Matrices & Determinants
Non-trivial solutions and determinants
Grade 12
Question:
<p>If the system of equations<br>\((a-t)x + by + cz = 0\)<br>\(bx + (c-t)y + az = 0\)<br>\(cx + ay + (b-t)z = 0\)<br>has non-trivial solution, then product of all possible values of \(t\) is</p>
<p>\(\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}\)</p>
<p>\(a + b + c\)</p>
<p>\(a^2 + b^2 + c^2\)</p>
<p>\(1\)</p>
Step-by-Step Solution
Key Concept: For a homogeneous system to have non-trivial solutions, the determinant of the coefficient matrix must be zero. The coefficient matrix has a special symmetric structure that allows factorization of its determinant.
<p><strong>Step 1:</strong> For non-trivial solutions, set the determinant equal to zero:</p><p>$$\begin{vmatrix} a-t & b & c \\ b & c-t & a \\ c & a & b-t \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Recognize the cyclic symmetric structure. Add all rows to the first row:</p><p>$$\begin{vmatrix} (a+b+c-t) & (a+b+c-t) & (a+b+c-t) \\ b & c-t & a \\ c & a & b-t \end{vmatrix} = 0$$</p><p><strong>Step 3:</strong> Factor out $(a+b+c-t)$ from the first row:</p><p>$$(a+b+c-t)\begin{vmatrix} 1 & 1 & 1 \\ b & c-t & a \\ c & a & b-t \end{vmatrix} = 0$$</p><p><strong>Step 4:</strong> Apply row operations ($R_2 - bR_1$ and $R_3 - cR_1$) and simplify to get:</p><p>$$(a+b+c-t)[-(t^2 - (a+b+c)t + ab + bc + ca)] = 0$$</p><p><strong>Step 5:</strong> This gives three values of $t$: one is $t_1 = a+b+c$ and two others from $t^2 - (a+b+c)t + ab + bc + ca = 0$</p><p><strong>Step 6:</strong> By Vieta's formulas for the quadratic, the product of all three roots equals:</p><p>$$t_1 \cdot t_2 \cdot t_3 = (a+b+c) \times (ab + bc + ca) = ab + bc + ca$$</p><p>∴ Answer: A</p>
Correct Answer: A