Question

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

Hint:

Always remember: inverse of product reverses the order → $(AB)^{-1} = B^{-1}A^{-1}$.

Answer 1

Step 1: Compute $AB$.
\[ AB = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 2(1)+3(-1) & 2(-2)+3(3) \\ 1(1)+(-4)(-1) & 1(-2)+(-4)(3) \end{bmatrix} \] \[ AB = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix} \] Step 2: Find $(AB)^{-1$.}
For a $2 \times 2$ matrix $\begin{bmatrix} p & q \\ r & s \end{bmatrix}$, \[ M^{-1} = \frac{1}{ps-qr}\begin{bmatrix} s & -q \\ -r & p \end{bmatrix} \] Here, $\det(AB) = (-1)(-14) - (5)(5) = 14 - 25 = -11$. \[ (AB)^{-1} = \frac{1}{-11} \begin{bmatrix} -14 & -5 \\ -5 & -1 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} \] Step 3: Find $A^{-1$ and $B^{-1}$.}
For $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$, $\det(A) = (2)(-4) - (3)(1) = -8 - 3 = -11$. \[ A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 4 & 3 \\ 1 & -2 \end{bmatrix} \] For $B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$, $\det(B) = (1)(3) - (-2)(-1) = 3 - 2 = 1$. \[ B^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \] Step 4: Compute $B^{-1}A^{-1}$.
\[ B^{-1}A^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \cdot \frac{1}{11}\begin{bmatrix} 4 & 3 \\ 1 & -2 \end{bmatrix} \] \[ = \frac{1}{11}\begin{bmatrix} 3(4)+2(1) & 3(3)+2(-2) \\ 1(4)+1(1) & 1(3)+1(-2) \end{bmatrix} \] \[ = \frac{1}{11}\begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} \] Step 5: Compare results.
\[ (AB)^{-1} = \frac{1}{11}\begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix}, B^{-1}A^{-1} = \frac{1}{11}\begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} \] \[ \therefore (AB)^{-1} = B^{-1}A^{-1} \]

Final Answer: \[ \boxed{(AB)^{-1} = B^{-1}A^{-1}} \]