \times 4) - (2 \times 3) = 8 - 6 = 2$.
The inverse of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $A^{-1} = \dfrac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Substituting the values $a=2, b=2, c=3, d=4$ and $|A|=2$, we get:
$A^{-1} = \dfrac{1}{2} \begin{bmatrix} 4 & -2 \\ -3 & 2 \end{bmatrix}$
$A^{-1} = \begin{bmatrix} \dfrac{4}{2} & \dfrac{-2}{2} \\ \dfrac{-3}{2} & \dfrac{2}{2} \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3/2 & 1 \end{bmatrix}$.
Answer: $\left[\begin{array}{cc}2 & -1 \\ -3 / 2 & 1\end{array}\right]$
Hence, the correct answer is option: 'D'.