Let - - a a-b a-b-c- 3a 4a-3b 5a-4b-3c- 6a 9a-6b 11a-9b-6c- - -where a- i- b- and c-2- then prove that equals i-

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Byju's Answer

Standard XII

Mathematics

Definition of a Determinant

Let = |[ ...

Question

Let $△ =$ $∣ ∣ ∣ ∣ \begin{matrix} a & a + b & a + b + c 3 a & 4 a + 3 b & 5 a + 4 b + 3 c 6 a & 9 a + 6 b & 11 a + 9 b + 6 c \end{matrix} ∣ ∣ ∣ ∣$

where a= i, $b = ω$ , and $c = ω^{2}$ , then prove that $△$ equals i.

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Solution

Apply $R_{3} - 2 R_{2}, R_{2} - 3 R_{1}$ thus making two zeros and expand.
$△ = - a^{3} = - (i)^{3} = i$

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Similar questions

△ = ∣ ∣ ∣ ∣ \begin{matrix} a & a + b & a + b + c 3 a & 4 a + 3 b & 5 a + 4 b + 3 c 6 a & 9 a + 6 b & 11 a + 9 b + 6 c \end{matrix} ∣ ∣ ∣ ∣

where

a = i

b = ω

c = ω^{2}

, then

△

is equal to

Q. Let

Δ^{=}

∣ ∣ ∣ ∣ \begin{matrix} a & a + b & a + b + c 3 a & 4 a + 3 b & 5 a + 4 b + 3 c 6 a & 9 a + 6 b & 11 a + 9 b + 6 c \end{matrix} ∣ ∣ ∣ ∣

where

a = 1, b = ω

and

c = ω^{2}

then

Δ^{=}

Q. If

\frac{1}{a + ω} + \frac{1}{b + ω} + \frac{1}{c + ω} + \frac{1}{d + ω} = 2 ω^{2}

and

\frac{1}{a + ω^{2}} + \frac{1}{b + ω^{2}} + \frac{1}{c + ω^{2}} + \frac{1}{d + ω^{2}} = 2 ω

then prove that

\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} + \frac{1}{d + 1} = 2

Q. Assertion :Let

A = (\begin{matrix} i & 0 0 & i \end{matrix})

B = (\begin{matrix} 0 & - i - i & 0 \end{matrix})

,where

i = \sqrt{- 1}

then

(A + B) (A - B)

equals

A^{2} - B^{2}

Reason:

A B = B A

If $ω$ , $ω^{2}$ are imaginary cube roots of unity and

$\frac{1}{a + ω}$ + $\frac{1}{b + ω}$ + $\frac{1}{c + ω}$ = 2 $ω^{2}$ and $\frac{1}{a + ω^{2}}$ + $\frac{1}{b + ω^{2}}$ + $\frac{1}{c + ω^{2}}$ = 2 $ω$ , then $\frac{1}{a + 1}$ + $\frac{1}{b + 1}$ + $\frac{1}{c + 1}$ is equal to:

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Let △= ∣∣ ∣∣aa+ba+b+c3a4a+3b5a+4b+3c6a9a+6b11a+9b+6c∣∣ ∣∣where a= i, b=ω, and c=ω2, then prove that △ equals i.

Apply R3−2R2,R2−3R1 thus making two zeros and expand.△=−a3=−(i)3=i

Let $△ =$ $∣ ∣ ∣ ∣ \begin{matrix} a & a + b & a + b + c 3 a & 4 a + 3 b & 5 a + 4 b + 3 c 6 a & 9 a + 6 b & 11 a + 9 b + 6 c \end{matrix} ∣ ∣ ∣ ∣$

where a= i, $b = ω$ , and $c = ω^{2}$ , then prove that $△$ equals i.

Apply $R_{3} - 2 R_{2}, R_{2} - 3 R_{1}$ thus making two zeros and expand.
$△ = - a^{3} = - (i)^{3} = i$