Q.Prove that $(AB)^{-1} = B^{-1}A^{-1}$, where $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 1 & 9 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 5 & 3 \\ 3 & 1 & 2 \\ 1 & 2 & 1 \end{bmatrix}$.

Concept and Intuition

The inverse of a square matrix $M$, denoted $M^{-1}$, is defined such that when $M$ is multiplied by $M^{-1}$, the result is the identity matrix $I$. That is, $MM^{-1} = M^{-1}M = I$. The identity matrix acts like the number $1$ in scalar multiplication: $MI = IM = M$.

When we consider a scalar multiple of a matrix, say $kA$, we are essentially scaling every element of the matrix $A$ by the scalar $k$. If we want to find the inverse of this new matrix $(kA)$, we need to find a matrix that, when multiplied by $kA$, yields the identity matrix $I$.

Intuitively, if $A^{-1}$ "undoes" the operation of $A$, and $k$ "scales" $A$, then to "undo" $kA$, we would need to "un-scale" by $\frac{1}{k}$ and then "un-matrix" by $A^{-1}$. This suggests that the inverse of $kA$ should involve $\frac{1}{k}$ and $A^{-1}$.

Let's verify this intuition formally.

Step-by-Step Derivation

1. Recall the definition of an inverse matrix:

   For any invertible square matrix $M$, its inverse $M^{-1}$ satisfies the property $MM^{-1} = I$, where $I$ is the identity matrix of the same dimension.

2. Apply the definition to $(3A)$:

   We are looking for $(3A)^{-1}$. Let's denote this unknown inverse as $X$. By definition, $X$ must satisfy:

   $(3A)X = I$

3. Propose a form for $X$ based on intuition:

   As discussed in the concept section, we expect $X$ to be of the form $c A^{-1}$ for some scalar $c$. Let's substitute this into the equation:

   $(3A)(cA^{-1}) = I$

4. Use properties of scalar and matrix multiplication:

   For any scalars $k_1, k_2$ and matrices $M_1, M_2$, we know that $(k_1 M_1)(k_2 M_2) = (k_1 k_2)(M_1 M_2)$. Applying this property:

   $(3c)(AA^{-1}) = I$

5. Substitute $AA^{-1} = I$: …