By using properties of determinants- show that- 1 x x2- x2 1 x- x x2 1 -1-x32

Applying R1 → R1 + R2 + R3, we have: Applying C2 → C2 − C1 and C3 → C3 − C1, we have: Expanding along R1, we have: Hence, the given result is proved. ...

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

Byju's Answer

Standard XII

Mathematics

Method of Intervals

By using prop...

Question

By using properties of determinants, show that:

Open in App

Solution

Applying R₁ → R₁ + R₂+ R₃, we have:

Applying C₂ → C₂ − C₁and C₃ → C₃ − C₁, we have:

Expanding along R₁, we have:

Hence, the given result is proved.

Suggest Corrections

Similar questions

Q. Prove that

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} x^{2} & 1 & x x & x^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 - x^{3})^{2}

Using the properties of determinants, show that:

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} x^{2} & 1 & x x & x^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 - x^{3})^{2}

Q. Using properties of determinants, show that

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} x^{2} & 1 & x x & x^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = {(1 - x^{3})}^{2}

Using properties of determinant, prove that:

∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & 1 + {p x}^{3} y & y^{2} & 1 + {p y}^{3} z & z^{2} & 1 + {p z}^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣

= (1 + p x y z) (x - y) (y - z) (z - x)

Q. Using properties of determinants prove the following:

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} x^{2} & 1 & x x & x^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 - x^{3})^{2}

Join BYJU'S Learning Program