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: Approach
Check each option by multiplying with $A$ and looking for the identity.

Step 2: Test option (D)
Take $B=-\dfrac16\begin{bmatrix}4&5\\2&1\end{bmatrix}$. The first row of $A$ is $(1,-5)$.
Row 1 times column 1 of $B$: $-\dfrac16(4-10)=1$. Row 1 times column 2: $-\dfrac16(5-5)=0$.
Row 2 of $A$ is $(-2,4)$. Row 2 times column 1: $-\dfrac16(-8+8)=0$. Row 2 times column 2: $-\dfrac16(-10+4)=1$.

Step 3: Conclusion
$AB=I$, so $B=A^{-1}$. Option (D).

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