If the determinant D - - - 1 1 1- - - - - 2 - - 2 2- - - - 2- - 2 - - 2 - and D1 - 1 0 0 0 - - 0 - - then find the determinant of D2 such that D2 - DD1-

1 amp; 1 amp; 1 α +β #160; amp; α #160; ^ 2 +β #160; ^ 2 amp; 2αβ #160; α +β #160; amp; 2αβ #160; amp; α #160; ^ ...

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Byju's Answer

Standard XII

Mathematics

Properties of Determinants

If the determ...

Question

If the determinant $D =$ $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 α + β & α^{2} + β^{2} & 2 α β α + β & 2 α β & α^{2} + β^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ and $D_{1} =$ $∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & α & β 0 & β & α \end{matrix} ∣ ∣ ∣ ∣$ , then find the determinant of $D_{2}$ such that $D_{2} = \frac{D}{D_{1}}$ .

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Solution

$∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 α + β & α^{2} + β^{2} & 2 α β α + β & 2 α β & α^{2} + β^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ a n d$

$∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & α & β 0 & β & α \end{matrix} ∣ ∣ ∣ ∣ t o f i n d D_{2} = \frac{D}{D_{1}}$

Now,

$D = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 α + β & α^{2} + β^{2} & 2 α β α + β & 2 α β & α^{2} + β^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$

= $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 - (α + β) & 1 α + β & 2 α β & 2 α β α + β & α + β & α^{2} + β^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ {c_{2} ⟶ c_{2} - (α + β) c_{1}}$

= $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & - (α + β) & 1 α + β & 0 & 2 α β α + β & 0 & α^{2} + β^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ {c_{2} ⟶ c_{2} - c_{3}}$ \quad

= $- {- (α + β) ∣ ∣ ∣ \begin{matrix} α + β & 2 α β α + β & α^{2} + β^{2} \end{matrix} ∣ ∣ ∣}, e x p a n d i g w i t h R_{1}$

= $(α + β) ∣ ∣ ∣ \begin{matrix} α + β & 2 α β α + β & α^{2} + β^{2} \end{matrix} ∣ ∣ ∣$

$= (α + β) ∣ ∣ ∣ \begin{matrix} α + β & 2 α β 0 & α^{2} + β^{2} - 2 α β \end{matrix} ∣ ∣ ∣$ ${R_{2} \to R_{2} - R_{1}}$

$= (α + β) ∣ ∣ ∣ \begin{matrix} α + β & 2 α β 0 & {(α - β)}^{2} \end{matrix} ∣ ∣ ∣$

$= (α + β) \times (α + β) {(α - β)}^{2}$ , expanding the determinant

$= {(α + β)}^{2} {(α - β)}^{2}$

$= {[(α + β) (α - β)]}^{2}$

$= [α^{2} - β^{2}]$

Again,

$D_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & α & β 0 & β & α \end{matrix} ∣ ∣ ∣ ∣$

$= 1 (α^{2} - β^{2})$ , expanding by $R_{1}$

$= (α^{2} - β^{2})$

Now, $\frac{D}{D_{1}} = \frac{{(α^{2} - β^{2})}^{2}}{α^{2} - β^{2}}$

$= α^{2} - β^{2}$

$D_{2} = α^{2} - β^{2}$

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A body of density d and volume V floats with volume $V_{1}$ of it immersed in a liquid of density $d_{1}$ and rest of the volume $V_{2}$ immersed in another liquid of density $d_{2}$ (< $d_{1}$ ) then $V_{1}$ is

Q. Assertion :If

D = d i a g [d_{1}, d_{2}, . . . ., d_{n}]

, then

D^{- 1} = d i a g [d_{1}^{- 1}, d_{2}^{- 1} . . . . ., d_{n}^{- 1}]

Reason: If

D = d i a g [d_{1}, d_{2}, . . . . . . d_{n}],

then

D^{n} = d i a g [d_{1}^{n}, d_{2}^{n} . . . . . ., d_{n}^{n}] .

Q. If the two determinants

D_{1}

and

D_{2}

given below be equal, then prove that

c o s^{2} θ + c o s^{2} ϕ + c o s^{2} ψ = 1

where

D_{1}

∣ ∣ ∣ ∣ \begin{matrix} 1 & c o s θ & c o s ϕ c o s θ & 1 & c o s ψ c o s ϕ & c o s ψ & 1 \end{matrix} ∣ ∣ ∣ ∣

D_{2}

∣ ∣ ∣ ∣ \begin{matrix} 0 & c o s θ & c o s ϕ c o s θ & 0 & c o s ψ c o s ϕ & c o s ψ & 0 \end{matrix} ∣ ∣ ∣ ∣

Q. let

D_{1} = ∣ ∣ ∣ ∣ \begin{matrix} a & b & a + b c & d & c + d a & b & a - b \end{matrix} ∣ ∣ ∣ ∣

and

D_{2} = ∣ ∣ ∣ ∣ \begin{matrix} a & c & a + c b & d & b + d a & c & a + b + c \end{matrix} ∣ ∣ ∣ ∣

then the value of

\frac{D_{1}}{D_{2}}

where

b \neq 0

and

a d \neq b c

,is

Q. If

α, β

are the roots of

a x^{2} + b x + c = 0

;

α + h, β + h

are the roots of

p x^{2} + q x + r = 0

; and

D_{1}, D_{2}

are the respective discriminants of these equations, then

D_{1} : D_{2}

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If the determinant D= ∣∣ ∣ ∣∣111α+βα2+β22αβα+β2αβα2+β2∣∣ ∣ ∣∣ and D1= ∣∣ ∣∣1000αβ0βα∣∣ ∣∣, then find the determinant of D2 such that D2=DD1.