Question

if w is imaginary cube root of 1, then show that (a+bw+cw^2) is a factor of ∆= 3×3 matrix
| a b c|
|b c a |
|c a b |

please answer in detailed and understandable form. please help it's urgent.
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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO PROVE

$\sf{ a + b \omega + c { \omega}^{2} \: } \: is \: a \: factor \: of \:$

$\displaystyle\begin{vmatrix} a & b & c\\ b & c & a \\ c & a & b \end{vmatrix}$

PROOF

Here neither of the two rows nor of the two columns are equal

So the determinant value is non zero

$\sf{Since \: \omega \: is \: a \: cube \: root \: of \: unity \: so} \: \: { \omega}^{3} = 1$

Now

$\displaystyle\begin{vmatrix} a & b & c\\ b & c & a \\ c & a & b \end{vmatrix}$

$= \sf{ \frac{1}{ { \omega}^{2}}. \frac{1}{ \omega }\: \displaystyle\begin{vmatrix} a & b \omega & c{ \omega}^{2}\\ b & c \omega & a{ \omega}^{2} \\ c & a \omega & b{ \omega}^{2} \end{vmatrix} }$

$= \displaystyle\begin{vmatrix} a & b \omega & c{ \omega}^{2}\\ b & c \omega & a{ \omega}^{2} \\ c & a \omega & b{ \omega}^{2} \end{vmatrix}$

$= \displaystyle\begin{vmatrix} a + b \omega + c{ \omega}^{2} & b \omega & c{ \omega}^{2}\\ b + c \omega + a{ \omega}^{2} & c \omega & a{ \omega}^{2} \\ c + a \omega + b{ \omega}^{2} & a \omega & b{ \omega}^{2} \end{vmatrix} \: \: \: ( \: C_1 '= C_1+ C_2 +C_3 \: )$

$= \displaystyle\begin{vmatrix} a + b \omega + c{ \omega}^{2} & b \omega & c{ \omega}^{2}\\ b{ \omega}^{3} + c { \omega}^{4} + a{ \omega}^{2} & c \omega & a{ \omega}^{2} \\ c { \omega}^{3} + a \omega + b{ \omega}^{2} & a \omega & b{ \omega}^{2} \end{vmatrix} \: \: \: ( \: \because \: { \omega}^{3} = 1 \: \: )$

$= \displaystyle\begin{vmatrix} a + b \omega + c{ \omega}^{2} & b \omega & c{ \omega}^{2}\\ a + b \omega + c{ \omega}^{2} & c \omega & a{ \omega}^{2} \\ a + b \omega + c{ \omega}^{2} & a \omega & b{ \omega}^{2} \end{vmatrix}$

$\sf{taking \: common \: \: \omega \: \: and \: \: { \omega}^{2} \: from \: 3rd \: and \: 2nd \: row \: respectively}$

$= ( \: a + b \omega + c{ \omega}^{2} \: ) \displaystyle\begin{vmatrix} 1 & b \omega & c{ \omega}^{2}\\ 1 & c \omega & a{ \omega}^{2} \\ 1 & a \omega & b{ \omega}^{2} \end{vmatrix}$

$\sf{This \: shows \: that \: ( a + b \omega + c{ \omega}^{2} ) \: is \: factor \: of}$

$\displaystyle\begin{vmatrix} a & b & c\\ b & c & a \\ c & a & b \end{vmatrix}$

Hence proved

━━━━━━━━━━━━━━━━

LEARN MORE FROM BRAINLY

For a square matrix A and a non-singular

matrix B of the same order, the value of

det(B inverse AB)

https://brainly.in/question/23783093

Answer 2

Step-by-step explanation:

\displaystyle\huge\red{\underline{\underline{Solution}}}

Solution

TO PROVE

\sf{ a + b \omega + c { \omega}^{2} \: } \: is \: a \: factor \: of \:a+bω+cω

2

isafactorof

\begin{gathered}\displaystyle\begin{vmatrix} a & b & c\\ b & c & a \\ c & a & b \end{vmatrix}\end{gathered}

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PROOF

Here neither of the two rows nor of the two columns are equal

So the determinant value is non zero

\sf{Since \: \omega \: is \: a \: cube \: root \: of \: unity \: so} \: \: { \omega}^{3} = 1Sinceωisacuberootofunitysoω

3

=1

Now

\begin{gathered}\displaystyle\begin{vmatrix} a & b & c\\ b & c & a \\ c & a & b \end{vmatrix}\end{gathered}

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a

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a

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a

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\begin{gathered}= \sf{ \frac{1}{ { \omega}^{2}}. \frac{1}{ \omega }\: \displaystyle\begin{vmatrix} a & b \omega & c{ \omega}^{2}\\ b & c \omega & a{ \omega}^{2} \\ c & a \omega & b{ \omega}^{2} \end{vmatrix} }\end{gathered}

=

ω

2

1

.

ω

1

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a

b

c

bω

cω

aω

cω

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aω

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bω

2

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\begin{gathered}= \displaystyle\begin{vmatrix} a & b \omega & c{ \omega}^{2}\\ b & c \omega & a{ \omega}^{2} \\ c & a \omega & b{ \omega}^{2} \end{vmatrix}\end{gathered}

=

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a

b

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bω

cω

aω

cω

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aω

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bω

2

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\begin{gathered}= \displaystyle\begin{vmatrix} a + b \omega + c{ \omega}^{2} & b \omega & c{ \omega}^{2}\\ b + c \omega + a{ \omega}^{2} & c \omega & a{ \omega}^{2} \\ c + a \omega + b{ \omega}^{2} & a \omega & b{ \omega}^{2} \end{vmatrix} \: \: \: ( \: C_1 '= C_1+ C_2 +C_3 \: )\end{gathered}

=

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a+bω+cω

2

b+cω+aω

2

c+aω+bω

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bω

cω

aω

cω

2

aω

2

bω

2

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(C

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′

=C

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+C

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+C

3

)

\begin{gathered}= \displaystyle\begin{vmatrix} a + b \omega + c{ \omega}^{2} & b \omega & c{ \omega}^{2}\\ b{ \omega}^{3} + c { \omega}^{4} + a{ \omega}^{2} & c \omega & a{ \omega}^{2} \\ c { \omega}^{3} + a \omega + b{ \omega}^{2} & a \omega & b{ \omega}^{2} \end{vmatrix} \: \: \: ( \: \because \: { \omega}^{3} = 1 \: \: )\end{gathered}

=

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a+bω+cω

2

bω

3

+cω

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+aω

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cω

3

+aω+bω

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bω

cω

aω

cω

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aω

2

bω

2

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(∵ω

3

=1)

\begin{gathered}= \displaystyle\begin{vmatrix} a + b \omega + c{ \omega}^{2} & b \omega & c{ \omega}^{2}\\ a + b \omega + c{ \omega}^{2} & c \omega & a{ \omega}^{2} \\ a + b \omega + c{ \omega}^{2} & a \omega & b{ \omega}^{2} \end{vmatrix}\end{gathered}

=

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a+bω+cω

2

a+bω+cω

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a+bω+cω

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bω

cω

aω

cω

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aω

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bω

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\sf{taking \: common \: \: \omega \: \: and \: \: { \omega}^{2} \: from \: 3rd \: and \: 2nd \: row \: respectively}takingcommonωandω

2

from3rdand2ndrowrespectively

\begin{gathered}= ( \: a + b \omega + c{ \omega}^{2} \: ) \displaystyle\begin{vmatrix} 1 & b \omega & c{ \omega}^{2}\\ 1 & c \omega & a{ \omega}^{2} \\ 1 & a \omega & b{ \omega}^{2} \end{vmatrix}\end{gathered}

=(a+bω+cω

2

)

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1

bω

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aω

cω

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\sf{This \: shows \: that \: ( a + b \omega + c{ \omega}^{2} ) \: is \: factor \: of}Thisshowsthat(a+bω+cω

2

)isfactorof

\begin{gathered}\displaystyle\begin{vmatrix} a & b & c\\ b & c & a \\ c & a & b \end{vmatrix}\end{gathered}

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Hence proved

━━━━━━━━━━━━━━━━

LEARN MORE FROM BRAINLY

For a square matrix A and a non-singular

matrix B of the same order, the value of

det(B inverse AB)

https://brainly.in/question/23783093