Matrices & Determinants
Determinant Properties
Grade 12
Question:
<p>If <span class="math">f(x) = \begin{vmatrix} (1-x)a & (1-2x)b & 1 \\ 1 & (1-x)a & (1-2x)b \\ b & a & (1-2x) \end{vmatrix}</span>, where <span class="math">a, b</span> are positive integers, then:</p>
<p>(a) constant term in <span class="math">f(x)</span> is 4</p>
<p>(b) coefficient of <span class="math">x</span> in <span class="math">f(x)</span> is 0</p>
<p>(c) constant term in <span class="math">f(x)</span> is <span class="math">(a+b)</span></p>
<p>(d) constant term in <span class="math">f(x)</span> is <span class="math">(a-b)</span></p>
Step-by-Step Solution
Key Concept: Use substitution of specific values and differentiation to extract Taylor coefficients of the determinant function.
<p><strong>Step 1:</strong> Let <span class="math">f(x) = A - Bx - Cx^2 - \ldots</span></p><p><strong>Step 2:</strong> Put <span class="math">x = 0</span>: <span class="math">f(0) = \begin{vmatrix} a & b & 1 \\ 1 & a & b \\ b & a & 1 \end{vmatrix} = A</span></p><p><strong>Step 3:</strong> Expanding: <span class="math">f(0) = a(a - ab) - b(1 - b^2) + 1(a - ab) = 0</span></p><p>So <span class="math">A = 0</span>.</p><p><strong>Step 4:</strong> Differentiate with respect to <span class="math">x</span> and put <span class="math">x = 0</span>:</p><p><span class="math">\begin{vmatrix} -a & -2b & 0 \\ 1 & -a & -2b \\ 0 & 0 & -2 \end{vmatrix} - \begin{vmatrix} 0 & 0 & 1 \\ 0 & 0 & b \\ 0 & a & 1 \end{vmatrix} = B</span></p><p>Computing both determinants yields <span class="math">B = 0</span>.</p><p>∴ The constant term in <span class="math">f(x)</span> is zero and the coefficient of <span class="math">x</span> in <span class="math">f(x)</span> is 0.</p><p>Answer is (b).</p>
Correct Answer: B