Question

If, \(f(x) = \begin{bmatrix}0 & x-a & x-b \\[0.3em]x+a&o & x-c \\[0.3em]x+b & x+c & 0\\[0.3em] \end{bmatrix}\), then \(f(0)\) is:

• A. 3abc
• B. - abc
• C. abc
• D. 0

Answer 1

The Correct Option is D

To determine \(f(0)\), substitute \(x = 0\) into the given function \(f(x)\). The matrix is then defined as:

\(f(0) = \begin{bmatrix}0 & 0-a & 0-b \\[0.3em]0+a&0 & 0-c \\[0.3em]0+b & 0+c & 0\\[0.3em] \end{bmatrix} = \begin{bmatrix}0 & -a & -b \\[0.3em]a&0 & -c \\[0.3em]b & c & 0\\ \end{bmatrix}\)

We now calculate the determinant of this matrix to find \(f(0)\):

\[\det \left( \begin{bmatrix}0 & -a & -b \\[0.3em]a&0 & -c \\[0.3em]b & c & 0\\[0.3em] \end{bmatrix} \right)\]

To compute the determinant of a 3x3 matrix:

\[\det(A) = a(ei-fh) - b(di-fg) + c(dh-eg)\]

For our matrix, \(a=0\), \(b=-a\), \(c=-b\), \(d=a\), \(e=0\), \(f=-c\), \(g=b\), \(h=c\), \(i=0\).

Substituting these into the formula:

\[\det(A) = 0(0 - (-c) \cdot c) - (-a)(a \cdot 0 - (-c) \cdot b) + (-b)(a \cdot c - 0)\]

\[\Rightarrow\ 0 - a \cdot (c \cdot b) - b \cdot (a \cdot c)\]

\[\Rightarrow\ -abc - abc =-2abc\]

However, recognizing the structure of \(f(0)\) as a skew-symmetric matrix (where \(A^T = -A\) and determinant of odd-sized skew-symmetric matrices is zero), mathematically simplifies the determinant to:

\[\det(f(0))=0\]

Thus, the answer is the option 0.