If -1-a b c x y z p q r and -2-y b q x a p z c r then -1 is equal to

The correct option is A Δ_2 Δ #160;_ 1 = a amp; b amp; c x amp; y amp; z p amp; q amp; r Interchanging rows by column ...

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Byju's Answer

Standard XII

Mathematics

Properties of Determinants

If Δ1=a b...

Question

If $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} a & b & c x & y & z p & q & r \end{matrix} ∣ ∣ ∣ ∣$ and $Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} y & b & q x & a & p z & c & r \end{matrix} ∣ ∣ ∣ ∣$ then $Δ_{1}$ is equal to

2 Δ_{2}

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Δ_{2}

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- Δ_{2}

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none of these

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Solution

The correct option is A $Δ_{2}$
$Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} a & b & c x & y & z p & q & r \end{matrix} ∣ ∣ ∣ ∣$
Interchanging rows by columns
$Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} a & x & p b & y & q c & z & r \end{matrix} ∣ ∣ ∣ ∣$
Replacing $C_{1}$ by $C_{2}$ , we get
$Δ_{1} = - ∣ ∣ ∣ ∣ \begin{matrix} x & a & p y & b & q y & c & r \end{matrix} ∣ ∣ ∣ ∣$
Replacing $R_{1}$ by $R_{2}$ , we get
$Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} y & b & q x & a & p z & c & r \end{matrix} ∣ ∣ ∣ ∣ = Δ_{2}$

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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. 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})

Q. If

Δ_{1} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a_{1}^{2} + b_{1} + c_{1} & a_{1} a_{2} + b_{2} + c_{2} & a_{1} a_{3} + b_{3} + c_{3} b_{1} b_{2} + c_{1} & b_{2}^{2} + c_{2} & b_{2} b_{1} + c_{3} c_{3} c_{1} & c_{3} c_{2} & c_{3}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

and

Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

, then

\frac{Δ_{1}}{Δ_{2}}

is equal to

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

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If Δ1=∣∣ ∣∣abcxyzpqr∣∣ ∣∣ and Δ2=∣∣ ∣∣ybqxapzcr∣∣ ∣∣ then Δ1 is equal to

The correct option is A Δ2Δ1=∣∣ ∣∣abcxyzpqr∣∣ ∣∣Interchanging rows by columnsΔ1=∣∣ ∣∣axpbyqczr∣∣ ∣∣Replacing C1 by C2, we getΔ1=−∣∣ ∣∣xapybqycr∣∣ ∣∣Replacing R1 by R2, we getΔ1=∣∣ ∣∣ybqxapzcr∣∣ ∣∣=Δ2

If $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} a & b & c x & y & z p & q & r \end{matrix} ∣ ∣ ∣ ∣$ and $Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} y & b & q x & a & p z & c & r \end{matrix} ∣ ∣ ∣ ∣$ then $Δ_{1}$ is equal to