Q.Let $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$. Verify that $(AB)^{-1} = B^{-1} A^{-1}$.

For invertible $A,B$: $(AB)^{-1}=B^{-1}A^{-1}$. Verify by computing both sides.

$AB$: $\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}.$

$(AB)^{-1}$: $\det(AB)=67\cdot61-87\cdot47 = 4087-4089 = -2$, so $(AB)^{-1} = \dfrac{1}{-2}\begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \begin{bmatrix} -\frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & -\frac{67}{2} \end{bmatrix}.$

$A^{-1},B^{-1}$: $\det A=1 \Rightarrow A^{-1}=\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$; $\det B=-2 \Rightarrow B^{-1}=\dfrac{1}{-2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix}=\begin{bmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix}.$ …

The rule $(AB)^{-1}=B^{-1}A^{-1}$ reverses the order (the "socks and shoes" idea), because $(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AA^{-1}=I$. Let us check it on the given matrices.

1. Compute $AB$

2. Invert $AB$

For $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ the inverse is $\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

3. Invert $A$ and $B$ separately

$\det A = 15-14 = 1$, so $A^{-1} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}.$

$\det B = 54-56 = -2$, so $B^{-1} = \frac{1}{-2}\begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \begin{bmatrix} -\frac{9}{2} & 4 \\ \frac{7}{2} & -3 \end{bmatrix}.$

4. Multiply $B^{-1}A^{-1}$ (reverse order) …

Method: Verifying $(AB)^{-1} = B^{-1}A^{-1}$

This method verifies the "reversal rule" for the inverse of a product by computing both sides independently and checking they agree.

Steps

Step 1: Compute the product $AB$

Multiply the two given matrices in the order given — order matters, since matrix multiplication does not commute.

Step 2: Invert $AB$ directly

For a $2\times2$ matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, use

Check $\det(AB)\neq0$ before proceeding.

Step 3: Invert $A$ and $B$ separately, using the same $2\times2$ inverse formula …

Common Mistakes

Mistake 1: Computing $A^{-1}B^{-1}$ instead of $B^{-1}A^{-1}$

Why it's wrong: matrix multiplication does not commute, so $(AB)^{-1}=B^{-1}A^{-1}$ — the order must reverse, not stay the same. Multiplying in the original order gives a matrix that, in general, does not equal $(AB)^{-1}$. Correct approach: always write the inverse of a product with the factors in reverse order, and verify the "socks and shoes" logic ($(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AA^{-1}=I$) if unsure which order is right.

Mistake 2: An arithmetic slip in the $2\times2$ matrix multiplication, especially with fractions …