Question

If \(A=\begin{bmatrix}2&3\\1&-4\end{bmatrix},\ B=\begin{bmatrix}1&-2\\-1&3\end{bmatrix}\), then find \((AB)^{-1}\).

Hint:

Find AB first, then apply the 2x2 inverse formula; or use (AB)⁻¹=B⁻¹A⁻¹.

Answer 1

Step 1: Key Formula or Approach:
First compute the product matrix \(AB\), then use the \(2\times2\) inverse formula: for \(M=\begin{bmatrix}p&q\\r&s\end{bmatrix}\), \(M^{-1} = \dfrac{1}{ps-qr}\begin{bmatrix}s&-q\\-r&p\end{bmatrix}\).

Step 2: Computing \(AB\):
\[ AB = \begin{bmatrix}2&3\\1&-4\end{bmatrix}\begin{bmatrix}1&-2\\-1&3\end{bmatrix} = \begin{bmatrix}2(1)+3(-1) & 2(-2)+3(3)\\1(1)+(-4)(-1) & 1(-2)+(-4)(3)\end{bmatrix} = \begin{bmatrix}-1&5\\5&-14\end{bmatrix} \]

Step 3: Finding the determinant of \(AB\):
\[ \det(AB) = (-1)(-14) - (5)(5) = 14-25 = -11 \]

Step 4: Applying the inverse formula:
\[ (AB)^{-1} = \frac{1}{-11}\begin{bmatrix}-14&-5\\-5&-1\end{bmatrix} = \begin{bmatrix}14/11&5/11\\5/11&1/11\end{bmatrix} \]

Final Answer:
\[ (AB)^{-1} = \begin{bmatrix}14/11&5/11\\5/11&1/11\end{bmatrix} \] \[ \boxed{(AB)^{-1}=\begin{bmatrix}14/11&5/11\\5/11&1/11\end{bmatrix}} \]