If $A$ is a skew-symmetric matrix of order $3$, then $|A|$ is:

Question

If $\alpha \neq a$, $\beta \neq b$, $\gamma \neq c$ and \[ \begin{vmatrix} \alpha & b & c \\ a & \beta & c \\ a & b & \gamma \end{vmatrix} = 0,\] then \[ \frac{a}{\alpha - a} + \frac{b}{\beta - b} + \frac{\gamma}{\gamma - c} \] is equal to:

Answer 1

Understanding the Concept: A skew-symmetric matrix satisfies: \[ A^T=-A \] An important property is: \[ |A^T|=|A| \] Also, \[ |-A|=(-1)^n|A| \] where $n$ is the order of the matrix.
Step 1: Use the skew-symmetric property.
Given: \[ A^T=-A \] Taking determinants: \[ |A^T|=|-A| \] Using determinant properties: \[ |A|=(-1)^n|A| \] Since order is $3$: \[ |A|=(-1)^3|A| \] \[ |A|=-|A| \]
Step 2: Simplify the equation.
Adding $|A|$ to both sides: \[ 2|A|=0 \] Therefore, \[ |A|=0 \] Hence, \[ \boxed{0} \]

If $A$ is a skew-symmetric matrix of order $3$, then $|A|$ is:

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The Correct Option is B

Solution and Explanation

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