Using properties of determinants- prove that- 3 a -a-b -a-c- -b-a 3 b -b-c- -c-a -c-b 3 c -3a-b-ca b-b c-c a

Expanding along C1, we have: Hence, the given result is proved. ...

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Standard XII

Mathematics

Invertible Element Binary Operation

Using propert...

Question

Using properties of determinants, prove that:

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Solution

Expanding along C₁, we have:

Hence, the given result is proved.

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Similar questions

Using properties of determinant, prove that:

∣ ∣ ∣ ∣ \begin{matrix} 3 a & - a + b & - a + c - b + a & 3 b & - b + c - c + a & - c + b & 3 c \end{matrix} ∣ ∣ ∣ ∣

= 3 (a + b + c) (a b + b c + c a)

Using properties of determinants prove the following questions.

$∣ ∣ ∣ ∣ \begin{matrix} 3 a & - a + b & - a + c - b + a & 3 b & - b + c - c + a & - c + b & 3 c \end{matrix} ∣ ∣ ∣ ∣ = 3 (a + b + c) (a b + b c + c a)$ .

Q. Prove the following identities:

\frac{a^{3} (b + c)}{(a - b) (a - c)} + \frac{b^{3} (c + a)}{(b - c) (b - a)} + \frac{a^{3} (a + b)}{(c - a) (c - b)} = b c + c a + a b

Q. Using properties of determinants, prove the following :

⎡ ⎢ ⎣ \begin{matrix} a & a^{2} & b c b & b^{2} & c a c & c^{2} & a b \end{matrix} ⎤ ⎥ ⎦

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

Using the property of determinants and without expanding, prove that

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a (b + c) 1 & c a & b (c + a) 1 & a b & c (a + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

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