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Byju's Answer

Standard XII

Mathematics

Adjoint of a Matrix

Show that a...

Question

Show that $∣ ∣ ∣ ∣ ∣ \begin{matrix} (a - a_{1})^{- 2} & (a - a_{1})^{- 1} & a_{1}^{- 1} (a - a_{2})^{- 2} & (a - a_{2})^{- 1} & a_{2}^{- 1} (a - a_{3})^{- 2} & (a - a_{3})^{- 1} & a_{3}^{- 1} \end{matrix} ∣ ∣ ∣ ∣ ∣ = \pm \frac{a^{2} \prod (a_{i} - a_{j})}{\prod a_{i} \prod (a - a_{i})^{2}}$ Write out the terms of the product in the numerator and give the resulting expression its correct sign.

- a^{2} (a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1})

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a^{2} (a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1})

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- a^{2} (a_{1} + a_{2}) (a_{2} + a_{3}) (a_{3} - a_{1})

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- a^{2} (a_{1} + a_{2}) (a_{2} + a_{3}) (a_{3} + a_{1})

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Solution

The correct option is A $- a^{2} (a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1})$
$L . H . S . = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (a - a_{1})^{- 2} & (a - a_{1})^{- 1} & a_{1}^{- 1} (a - a_{2})^{- 2} & (a - a_{2})^{- 1} & a_{2}^{- 1} (a - a_{3})^{- 2} & (a - a_{3})^{- 1} & a_{3}^{- 1} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow (a - a_{1})^{- 2} (a - a_{2})^{- 2} (a - a_{3})^{- 2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & (a - a_{1}) & a_{1}^{- 1} (a - a_{1})^{2} 1 & (a - a_{2}) & a_{2}^{- 2} (a - a_{2})^{2} 1 & (a - a_{3}) & a_{3}^{- 1} (a - a_{3})^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$

Apply $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$

$= \frac{1}{\prod (a - a_{i})^{2}} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & (a - a_{1}) & a_{1}^{- 1} (a - a_{1})^{2} 0 & (a_{1} - a_{2}) & \frac{((a^{2} - a_{1} a_{2}) (a_{1} - a_{2})}{a_{1} a_{2}} 0 & (a_{1} - a_{3}) & \frac{(a^{2} - a_{1} a_{3}) (a_{1} - a_{3})}{a_{1} a_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$

Expanding w.r.to 1st row, we have

$\Rightarrow \frac{1}{\prod (a - a_{i})^{2}} ∣ ∣ ∣ ∣ ∣ \begin{matrix} (a_{1} - a_{2}) & \frac{(a^{2} - a_{1} a_{2}) (a_{1} - a_{2})}{a_{1} a_{2}} (a_{1} - a_{3}) & \frac{(a^{2} - a_{1} a_{3}) (a_{1} - a_{3})}{a_{1} a_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow \frac{(a_{1} - a_{2}) (a_{1} - a_{3})}{\prod (a - a_{i})^{2}} ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & \frac{a^{2} - a_{1} a_{2}}{a_{1} a_{2}} 1 & \frac{a^{2} - a_{1} a_{3}}{a_{1} a_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow \frac{(a_{1} - a_{2}) (a_{1} - a_{3}) a^{2} (a_{2} - a_{3})}{a_{1} a_{2} a_{2} \prod (a - a_{i})^{2}} = \frac{- a^{2} \prod (a_{i} - a_{j})}{\prod a_{i} \prod (a - a_{i})^{2}}$

Numerator $= - a^{2} (a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1})$
The resulting expression has negative sign.

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Similar questions

Q. If

a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} \in R

and are such that

a_{i} b_{j} \neq 1

for

1 \leq i, j \leq 3

, then

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1 - a_{1}^{3} b_{1}^{3}}{1 - a_{1} b_{1}} & \frac{1 - a_{1}^{3} b_{2}^{3}}{1 - a_{1} b_{2}} & \frac{1 - a_{1}^{3} b_{3}^{3}}{1 - a_{1} b_{3}} \frac{1 - a_{2}^{3} b_{1}^{3}}{1 - a_{2} b_{1}} & \frac{1 - a_{2}^{3} b_{2}^{3}}{1 - a_{2} b_{2}} & \frac{1 - a_{2}^{3} b_{3}^{3}}{1 - a_{2} b_{3}} \frac{1 - a_{3}^{3} b_{1}^{3}}{1 - a_{3} b_{1}} & \frac{1 - a_{3}^{3} b_{2}^{3}}{1 - a_{3} b_{2}} & \frac{1 - a_{3}^{3} b_{3}^{3}}{1 - a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

> 0 Provided either

a_{1} < a_{2} < a_{3}

and

b_{1} < b_{2} < b_{3}

, or

a_{1} > a_{2} a_{3}

and

b_{1} > b_{2} > b_{3}

then show

(a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1}) (b_{1} - b_{2}) (b_{2} - b_{3}) (b_{3} - b_{1}) < 0

Q. If

a_{1}, a_{2}, a_{3} \dots \dots \dots, a_{n}

are in

H . P .

then

\frac{a_{1}}{a_{2} + a_{3} + \dots \dots \dots \dots + a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + \dots \dots \dots + a_{n}} \dots \dots \frac{a_{n}}{a_{1} + a_{2} + \dots \dots + a_{n - 1}}

are in

Q. If in the determinant

Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣, A_{1}, B_{1}, C_{1}

etc. be the co-factors of

a_{1}, b_{1}, c_{1}

etc., then which of the following relations is incorrect

Given that a_i $>$ 0 and i belongs to a set of natural numbers. If $a_{1}, a_{2}, a_{3} . . . . . a_{2 n}$ are in AP, then find the value of $\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n} - 1}{{\sqrt{a}}_{2} + \sqrt{a_{3}}} . . . . . . . . . . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}$

Q. State true or false:

a_{1}, a_{2}, a_{3} \in + R

then,

(a_{1} + a_{2} + a_{3}) (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}}) \geq 9

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Show that ∣∣ ∣ ∣∣(a−a1)−2(a−a1)−1a−11(a−a2)−2(a−a2)−1a−12(a−a3)−2(a−a3)−1a−13∣∣ ∣ ∣∣=±a2∏(ai−aj)∏ai∏(a−ai)2 Write out the terms of the product in the numerator and give the resulting expression its correct sign.