Matrices & Determinants
Matrices and Determinants
Allen Star Batch
Grade 12
Question:
If $a, b, c$ are non-zero real numbers such that $$\begin{vmatrix} bc & ca & ab \\ ca & ab & bc \\ ab & bc & ca \end{vmatrix} = 0$$, then:
$\frac{1}{a} + \frac{1}{bo} + \frac{1}{co^2} = 0$
$\frac{1}{a} + \frac{1}{bo^2} + \frac{1}{co} = 0$
$\frac{1}{ao} + \frac{1}{bo^2} + \frac{1}{c} = 0$
None of these
Step-by-Step Solution
Key Concept: The determinant equals zero when the rows form a cyclic pattern with bc, ca, ab. This occurs when (bc)³ + (ca)³ + (ab)³ - 3(bc)(ca)(ab) = 0, which factors using the identity x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx) and relates to cube roots of unity ω where 1 + ω + ω² = 0.
For the determinant $\begin{vmatrix} bc & ca & ab \\ ca & ab & bc \\ ab & bc & ca \end{vmatrix} = 0$, we use the identity $(ab)^3 + (bc)^3 + (ca)^3 - 3(ab)(bc)(ca) = 0$ which factors as $(ab + bc + ca)^3 - 3(ab)(bc)(ca) = 0$. This can be rewritten as three separate equations: $ab + bcw^2 + caw = 0$, $abw + bc + ca^2 = 0$, and $abw^2 + bcw + ca = 0$, where $w$ is a cube root of unity. Dividing by $abc$ gives $\frac{1}{cw^2} + \frac{1}{a} + \frac{1}{bw} = 0$ and cyclic permutations.
Correct Answer: 1,2,3