Using properties of determinants- prove that- x x2 1-p x3- y y2 1-p y3- z z2 1-p z3 -1-p x y zx-yy-zz-x

Expanding along R3, we have: Hence, the given result is proved. ...

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Standard XII

Mathematics

Direct Method of Validation

Using propert...

Question

Using properties of determinants, prove that:

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Solution

Expanding along R₃, we have:

Hence, the given result is proved.

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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 that

ω ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & y & z x^{2} & y^{2} & z^{2} y + z & z + x & x + y \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x) (x + y + z)

Using properties of determinants prove the following questions.

$∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & 1 + p x^{3} y & y^{2} & 1 + p y^{3} z & z^{2} & 1 + p z^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 + p x y z) (x - y) (y - z) (z - x)$ , where p is any scalar.

By using properties of determinants, show that:

Using the properties of determinants, show that:

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

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