Matrices & Determinants
Matrices and Determinants
Allen Star Batch
Grade 12
Question:
If $ax + by + cz = 0$, $bx + cy + az = 0$, $cx + ay + bz = 0$, $a,b,c \in \mathbb{R}^+$ then:
System have only trivial solution if $a^3 + b^3 + c^3 \neq 3abc$
System will have non trivial solution only if $a = b = c$
System have no solution if $a^3 + b^3 + c^3 = 3abc$
If system have non trivial solution then minimum value of $(x-1)^2 + (y-2)^2 + (z-3)^2$ is 12
Step-by-Step Solution
Key Concept: For a homogeneous linear system to have non-trivial solutions, the coefficient matrix determinant must equal zero. Here, det = 3abc - a³ - b³ - c³ = 0, which factors as -(a³ + b³ + c³ - 3abc) = -(a + b + c)(a² + b² + c² - ab - bc - ca). Since a,b,c ∈ ℝ⁺, non-trivial solutions exist only when a = b = c, giving the plane x + y + z = 0.
The determinant $|ABD| = 3abc - a^3 - b^3 - c^3 ≠ 0$ for only trivial solution. For non-trivial solution, $a = b = c$, which makes all three equations become $x + y + z = 0$. The distance from point $(1, 2, 3)$ to this plane is $\frac{|1+2+3|}{\sqrt{1^2+1^2+1^2}} = \frac{6}{\sqrt{3}}$. The minimum value of $(x-1)^2 + (y-2)^2 + (z-3)^2$ is the square of this distance: $\left(\frac{6}{\sqrt{3}}\right)^2 = 12$.
Correct Answer: 1,2,4