Matrices & Determinants
Properties of Determinants
Grade 12
Question:
<p><strong>Example 26 (Statement-1):</strong> Consider the determinant \[f(x) = \begin{vmatrix} x^2-a & x^3 & 0 \\ x^2+a & 0 & x^2+c \\ x+b & x+c & 0 \end{vmatrix}\] Then \(f(x) = 0\) has one root \(x = 0\).</p><p><strong>Statement-2:</strong> The value of skew-symmetric determinant of odd order is always zero.</p>
<p>(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1</p>
<p>(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1</p>
<p>(c) Statement-1 is true, Statement-2 is false</p>
<p>(d) Statement-1 is false, Statement-2 is true</p>
Step-by-Step Solution
Key Concept: A skew-symmetric matrix of odd order always has determinant zero; this property directly proves that x=0 is a root when the determinant becomes skew-symmetric.
<p><strong>Solution:</strong> For $x = 0$, the determinant reduces to: $$f(0) = \begin{vmatrix} -a & 0 & 0 \\ a & 0 & c \\ b & c & 0 \end{vmatrix}$$ This is a skew-symmetric determinant of odd order (order 3), which is always zero by the property that $\det(A) = \det(-A^T) = (-1)^n \det(A^T) = (-1)^n \det(A)$. For odd $n$, this gives $\det(A) = -\det(A)$, hence $\det(A) = 0$. Therefore, $x = 0$ is indeed a root, making Statement-1 true, and Statement-2 provides the correct explanation.</p><p>∴ Answer is (a).</p>
Correct Answer: A