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

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> [!TLDR]
> The inverse of a product equals the product of inverses in reverse order: $(AB)^{-1} = B^{-1}A^{-1}$. For the given matrices, we compute $AB$, then its inverse, and separately compute $B^{-1}$ and $A^{-1}$ to verify the equality holds.

This problem tests a fundamental property of matrix inverses — the "reversal law." When you multiply two invertible matrices $A$ and $B$, the inverse of their product is not $A^{-1}B^{-1}$ but the reverse: $B^{-1}A^{-1}$. Why? Because matrix multiplication is not commutative. To undo $AB$, you must first undo $B$ (with $B^{-1}$), then undo $A$ (with $A^{-1}$). Let's verify this step by step.

1. **Compute $AB$**  
   Multiply $A$ and $B$:
   $$AB = \begin{bmatrix} 2 & 3 \ 1 & -4 \end{bmatrix} \begin{bmatrix} 1 & -2 \ -1 & 3 \end{bmatrix} = \begin{bmatrix} 2(1)+3(-1) & 2(-2)+3(3) \ 1(1)+(-4)(-1) & 1(-2)+(-4)(3) \end{bmatrix} = \begin{bmatrix} -1 & 5 \ 5 & -14 \end{bmatrix}.$$

2. **Find $(AB)^{-1}$**  
   For a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$, the inverse is $\frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$, provided the determinant is non-zero.  
   Here, $\det(AB) = (-1)(-14) - (5)(5) = 14 - 25 = -11$.  
   So,
   $$(AB)^{-1} = \frac{1}{-11} \begin{bmatrix} -14 & -5 \ -5 & -1 \end{bmatrix} = \begin{bmatrix} \frac{14}{11} & \frac{5}{11} \ \frac{5}{11} & \frac{1}{11} \end{bmatrix}.$$

3. **Find $A^{-1}$ and $B^{-1}$ individually**  
   - For $A$: $\det(A) = (2)(-4) - (3)(1) = -8 - 3 = -11$.  
     $$A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{4}{11} & \frac{3}{11} \ \frac{1}{11} & -\frac{2}{11} \end{bmatrix}.$$
   - For $B$: $\det(B) = (1)(3) - (-2)(-1) = 3 - 2 = 1$.  
     $$B^{-1} = \frac{1}{1} \begin{bmatrix} 3 & 2 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 \ 1 & 1 \end{bmatrix}.$$

4. **Compute $B^{-1}A^{-1}$**  
   Multiply $B^{-1}$ and $A^{-1}$ in that order:
   $$B^{-1}A^{-1} = \begin{bmatrix} 3 & 2 \ 1

Q.If $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$ , then verify that $(AB)^{-1} = B^{-1}A^{-1}$ .

Concept understanding — Inverse of a Product

Inverse of a Product: The "Socks and Shoes" Principle

Why the order must reverse

A concrete example with numbers

Method: Verifying the Reversal Law for the Inverse of a Product

Steps

Common Mistakes

Q.If A=[231−4]A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}A=[21​3−4​] and B=[1−2−13]B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}B=[1−1​−23​], then verify that (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}(AB)−1=B−1A−1.

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Q.If $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$ , then verify that $(AB)^{-1} = B^{-1}A^{-1}$ .

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