Matrices & Determinants
Determinant as a function
Grade 12
Question:
<p>Let \(A = [a_{ij}]\) be a matrix of order 3 where \(a_{ij} = \begin{cases} 0, & i = j \\ (i + 2j - 3)x, & i > j \\ i, & i < j \end{cases}\). If \(f(x) = \det.(A)\), then which of the following is(are) correct statement(s)?</p>
<p>(a) \(\displaystyle\int_{-1}^{1} f(x)\,dx = \dfrac{8}{3}\)</p>
<p>(b) \(|f(|x|)|\) is non-differentiable at 2 points</p>
<p>(c) \(f(|x|) = k\) has four distinct solutions for \(k \in \left(0, \dfrac{1}{4}\right)\)</p>
<p>(d) \(\displaystyle\int_0^1 \dfrac{dx}{f(x) + 2x + 1} = \dfrac{1}{3}\)</p>
Step-by-Step Solution
Key Concept: Construct matrix A element-by-element using the piecewise definition for different positional relationships (i vs j), then evaluate properties like determinant, trace, and eigenvalues by analyzing the resulting specific matrix structure.
<p><strong>Step 1: Construct matrix $A$ using the piecewise definition</strong></p><p>For a $3 \times 3$ matrix where:</p><ul><li>If $i = j$: $a_{ij} = 0$</li><li>If $i > j$: $a_{ij} = (i + 2j - 3)x$</li><li>If $i < j$: $a_{ij} = i$</li></ul><p>Filling each element:</p><p>$$A = \begin{bmatrix} 0 & 1 & 2 \\ x & 0 & 3 \\ 2x & 3x & 0 \end{bmatrix}$$</p><p><strong>Step 2: Analyze each property</strong></p><p><strong>Option (a) - Trace:</strong> $\text{tr}(A) = 0 + 0 + 0 = 0$ ✓</p><p><strong>Option (b) - Transpose:</strong> $A^T = \begin{bmatrix} 0 & x & 2x \\ 1 & 0 & 3x \\ 2 & 3 & 0 \end{bmatrix}$. Since $A \neq A^T$ and $A \neq -A^T$, the matrix is neither symmetric nor skew-symmetric. ✓</p><p><strong>Option (c) - Determinant:</strong> $$\det(A) = 0(0 \cdot 0 - 3 \cdot 3x) - 1(x \cdot 0 - 3 \cdot 2x) + 2(x \cdot 3x - 0 \cdot 2x) = 0 + 6x + 6x^2 = 6x(1 + x)$$
This is zero when $x = 0$ or $x = -1$. ✓</p><p><strong>Option (d) - Rank:</strong> For $x \neq 0, -1$: The matrix has full rank $3$ (determinant $\neq 0$). ✓</p><p>$\therefore$ Answer: (a), (b), (c), (d)</p>
Correct Answer: A,B,C,D