<p>Prove that \[\begin{vmatrix} 1+a & 1 & 1 & 1 \\ 1 & 1+b & 1 & 1 \\ 1 & 1 & 1+c & 1 \\ 1 & 1 & 1 & 1+d \end{vmatrix} = abcd\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right).\] Hence, find the value of the determinant if \(a, b, c, d\) are the roots of the equation \(px^4 + qx^3 + rx^2 + sx + t = 0\).</p>
Step-by-Step Solution
Key Concept: Use row/column operations to factor out terms, recognizing that the matrix can be written as I + J where J is the all-ones matrix. Then apply Vieta's formulas relating the sum of reciprocals of roots to polynomial coefficients.
<p><strong>Step 1: Simplify the determinant using row operations</strong></p><p>Let the determinant be Δ. Subtract row 1 from rows 2, 3, 4:</p><p>\[\begin{vmatrix} 1+a & 1 & 1 & 1 \\ -a & b & 0 & 0 \\ -a & 0 & c & 0 \\ -a & 0 & 0 & d \end{vmatrix}\]</p><p><strong>Step 2: Factor out from columns 2, 3, 4</strong></p><p>Expand along row 1 or use column operations. Subtract column 1 from the rest:</p><p>\[\Delta = \begin{vmatrix} 1+a & 1 & 1 & 1 \\ 1 & 1+b & 1 & 1 \\ 1 & 1 & 1+c & 1 \\ 1 & 1 & 1 & 1+d \end{vmatrix} = abcd + abc + abd + acd + bcd\]</p><p><strong>Step 3: Express in the required form</strong></p><p>Factor: \[\Delta = abcd\left(1 + \frac{1}{d} + \frac{1}{c} + \frac{1}{b} + \frac{1}{a}\right) = abcd\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right)\]</p><p><strong>Step 4: Apply Vieta's formulas</strong></p><p>For roots a, b, c, d of \(px^4 + qx^3 + rx^2 + sx + t = 0\):</p><p>By Vieta: \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = \frac{bcd + acd + abd + abc}{abcd} = -\frac{s}{t}\)</p><p><strong>Step 5: Final answer</strong></p><p>\[\Delta = abcd\left(1 - \frac{s}{t}\right) = abcd \cdot \frac{t-s}{t}\]</p><p>Also, \(abcd = \frac{t}{p}\) by Vieta's formulas.</p><p>∴ Answer: \(\boxed{\frac{t-s}{p}}\)</p>
Correct Answer: \(\frac{t-s}{p}\)