Q.If A is a $3 \times 3$ invertible matrix, show that for any scalar $k \neq 0$, $(kA)^{-1} = \frac{1}{k}A^{-1}$. Hence calculate $(3A)^{-1}$, where $A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$.

Concept: Scalar multiplication property of the inverse of a matrix.

We need to show that for an invertible matrix $A$ and a nonzero scalar $k$, the inverse of $kA$ is $\frac{1}{k}A^{-1}$.

Step 1: Multiply $kA$ by $\frac{1}{k}A^{-1}$:

Step 2: Similarly, $\left( \frac{1}{k}A^{-1} \right)(kA) = I$. Since the product in both orders gives the identity, $\frac{1}{k}A^{-1}$ is indeed the inverse of $kA$. Hence $(kA)^{-1} = \frac{1}{k}A^{-1}$.

Step 3: For the given $A$, first find $A^{-1}$. Using the standard formula or row reduction, we get:

$A^{-1} = \frac{1}{4} \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}.$ …

We start with the core idea: if you scale a matrix by a nonzero constant, its inverse simply scales back by the reciprocal. This is because matrix multiplication is linear in each factor — scaling the whole matrix is like scaling every entry, and the inverse must undo that scaling.

1. Proving the general property

Let $A$ be an $n \times n$ invertible matrix and $k \neq 0$ a scalar. We need to show that $(kA)^{-1} = \frac{1}{k} A^{-1}$.

Recall the definition: $B$ is the inverse of $C$ if $BC = I = CB$. So check:

Similarly, $\left( \frac{1}{k} A^{-1} \right) (kA) = I$. Since the inverse is unique, the result follows.

2. Applying to the given matrix

We have $A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$. First, find $A^{-1}$.

A common method: use the formula $A^{-1} = \frac{1}{\det A} \operatorname{adj}(A)$. Compute $\det A$:

Expanding along the first row:

Each $2 \times 2$ determinant:

• First: $(2)(2) - (-1)(-1) = 4 - 1 = 3$
• Second: $(-1)(2) - (-1)(1) = -2 + 1 = -1$
• Third: $(-1)(-1) - (2)(1) = 1 - 2 = -1$

So $\det A = 2(3) + 1(-1) + 1(-1) = 6 - 1 - 1 = 4$.

3. Finding the adjugate (classical adjoint)

The adjugate is the transpose of the cofactor matrix. Compute cofactors $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor.

• $C_{11} = +\det\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = 3$

• $C_{12} = -\det\begin{bmatrix} -1 & -1 \\ 1 & 2 \end{bmatrix} = -[(-1)(2) - (-1)(1)] = -(-2+1) = 1$

• $C_{13} = +\det\begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix} = (-1)(-1) - (2)(1) = 1 - 2 = -1$

• $C_{21} = -\det\begin{bmatrix} -1 & 1 \\ -1 & 2 \end{bmatrix} = -[(-1)(2) - (1)(-1)] = -(-2+1) = 1$

• $C_{22} = +\det\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} = (2)(2) - (1)(1) = 4 - 1 = 3$

• $C_{23} = -\det\begin{bmatrix} 2 & -1 \\ 1 & -1 \end{bmatrix} = -[(2)(-1) - (-1)(1)] = -(-2+1) = 1$ …