Question

If \(A = [\begin{array}{cc}1 & -5 \\ -2 & 4\end{array}]\), then \(A^{-1} =\)

• A. \(-\frac{1}{6}[\begin{array}{cc}-4 & 5 \\ 2 & -1\end{array}]\)
• B. \(\frac{1}{14}[\begin{array}{cc}-1 & 5 \\ 2 & -4\end{array}]\)
• C. \(\frac{1}{14}[\begin{array}{cc}4 & 5 \\ 2 & 1\end{array}]\)
• D. \(-\frac{1}{6}[\begin{array}{cc}4 & 5 \\ 2 & 1\end{array}]\)

Hint:

Inverse of a 2 by 2 matrix: swap the diagonal, change signs of the others, divide by the determinant.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For \(A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\), the inverse is \(\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\).

Step 2: Determinant
\[ |A|=(1)(4)-(-5)(-2)=4-10=-6 \]

Step 3: Adjoint
Swap the diagonal entries and change the sign of the other two:
\[ \operatorname{adj}A=\begin{bmatrix}4&5\\2&1\end{bmatrix} \]

Step 4: Inverse
\[ A^{-1}=-\frac16\begin{bmatrix}4&5\\2&1\end{bmatrix} \]

Step 5: Check
Multiply \(A\) by this: first row first column is \((4-10)/(-6)=1\). The off-diagonal \((5-5)/(-6)=0\). So it is correct, option (D).

Final Answer:
The inverse is \(-\dfrac16\begin{bmatrix}4&5\\2&1\end{bmatrix}\), option (D). \[ \boxed{-\frac{1}{6}\begin{bmatrix}4&5\\2&1\end{bmatrix}} \]