$∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ =$

abc(a+b+c)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 a^{2} b^{2} c^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C \N
$Δ = (b - a) (b - a) . ∣ ∣ ∣ ∣ \begin{matrix} b & b - c & c a & a - b & b c & c - a & a \end{matrix} ∣ ∣ ∣ ∣ = (a - b)^{2} ∣ ∣ ∣ ∣ \begin{matrix} b & b & c a & a & b c & c & a \end{matrix} ∣ ∣ ∣ ∣ = 0, [b y C_{2} \to C_{2} + C_{3}]$ .

Suggest Corrections

Similar questions

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

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

∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

|\begin{matrix} b^{2} - a b & b - c & b c - a c \\ a b - a^{2} & a - b & b^{2} - a b \\ b c - c a & c - a & a b - a^{2} \end{matrix}|

a b c (b - c) (c - a) (a - b)

(b - c) (c - a) (a - b)

(a + b + c) (b - c) (c - a) (a - b)

∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

Join BYJU'S Learning Program