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: Understanding the Concept:
For \(A = \begin{bmatrix} p & q \\ r & s \end{bmatrix}\), \(A^{-1} = \dfrac{1}{ps - qr}\begin{bmatrix} s & -q \\ -r & p \end{bmatrix}\).

Step 2: Determinant:
\[ |A| = 3\cdot4 - (-2)(1) = 12 + 2 = 14 \]

Step 3: Adjoint and inverse:
Swap the diagonal entries and change the signs of the off-diagonal entries:
\[ \operatorname{adj}A = \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix} \]
\[ A^{-1} = \frac{1}{14}\begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} \tfrac27 & \tfrac17 \\ -\tfrac1{14} & \tfrac3{14} \end{bmatrix} \]

Step 4: Check:
Row 1 times column 1 of \(A\): \(\tfrac27\cdot3 + \tfrac17\cdot1 = \tfrac{6+1}{7} = 1\). This matches option (A); the other options have wrong signs or entries.

Final Answer:
The inverse is (1/14) times [[4, 2], [-1, 3]]. \[ \boxed{\text{(A) }\begin{bmatrix} 2/7 & 1/7 \\ -1/14 & 3/14 \end{bmatrix}} \]