Question

then the value of lambda is ??​

Answer 1

So we're given,

$\small\displaystyle\longrightarrow\Delta=\left|\begin{array}{ccc}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{array}\right|$

Performing the operation $\small\displaystyle C_1\to C_1-C_2,$

$\small\displaystyle\longrightarrow\Delta=\left|\begin{array}{ccc}0&1&1\\a^2-b^2&b^2&c^2\\a^3-b^3&b^3&c^3\end{array}\right|$

$\small\displaystyle\longrightarrow\Delta=\left|\begin{array}{ccc}0&1&1\\(a-b)(a+b)&b^2&c^2\\(a-b)(a^2+ab+b^2)&b^3&c^3\end{array}\right|$

Taking $\small\displaystyle a-b$ common from $\small\displaystyle C_1,$

$\small\displaystyle\longrightarrow\Delta=(a-b)\left|\begin{array}{ccc}0&1&1\\a+b&b^2&c^2\\a^2+ab+b^2&b^3&c^3\end{array}\right|$

Performing the operation $\small\displaystyle C_2\to C_2-C_3,$

$\small\displaystyle\longrightarrow\Delta=(a-b)\left|\begin{array}{ccc}0&0&1\\a+b&b^2-c^2&c^2\\a^2+ab+b^2&b^3-c^3&c^3\end{array}\right|$

$\small\displaystyle\longrightarrow\Delta=(a-b)\left|\begin{array}{ccc}0&0&1\\a+b&(b-c)(b+c)&c^2\\a^2+ab+b^2&(b-c)(b^2+bc+c^2)&c^3\end{array}\right|$

Taking $\small\displaystyle b-c$ common from $\small\displaystyle C_2,$

$\small\displaystyle\longrightarrow\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0&0&1\\a+b&b+c&c^2\\a^2+ab+b^2&b^2+bc+c^2&c^3\end{array}\right|$

Performing the operation $\small\displaystyle C_2\to C_2-C_1,$

$\small\displaystyle\longrightarrow\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0&0&1\\a+b&c-a&c^2\\a^2+ab+b^2&bc+c^2-a^2-ab&c^3\end{array}\right|$

$\small\displaystyle\longrightarrow\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0&0&1\\a+b&c-a&c^2\\a^2+ab+b^2&b(c-a)+(c+a)(c-a)&c^3\end{array}\right|$

$\small\displaystyle\longrightarrow\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0&0&1\\a+b&c-a&c^2\\a^2+ab+b^2&(c-a)(a+b+c)&c^3\end{array}\right|$

Taking $\small\displaystyle c-a$ common from $\small\displaystyle C_2,$

$\small\displaystyle\longrightarrow\Delta=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0&0&1\\a+b&1&c^2\\a^2+ab+b^2&a+b+c&c^3\end{array}\right|$

Performing the operation $\small\displaystyle C_1\to C_1-(a+b)C_2,$

$\small\displaystyle\longrightarrow\Delta=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0&0&1\\a+b-(a+b)&1&c^2\\a^2+ab+b^2-(a+b)(a+b+c)&a+b+c&c^3\end{array}\right|$

$\small\displaystyle\longrightarrow\Delta=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0&0&1\\0&1&c^2\\-ab-bc-ca&a+b+c&c^3\end{array}\right|$

$\small\displaystyle\longrightarrow\Delta=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0&0&1\\0&1&c^2\\-(ab+bc+ca)&a+b+c&c^3\end{array}\right|$

Now expanding along $\small\displaystyle R_1$ or $\small\displaystyle C_1$ we get,

$\small\displaystyle\longrightarrow\Delta=(a-b)(b-c)(c-a)(ab+bc+ca)$

$\small\text{ \displaystyle\Longrightarrow\underline{\underline{\lambda=1}} }$

Answer 2

Step-by-step explanation:

So we're given,

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=\left|\begin{array}{ccc}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{array}\right|$}\end{gathered}

⟶Δ=

∣

∣

∣

∣

∣

∣

1

a

2

a

3

1

b

2

b

3

1

c

2

c

3

∣

∣

∣

∣

∣

∣

Performing the operation \small\text{$\displaystyle C_1\to C_1-C_2,$}C

1

→C

1

−C

2

,

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=\left|\begin{array}{ccc}0&1&1\\a^2-b^2&b^2&c^2\\a^3-b^3&b^3&c^3\end{array}\right|$}\end{gathered}

⟶Δ=

∣

∣

∣

∣

∣

∣

0

a

2

−b

2

a

3

−b

3

1

b

2

b

3

1

c

2

c

3

∣

∣

∣

∣

∣

∣

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=\left|\begin{array}{ccc}0&1&1\\(a-b)(a+b)&b^2&c^2\\(a-b)(a^2+ab+b^2)&b^3&c^3\end{array}\right|$}\end{gathered}

⟶Δ=

∣

∣

∣

∣

∣

∣

0

(a−b)(a+b)

(a−b)(a

2

+ab+b

2

)

1

b

2

b

3

1

c

2

c

3

∣

∣

∣

∣

∣

∣

Taking \small\text{$\displaystyle a-b$}a−b common from \small\text{$\displaystyle C_1,$}C

1

,

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=(a-b)\left|\begin{array}{ccc}0&1&1\\a+b&b^2&c^2\\a^2+ab+b^2&b^3&c^3\end{array}\right|$}\end{gathered}

⟶Δ=(a−b)

∣

∣

∣

∣

∣

∣

0

a+b

a

2

+ab+b

2

1

b

2

b

3

1

c

2

c

3

∣

∣

∣

∣

∣

∣

Performing the operation \small\text{$\displaystyle C_2\to C_2-C_3,$}C

2

→C

2

−C

3

,

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=(a-b)\left|\begin{array}{ccc}0&0&1\\a+b&b^2-c^2&c^2\\a^2+ab+b^2&b^3-c^3&c^3\end{array}\right|$}\end{gathered}

⟶Δ=(a−b)

∣

∣

∣

∣

∣

∣

0

a+b

a

2

+ab+b

2

0

b

2

−c

2

b

3

−c

3

1

c

2

c

3

∣

∣

∣

∣

∣

∣

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=(a-b)\left|\begin{array}{ccc}0&0&1\\a+b&(b-c)(b+c)&c^2\\a^2+ab+b^2&(b-c)(b^2+bc+c^2)&c^3\end{array}\right|$}\end{gathered}

⟶Δ=(a−b)

∣

∣

∣

∣

∣

∣

0

a+b

a

2

+ab+b

2

0

(b−c)(b+c)

(b−c)(b

2

+bc+c

2

)

1

c

2

c

3

∣

∣

∣

∣

∣

∣

Taking \small\text{$\displaystyle b-c$}b−c common from \small\text{$\displaystyle C_2,$}C

2

,

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0&0&1\\a+b&b+c&c^2\\a^2+ab+b^2&b^2+bc+c^2&c^3\end{array}\right|$}\end{gathered}

⟶Δ=(a−b)(b−c)

∣

∣

∣

∣

∣

∣

0

a+b

a

2

+ab+b

2

0

b+c

b

2

+bc+c

2

1

c

2

c

3

∣

∣

∣

∣

∣

∣

Performing the operation \small\text{$\displaystyle C_2\to C_2-C_1,$}C

2

→C

2

−C

1

,

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0&0&1\\a+b&c-a&c^2\\a^2+ab+b^2&bc+c^2-a^2-ab&c^3\end{array}\right|$}\end{gathered}

⟶Δ=(a−b)(b−c)

∣

∣

∣

∣

∣

∣

0

a+b

a

2

+ab+b

2

0

c−a

bc+c

2

−a

2

−ab

1

c

2

c

3

∣

∣

∣

∣

∣

∣

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0&0&1\\a+b&c-a&c^2\\a^2+ab+b^2&b(c-a)+(c+a)(c-a)&c^3\end{array}\right|$}\end{gathered}

⟶Δ=(a−b)(b−c)

∣

∣

∣

∣

∣

∣

0

a+b

a

2

+ab+b

2

0

c−a

b(c−a)+(c+a)(c−a)

1

c

2

c

3

∣

∣

∣

∣

∣

∣

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0&0&1\\a+b&c-a&c^2\\a^2+ab+b^2&(c-a)(a+b+c)&c^3\end{array}\right|$}\end{gathered}

⟶Δ=(a−b)(b−c)

∣

∣

∣

∣

∣

∣

0

a+b

a

2

+ab+b

2

0

c−a

(c−a)(a+b+c)

1

c

2

c

3

∣

∣

∣

∣

∣

∣

Taking \small\text{$\displaystyle c-a$}c−a common from \small\text{$\displaystyle C_2,$}C

2

,

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0&0&1\\a+b&1&c^2\\a^2+ab+b^2&a+b+c&c^3\end{array}\right|$}\end{gathered}

⟶Δ=(a−b)(b−c)(c−a)

∣

∣

∣

∣

∣

∣

0

a+b

a

2

+ab+b

2

0

1

a+b+c

1

c

2

c

3

∣

∣

∣

∣

∣

∣

Performing the operation \small\text{$\displaystyle C_1\to C_1-(a+b)C_2,$}C

1

→C

1

−(a+b)C

2

,

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0&0&1\\a+b-(a+b)&1&c^2\\a^2+ab+b^2-(a+b)(a+b+c)&a+b+c&c^3\end{array}\right|$}\end{gathered}

⟶Δ=(a−b)(b−c)(c−a)

∣

∣

∣

∣

∣

∣

0

a+b−(a+b)

a

2

+ab+b

2

−(a+b)(a+b+c)

0

1

a+b+c

1

c

2

c

3

∣

∣

∣

∣

∣

∣

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0&0&1\\0&1&c^2\\-ab-bc-ca&a+b+c&c^3\end{array}\right|$}\end{gathered}

⟶Δ=(a−b)(b−c)(c−a)

∣

∣

∣

∣

∣

∣

0

0

−ab−bc−ca

0

1

a+b+c

1

c

2

c

3

∣

∣

∣

∣

∣

∣

\begin{gathered}\small\text{$\displaystyle\longrightarrow\Delta=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0&0&1\\0&1&c^2\\-(ab+bc+ca)&a+b+c&c^3\end{array}\right|$}\end{gathered}

⟶Δ=(a−b)(b−c)(c−a)

∣

∣

∣

∣

∣

∣

0

0

−(ab+bc+ca)

0

1

a+b+c

1

c

2

c

3

∣

∣

∣

∣

∣

∣

Now expanding along \small\text{$\displaystyle R_1$}R

1

or \small\text{$\displaystyle C_1$}C

1

we get,

\small\text{$\displaystyle\longrightarrow\Delta=(a-b)(b-c)(c-a)(ab+bc+ca)$}⟶Δ=(a−b)(b−c)(c−a)(ab+bc+ca)

\small\text{$\displaystyle\Longrightarrow\underline{\underline{\lambda=1}}$}⟹

λ=1