Matrices & Determinants
Determinant Properties
Grade 12
Question:
<p>The value of determinant <span class="math">\(\begin{vmatrix} (a^x+a^{-x})^2 & (a^x-a^{-x})^2 & 1 \\ (b^x+b^{-x})^2 & (b^x-b^{-x})^2 & 1 \\ (c^x+c^{-x})^2 & (c^x-c^{-x})^2 & 1 \end{vmatrix}\)</span> is</p>
<p>(a) <span class="math">\(0\)</span></p>
<p>(b) <span class="math">\(2abc\)</span></p>
<p>(c) <span class="math">\(a^2b^2c^2\)</span></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Simplify each entry using the algebraic identity (p+q)² - (p-q)² = 4pq, then observe that each row becomes linearly dependent on the other rows, making the determinant zero.
<p><strong>Step 1: Simplify each column entry</strong></p><p>Let's use the algebraic identity: (p+q)² - (p-q)² = 4pq</p><p>For the first row with p = a^x, q = a^{-x}:</p><p>(a^x+a^{-x})² - (a^x-a^{-x})² = 4·a^x·a^{-x} = 4·a^{x-x} = 4</p><p>Similarly, for any base u:</p><p>(u^x+u^{-x})² - (u^x-u^{-x})² = 4</p></p><p><strong>Step 2: Find the relationship between columns</strong></p><p>This means: Column 1 - Column 2 = 4·(Column 3)</p><p>Or equivalently: Column 1 = Column 2 + 4·(Column 3)</p><p>This relationship holds for EVERY row in the determinant.</p></p><p><strong>Step 3: Apply determinant properties</strong></p><p>Since one column can be expressed as a linear combination of the other columns, the columns are linearly dependent.</p><p>When the columns (or rows) of a matrix are linearly dependent, the determinant equals zero.</p></p><p><strong>Step 4: Verify the linear dependence</strong></p><p>For row 1: (a^x+a^{-x})² - (a^x-a^{-x})² = 4·1</p><p>For row 2: (b^x+b^{-x})² - (b^x-b^{-x})² = 4·1</p><p>For row 3: (c^x+c^{-x})² - (c^x-c^{-x})² = 4·1</p><p>Each row satisfies: C₁ - C₂ - 4C₃ = 0</p><p>This confirms linear dependence of columns.</p></p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a