Question

Inverse of \([\begin{array}{cc}3 & -2 \\ 1 & 4\end{array}]\) is

• A. \(\left[ \begin{array}{cc}\frac{2}{7} & \frac{1}{7} \\ -\frac{1}{14} & \frac{3}{14}\end{array} \right]\)
• B. \(\left[ \begin{array}{cc}\frac{3}{14} & -\frac{1}{7} \\ \frac{1}{14} & \frac{2}{7}\end{array} \right]\)
• C. \(\left[ \begin{array}{cc}\frac{2}{7} & -\frac{1}{7} \\ \frac{1}{14} & \frac{3}{14}\end{array} \right]\)
• D. \(\left[ \begin{array}{cc}-\frac{3}{14} & \frac{1}{7} \\ \frac{1}{14} & \frac{2}{7}\end{array} \right]\)

Hint:

Use A inverse = adj A / |A| for a 2 by 2 matrix.

Answer 1

The Correct Option is A

Step 1: Elementary check by product:
Try option (A) as a candidate and multiply $A\cdot A^{-1}$.

Step 2: Multiply:
Row 1: $3\cdot\tfrac27 + (-2)(-\tfrac1{14}) = \tfrac67 + \tfrac17 = 1$; $3\cdot\tfrac17 + (-2)\tfrac3{14} = \tfrac37 - \tfrac37 = 0$. Row 2: $1\cdot\tfrac27 + 4(-\tfrac1{14}) = 0$; $1\cdot\tfrac17 + 4\cdot\tfrac3{14} = \tfrac17+\tfrac67 = 1$.

Step 3: Conclusion:
The product is the identity, so option (A) is the inverse.

Final Answer:
The inverse is option A. \[ \boxed{\text{(A) }\begin{bmatrix} 2/7 & 1/7 \\ -1/14 & 3/14 \end{bmatrix}} \]