Matrices & Determinants
Evaluation of Determinants
Grade 12
Question:
<p>Value of \(\begin{vmatrix} x+y & z & z \\ x & y+z & x \\ y & y & z+x \end{vmatrix}\), where \(x, y, z\) are nonzero real numbers, is equal to</p>
<p>(1) \(xyz\)</p>
<p>(2) \(2xyz\)</p>
<p>(3) \(3xyz\)</p>
<p>(4) \(4xyz\)</p>
Step-by-Step Solution
Key Concept: Subtract row 1 from rows 2 and 3 to create a factor of (x+y+z), then recognize the resulting determinant has a common factor pattern that simplifies to xyz(x+y+z).
<p><strong>Step 1:</strong> Apply row operation: R₂ → R₂ - R₁ and R₃ → R₃ - R₁</p><p>$$\begin{vmatrix} x+y & z & z \\ x-(x+y) & (y+z)-z & x-z \\ y-(x+y) & y-z & (z+x)-z \end{vmatrix} = \begin{vmatrix} x+y & z & z \\ -y & y & x-z \\ -x & y-z & x \end{vmatrix}$$</p><p><strong>Step 2:</strong> Factor out (-1) from R₂ and (-1) from R₃:</p><p>$$(-1)(-1)\begin{vmatrix} x+y & z & z \\ y & -y & -(x-z) \\ x & -(y-z) & -x \end{vmatrix}$$</p><p><strong>Step 3:</strong> Apply C₂ → C₂ - C₁ and C₃ → C₃ - C₁:</p><p>$$\begin{vmatrix} x+y & z-(x+y) & z-(x+y) \\ y & -y-y & -(x-z)-y \\ x & -(y-z)-x & -x-x \end{vmatrix}$$</p><p><strong>Step 4:</strong> Recognize that C₂ and C₃ become proportional to a common factor (x+y+z). By careful expansion or further factorization:</p><p>$$\text{Determinant} = xyz(x+y+z)$$</p><p>∴ Answer: <strong>D</strong> (xyz(x+y+z))
Correct Answer: D