Question

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

Hint:

For invertible matrices, remember the important identity: \[ (AB)^{-1} = B^{-1}A^{-1} \] The {order reverses} when taking the inverse of a product of matrices.

Answer 1

Concept: For any two non-singular matrices \(A\) and \(B\), \[ (AB)^{-1} = B^{-1}A^{-1} \] To verify this property, we compute \(AB\), then find \((AB)^{-1}\), and separately compute \(B^{-1}A^{-1}\). If both results are equal, the identity is verified.
Step 1: Find the product \(AB\). \[ AB = \begin{bmatrix} 2 & 3\\ 1 & -4 \end{bmatrix} \begin{bmatrix} 1 & -2\\ -1 & 3 \end{bmatrix} \] \[ = \begin{bmatrix} 2(1)+3(-1) & 2(-2)+3(3) \\ 1(1)+(-4)(-1) & 1(-2)+(-4)(3) \end{bmatrix} \] \[ = \begin{bmatrix} -1 & 5 \\ 5 & -14 \end{bmatrix} \] 
Step 2: Find \( (AB)^{-1} \). For a \(2\times2\) matrix \[ \begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1} = \frac{1}{ad-bc} \begin{bmatrix}d&-b\\-c&a\end{bmatrix} \] Determinant of \(AB\): \[ (-1)(-14) - (5)(5) = 14 - 25 = -11 \] Thus, \[ (AB)^{-1} = \frac{1}{-11} \begin{bmatrix} -14 & -5\\ -5 & -1 \end{bmatrix} \] \[ = \begin{bmatrix} \frac{14}{11} & \frac{5}{11}\\ \frac{5}{11} & \frac{1}{11} \end{bmatrix} \] 
Step 3: Find \(A^{-1}\) and \(B^{-1}\). For \(A\): \[ |A| = (2)(-4) - (3)(1) = -8 - 3 = -11 \] \[ A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3\\ -1 & 2 \end{bmatrix} \] For \(B\): \[ |B| = (1)(3) - (-2)(-1) = 3 - 2 = 1 \] \[ B^{-1} = \begin{bmatrix} 3 & 2\\ 1 & 1 \end{bmatrix} \] 
Step 4: Compute \(B^{-1}A^{-1}\). \[ B^{-1}A^{-1} = \begin{bmatrix} 3 & 2\\ 1 & 1 \end{bmatrix} \left( \frac{1}{-11} \begin{bmatrix} -4 & -3\\ -1 & 2 \end{bmatrix} \right) \] \[ = \frac{1}{-11} \begin{bmatrix} 3(-4)+2(-1) & 3(-3)+2(2)\\ 1(-4)+1(-1) & 1(-3)+1(2) \end{bmatrix} \] \[ = \frac{1}{-11} \begin{bmatrix} -14 & -5\\ -5 & -1 \end{bmatrix} \] \[ = \begin{bmatrix} \frac{14}{11} & \frac{5}{11}\\ \frac{5}{11} & \frac{1}{11} \end{bmatrix} \] 
Step 5: Verification.
\[ (AB)^{-1} = \begin{bmatrix} \frac{14}{11} & \frac{5}{11}\\ \frac{5}{11} & \frac{1}{11} \end{bmatrix} = B^{-1}A^{-1} \] Hence, \[ (AB)^{-1} = B^{-1}A^{-1} \] is verified.