Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>\(a, b, c\) are distinct real numbers, not equal to one. If \(ax + y + z = 0\), \(x + by + z = 0\), and \(x + y + cz = 0\) have a non-trivial solution, then the value of \(\dfrac{1}{1-a} + \dfrac{1}{1-b} + \dfrac{1}{1-c}\) is equal to</p>
<p>(1) -1</p>
<p>(2) 1</p>
<p>(3) zero</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: For a homogeneous system to have non-trivial solutions, the coefficient matrix determinant must be zero. This constraint on a, b, c allows us to manipulate the determinant condition algebraically to find the required sum.
<p><strong>Step 1:</strong> For non-trivial solution, the coefficient matrix determinant must equal zero:</p><p>$$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Expand using R₁:</p><p>$$a(bc - 1) - 1(c - 1) + 1(1 - b) = 0$$</p><p>$$abc - a - c + 1 + 1 - b = 0$$</p><p>$$abc - a - b - c + 2 = 0$$</p><p><strong>Step 3:</strong> Rearrange as:</p><p>$$abc = a + b + c - 2$$</p><p><strong>Step 4:</strong> We need to find $\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}$</p><p>Finding common denominator:</p><p>$$= \frac{(1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b)}{(1-a)(1-b)(1-c)}$$</p><p><strong>Step 5:</strong> Numerator expansion:</p><p>$$= (1 - b - c + bc) + (1 - a - c + ac) + (1 - a - b + ab)$$</p><p>$$= 3 - 2(a+b+c) + (ab+bc+ca)$$</p><p><strong>Step 6:</strong> Denominator expansion:</p><p>$$(1-a)(1-b)(1-c) = 1 - (a+b+c) + (ab+bc+ca) - abc$$</p><p>$$= 1 - (a+b+c) + (ab+bc+ca) - (a+b+c-2)$$</p><p>$$= 3 - 2(a+b+c) + (ab+bc+ca)$$</p><p><strong>Step 7:</strong> Therefore:</p><p>$$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = \frac{\text{Numerator}}{\text{Denominator}} = \frac{3 - 2(a+b+c) + (ab+bc+ca)}{3 - 2(a+b+c) + (ab+bc+ca)} = 1$$</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: B