Question

$\left| \begin{matrix} 18 & 40 & 89 \\ 40 & 89 & 198 \\ 89 & 198 & 440 \end{matrix} \right|$ =

• A. 0
• B. 1
• C. 2
• D. –1

Answer 1

Answer: (D)

–1

Answer 2

R₂ → R₂ – 2R₁; R₃ → R₃ – 5R₁

= $\left| \begin{matrix} 18 & 40 & 89 \ 4 & 9 & 20 \

• 1 & - 2 & - 5 \end{matrix} \right|$⇒ ∆ =$\left| \begin{matrix} 18 & 40 & 9 \ 4 & 9 & 2 \
• 1 & - 2 & - 1 \end{matrix} \right|$

apply C₃ – 2C₂

∆ = $\left| \begin{matrix} 0 & 40 & 9 \\ 0 & 9 & 2 \\ 1 & - 2 & - 1 \end{matrix} \right|$C₁ → C₁ – 2C₃ = $\left| \begin{matrix} 40 & 9 \\ 9 & 2 \end{matrix} \right|$

= 80 – 81 = –1