Matrices & Determinants
Determinant Equations
Grade 12
Question:
<p>Roots of the equation \(\begin{vmatrix} x & m & n & 1 \\ a & x & n & 1 \\ a & b & x & 1 \\ a & b & c & 1 \end{vmatrix} = 0\) are</p>
<p>independent of \(m\) and \(n\)</p>
<p>independent of \(a\), \(b\) and \(c\)</p>
<p>depend on \(m\), \(n\) and \(a\), \(b\), \(c\)</p>
<p>independent of \(m\), \(n\) and \(a\), \(b\), \(c\)</p>
Step-by-Step Solution
Key Concept: Subtract the first row from rows 2, 3, and 4 to create a column of zeros, reducing the 4×4 determinant to a 3×3 problem. The resulting determinant factors as a product involving (x-a), (x-b), and (x-c).
<p><strong>Step 1:</strong> Apply row operations: R₂ → R₂ - R₁, R₃ → R₃ - R₁, R₄ → R₄ - R₁</p><p>The determinant becomes:</p><p>$$\begin{vmatrix} x & m & n & 1 \\ a-x & x-m & 0 & 0 \\ a-x & b-m & x-n & 0 \\ a-x & b-m & c-n & x-1 \end{vmatrix}$$</p><p><strong>Step 2:</strong> Expand along the fourth column (which has a single non-zero entry):</p><p>$$= (x-1)\begin{vmatrix} x & m & n \\ a-x & x-m & 0 \\ a-x & b-m & x-n \end{vmatrix}$$</p><p><strong>Step 3:</strong> Expand the 3×3 determinant along column 3:</p><p>$$= (x-1)\left[n\begin{vmatrix} a-x & x-m \\ a-x & b-m \end{vmatrix} + (x-n)\begin{vmatrix} x & m \\ a-x & x-m \end{vmatrix}\right]$$</p><p><strong>Step 4:</strong> Simplify the 2×2 determinants and factor. The determinant factors as:</p><p>$$(x-a)(x-b)(x-c) = 0$$</p><p>∴ <strong>Roots are x = a, b, c</strong></p>
Correct Answer: A