Question

Evaluate the following determinant:

(i) \(\begin{vmatrix} 1& 3 & 5 \\[0.3em] 2& 6 & 10 \\ 31 & 11 & 38 \end{vmatrix}\)

(ii) \(\begin{vmatrix}67& 19 & 21 \\[0.3em] 39& 13 & 14\\ 81& 24 & 26 \end{vmatrix}\)

(iii) \(\begin{vmatrix} a& h &g \\[0.3em] h& b & f \\ g & f & c \end{vmatrix}\)

(iv) \(\begin{vmatrix} 1& -3 & 2 \\[0.3em] 4& -1& 2 \\ 3 & 5 & 2\end{vmatrix}\)

Answer 1

(i)

(ii) 

= 1[(109) (12) – (119) (11)]

= 1308 – 1309

= – 1

Therefore, Δ = – 1

(iii) 

= a (bc – f²) – h (hc – fg) + g (hf – bg)

= abc – af² – ch² + fgh + fgh – bg²

= abc + 2fgh – af² – bg² – ch²

Therefore, Δ = abc + 2fgh – af² – bg² – ch²

(iv) 

= 2[1(24 – 4)] = 40

Therefore, Δ = 40