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Standard XII

Mathematics

Evaluation of a Determinant

If . Delta - ...

Question

If $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x & sin θ & cos θ - sin θ & - x & 1 cos θ & 1 & x \end{matrix} ∣ ∣ ∣ ∣$ and $Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} x & sin 2 θ & cos 2 θ - sin 2 θ & - x & 1 cos 2 θ & 1 & x \end{matrix} ∣ ∣ ∣ ∣$ , $x \neq 0$ ; then for all $θ \in (0, \frac{π}{2})$ :

Δ_{1} - Δ_{2} = - 2 x^{3}

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Δ_{1} + Δ_{2} = - 2 x^{3}

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Δ_{1} + Δ_{2} = - 2 (x^{3} + x - 1)

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Δ_{1} - Δ_{2} = x (cos 2 θ - cos 4 θ)

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Solution

The correct option is B $Δ_{1} + Δ_{2} = - 2 x^{3}$
On expanding the determinant , we have
$Δ_{1} = x (- x^{2} - 1) - sin θ (- x sin θ - cos θ) + cos θ (- sin θ + x cos θ) = - x^{3} + x ({sin}^{2} θ + {cos}^{2} θ) - x = - x^{3}$

Similarly, $Δ_{2} = - x^{3}$
$Δ_{1} + Δ_{2} = - 2 x^{3}$

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Similar questions

Q. If

∆_{1} = |\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix}|, ∆_{2} = |\begin{matrix} 1 & b c & a \\ 1 & c a & b \\ 1 & a b & c \end{matrix}|, then

(a)

∆_{1} + ∆_{2} = 0

(b)

∆_{1} + 2 ∆_{2} = 0

(c)

∆_{1} = ∆_{2}

(d) none of these

$Δ_{1}$ , $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣$ , $Δ_{2}$ = $∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a 1 & c a & b 1 & a b & c \end{matrix} ∣ ∣ ∣ ∣$ then

Q. Let

Δ_{1} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & cos α & cos β cos α & 1 & cos γ cos β & cos γ & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

and

Δ_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & cos α & cos β cos α & 0 & cos γ cos β & cos γ & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ_{1} = Δ_{2}

, find

{sin}^{2} α + {sin}^{2} β + {sin}^{2} γ

Q. If

Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 7 & x & 2 - 5 & x + 1 & 3 4 & x & 7 \end{matrix} ∣ ∣ ∣ ∣, Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} x & 2 & 7 x + 1 & 3 & - 5 x & 7 & 4 \end{matrix} ∣ ∣ ∣ ∣

then

Δ_{1} - Δ_{2} = 0

for

Q. If

Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 10 & 4 & 3 17 & 7 & 4 4 & - 5 & 7 \end{matrix} ∣ ∣ ∣ ∣, Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 4 & x + 5 & 3 7 & x + 12 & 4 - 5 & x - 1 & 7 \end{matrix} ∣ ∣ ∣ ∣

such that

Δ_{1} + Δ_{2} = 0

then

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If Δ1=∣∣ ∣∣xsinθcosθ−sinθ−x1 cosθ1x∣∣ ∣∣ and Δ2=∣∣ ∣∣xsin2θcos2θ−sin2θ−x1 cos2θ1x∣∣ ∣∣ , x≠0 ; then for all θ∈(0,π2) :

The correct option is B Δ1+Δ2=−2x3 On expanding the determinant , we have Δ1=x(−x2−1)−sinθ(−xsinθ−cosθ) +cosθ(−sinθ+xcosθ) =−x3+x(sin2θ+cos2θ)−x =−x3 Similarly, Δ2=−x3 Δ1+Δ2=−2x3