Matrices & Determinants
Determinant Properties
Grade 12
Question:
<p>If <span class="math">\(a, b\)</span> and <span class="math">\(c\)</span> are unequal, what is the condition that the value of the determinant, <span class="math">\(\Delta = \begin{vmatrix} a & a & a+1 \\ b & b & b+1 \\ c & c & c+1 \end{vmatrix}\)</span> is <span class="math">\(0\)</span>?</p>
<p>(a) <span class="math">\(1 + abc = 0\)</span></p>
<p>(b) <span class="math">\(a + b + c = 0\)</span></p>
<p>(c) Other conditions</p>
<p>(d) No condition</p>
Step-by-Step Solution
Key Concept: Use column operations to simplify the determinant by making columns linearly dependent, then recognize that the determinant is zero for ALL values of a, b, c (unequal), regardless of any specific condition.
<p><strong>Step 1:</strong> Write the determinant: $$\Delta = \begin{vmatrix} a & a & a+1 \\ b & b & b+1 \\ c & c & c+1 \end{vmatrix}$$</p><p><strong>Step 2:</strong> Perform column operation $C_3 \to C_3 - C_1$:</p><p>$$\Delta = \begin{vmatrix} a & a & (a+1)-a \\ b & b & (b+1)-b \\ c & c & (c+1)-c \end{vmatrix} = \begin{vmatrix} a & a & 1 \\ b & b & 1 \\ c & c & 1 \end{vmatrix}$$</p><p><strong>Step 3:</strong> Perform column operation $C_2 \to C_2 - C_1$:</p><p>$$\Delta = \begin{vmatrix} a & 0 & 1 \\ b & 0 & 1 \\ c & 0 & 1 \end{vmatrix}$$</p><p><strong>Step 4:</strong> Since the second column is entirely zero, expanding along $C_2$ gives:</p><p>$$\Delta = 0 \cdot (\text{minors}) = 0$$</p><p><strong>Step 5:</strong> The determinant equals zero for ALL unequal values of $a, b, c$. There is no additional condition required. The determinant is identically zero due to the linear dependence of columns.</p><p><strong>∴ Answer:</strong> a (No condition - the determinant is always zero)</p>
Correct Answer: a