If beginvmatrix x - k - x - k -2- x - k -3 1y - k - y - k -2- y - k -3 1z - k - z - k -2- z - k -3A- K-1B- k -2C- k -0D- k -1

The correct option is A K = 1x^k amp; x^k+2 amp; x^k+3 y^k amp; y^k+2 amp; y^k+3 z^k amp; z^k+2 amp; z^k+3 =x^k.y^k.z^k(x y)(y z)(z x)(xy+y ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Properties of Determinants

If beginvmatr...

Question

If $∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{k} & x^{k + 2} & x^{k + 3} y^{k} & y^{k + 2} & y^{k + 3} z^{k} & z^{k + 2} & z^{k + 3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x) (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ , then

k = 2

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

K = -1

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

k = 0

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

k = 1

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

Open in App

Solution

Suggest Corrections

Similar questions

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{k} & x^{k + 2} & x^{k + 3} y^{k} & y^{k + 2} & y^{k + 3} z^{k} & z^{k + 2} & z^{k + 3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x) (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{k} & x^{k + 2} & x^{k + 3} y^{k} & y^{k + 2} & y^{k + 3} z^{k} & z^{k + 2} & z^{k + 3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x) (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})

- k

\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k} a n d \frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}

2^{n d}, 3^{r d}, 4^{t h},

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{k} & x^{k + 1} & x^{k + 2} y^{k} & y^{k + 1} & y^{k + 2} z^{k} & z^{k + 1} & z^{k + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (r - 1)^{2} (1 - \frac{1}{r^{2}})

\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}

\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}

Join BYJU'S Learning Program