Question

Let \(X=\begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix}\), \(Y=\begin{bmatrix} -1 & 2 \\ 1 & -3 \end{bmatrix}\), then \((XY)^{-1}\) is equal to

• A. \(\begin{bmatrix} \frac{11}{14} & \frac{-2}{14} \\ \frac{-4}{14} & \frac{-2}{14} \end{bmatrix}\)
• B. \(\begin{bmatrix} \frac{-11}{14} & \frac{2}{14} \\ \frac{-4}{14} & \frac{-2}{14} \end{bmatrix}\)
• C. \(\begin{bmatrix} \frac{-11}{14} & \frac{-2}{14} \\ \frac{4}{14} & \frac{-2}{14} \end{bmatrix}\)
• D. \(\begin{bmatrix} \frac{-11}{14} & \frac{-2}{14} \\ \frac{-4}{14} & \frac{-2}{14} \end{bmatrix}\)

Hint:

Multiply the matrices first, then find the inverse using the adjoint and determinant.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a \(2\times2\) matrix \(M\), \(M^{-1}=\dfrac{1}{|M|}\,\text{adj}(M)\). First find the product \(XY\).

Step 2: Find the Product
\(XY=\begin{bmatrix}4(-1)+2(1)&4(2)+2(-3)\\-1(-1)+3(1)&-1(2)+3(-3)\end{bmatrix}=\begin{bmatrix}-2&2\\4&-11\end{bmatrix}\).

Step 3: Determinant and Adjoint
\(|XY|=(-2)(-11)-(2)(4)=22-8=14\).
\(\text{adj}(XY)=\begin{bmatrix}-11&-2\\-4&-2\end{bmatrix}\).

Step 4: Result
\((XY)^{-1}=\dfrac{1}{14}\begin{bmatrix}-11&-2\\-4&-2\end{bmatrix}\), which is option (D).

Final Answer:
The inverse is one fourteenth of the matrix with rows (-11, -2) and (-4, -2). \[ \boxed{\text{(D) }\begin{bmatrix} \dfrac{-11}{14} & \dfrac{-2}{14} \\ \dfrac{-4}{14} & \dfrac{-2}{14} \end{bmatrix}} \]