= px4 + qx3 + rx2 + sx + t, then t =

Engineering

Mathematics

Introduction to Determinants

Question

$∣ \begin{array}{ccc} x^{2} + 3 & x - 1 & x + 3 \\ x + 3 & - 2 x & x - 4 \\ x - 3 & x + 4 & 3 x \end{array} ∣$ = px⁴ + qx³ + rx² + sx + t, then t =

– 48

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On expanding the determinant we get,
(x² + 3) (– 6x² – x² + 16) – (x – 1) (3x² + 9x – x² + 7x – 12) + (x + 3) (x² + 7x + 12 – 2x² + 6x)

Let us consider only the constant terms,
since t is constant term

∴ (16) (3) – (– 1) (– 12) + 3(12) = 72

Hence, t = 72

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