Without expanding the determinant, prove that

Open in App

Solution

Hence, the given result is proved.

Suggest Corrections

Similar questions

∣ ∣ ∣ ∣ ∣ \begin{matrix} a & a^{2} & b c b & b^{2} & c a c & c^{2} & a b \end{matrix} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a^{2} & a^{3} 1 & b^{2} & b^{3} 1 & c^{2} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} a & p & ℓ b & q & m c & r & n \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} r & n & c q & m & b p & ℓ & a \end{matrix} ∣ ∣ ∣ ∣ = 0

∣ ∣ ∣ ∣ ∣ \begin{matrix} a & a^{2} & b c b & b^{2} & c a c & c^{2} & a b \end{matrix} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a^{2} & a^{3} 1 & b^{2} & b^{3} 1 & c^{2} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} a & b & c x & y & z p & q & r \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} y & b & q x & a & p z & c & r \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} x & y & z p & q & r a & b & c \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} 2 & 7 & 65 3 & 8 & 75 5 & 9 & 86 \end{matrix} ∣ ∣ ∣ ∣ = 0

Join BYJU'S Learning Program