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: Use the identity $(AB)^{-1}=B^{-1}A^{-1}$ instead of multiplying first, as a cross-check route:
$\det(A)=2(-4)-3(1)=-8-3=-11$, so $A^{-1}=\dfrac{1}{-11}\begin{bmatrix}-4&-3\\-1&2\end{bmatrix}$. $\det(B)=1(3)-(-2)(-1)=3-2=1$, so $B^{-1}=\begin{bmatrix}3&2\\1&1\end{bmatrix}$.

Step 2: Multiply $B^{-1}A^{-1}$:
$B^{-1}A^{-1} = \begin{bmatrix}3&2\\1&1\end{bmatrix}\cdot\dfrac{1}{-11}\begin{bmatrix}-4&-3\\-1&2\end{bmatrix} = \dfrac{1}{-11}\begin{bmatrix}3(-4)+2(-1) & 3(-3)+2(2)\\1(-4)+1(-1) & 1(-3)+1(2)\end{bmatrix}$

Step 3: Simplify the entries:
$=\dfrac{1}{-11}\begin{bmatrix}-14&-5\\-5&-1\end{bmatrix}=\begin{bmatrix}14/11&5/11\\5/11&1/11\end{bmatrix}$, matching the direct-multiplication route exactly.

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