If `A=[a_(ij)]` is a scalar matrix of order `nxxn` and k is a scalar, then `abs(kA)=`

Options

A. `k^nabsA`
B. `kabsA`
C. `k^(n-1)absA`
D. 0

Answer

To solve the problem, we need to find the absolute value (or magnitude) of the product of a scalar \( k \) and a scalar matrix \( A \) of order \( n \times n \).

### Step-by-Step Solution:

1. **Understanding the Scalar Matrix**:
A scalar matrix \( A \) is a special type of diagonal matrix where all the diagonal elements are equal to a scalar \( a \) and all off-diagonal elements are zero. Thus, we can represent \( A \) as:
\[ A = aI_n = \begin{pmatrix} a & 0 & \cdots & 0 \\ 0 & a & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a \end{pmatrix} \]
where \( I_n \) is the identity matrix of order \( n \).

2. **Multiplying the Scalar with the Matrix**:
When we multiply the scalar \( k \) with the matrix \( A \), we get:
\[ kA = k \begin{pmatrix} a & 0 & \cdots & 0 \\ 0 & a & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a \end{pmatrix} = \begin{pmatrix} ka & 0 & \cdots & 0 \\ 0 & ka & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & ka \end{pmatrix} \]

3. **Finding the Magnitude of the Matrix**:
The magnitude (or absolute value) of a matrix, particularly for a scalar matrix, can be defined as the product of its eigenvalues. The eigenvalues of a scalar matrix \( A \) are all equal to \( a \). Therefore, the eigenvalues of \( kA \) are all equal to \( ka \).

4. **Calculating the Absolute Value**:
The absolute value of the matrix \( kA \) can be calculated as:
\[ |kA| = |ka|^n = (|k| \cdot |a|)^n = |k|^n \cdot |a|^n \]
Thus, we have:
\[ |kA| = |k|^n \cdot |A| \]

5. **Conclusion**:
Therefore, the final answer is:
\[ |kA| = |k|^n \cdot |A| \]