If - 1 x x 2 x x 2 1 x 2 1 x - -7 and -1- x 3 -1 0 x-x 4 0 x-x 4 x 1 -1 x-x 4 x 3 -1 0 - then

The correct option is D Δ_1=49 If all elements of a determinant ( Δ #160; ) are replaced by their corresponding co factors then value new determina ...

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Byju's Answer

Standard XII

Mathematics

Adjoint of a Matrix

If Δ=| 1 ...

Question

If $Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} x & x^{2} & 1 x^{2} & 1 & x \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 7$ and $Δ_{1} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{3} - 1 & 0 & x - x^{4} 0 & x - x^{4} & x^{1} - 1 x - x^{4} & x^{3} - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣$ then

Δ_{1} = 7

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Δ_{1} = 343

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Δ_{1} = - 49

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Δ_{1} = 49

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Solution

The correct option is D $Δ_{1} = 49$
If all elements of a determinant $(Δ)$ are replaced by their corresponding co-factors then value new determinant will be $(Δ^{n - 1})$ when n is order to determinant
$\begin{matrix} = {(- 7)}^{2} = 49 \end{matrix}$
Option D is correct.

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Similar questions

If ∆ = |\begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix}|, ∆_{1} = |\begin{matrix} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{matrix}|, then prove that ∆ + ∆_{1} = 0 .

Q. If

Δ =

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

and

Δ_{1}

∣ ∣ ∣ ∣ \begin{matrix} a_{1} + p b_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} a_{2} + p b_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} a_{3} + p b_{3} & b_{3} + q c_{3} & c_{3} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣

then

Δ_{1} =

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

Q. 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

Q. If

Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{11} & a_{12} & a_{13} a_{21} & a_{22} & a_{23} a_{31} & a_{32} & a_{33} \end{matrix} ∣ ∣ ∣ ∣

and

c_{i j} = {(- 1)}^{i + j}

(determinant obtained by deleting ith row and jth column), then

∣ ∣ ∣ ∣ \begin{matrix} c_{11} & c_{12} & c_{13} c_{21} & c_{22} & c_{23} c_{31} & c_{32} & c_{33} \end{matrix} ∣ ∣ ∣ ∣ = Δ^{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} x & x^{2} & 1 x^{2} & 1 & x \end{matrix} ∣ ∣ ∣ ∣ ∣ = 7

and

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{3} - 1 & 0 & x - x^{4} 0 & x - x^{4} & x^{3} - 1 x - x^{4} & x^{3} - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

, then

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If Δ=∣∣ ∣ ∣∣1xx2xx21x21x∣∣ ∣ ∣∣=−7 and Δ1=∣∣ ∣ ∣∣x3−10x−x40x−x4x1−1x−x4x3−10∣∣ ∣ ∣∣ then

The correct option is D Δ1=49If all elements of a determinant (Δ) are replaced by their corresponding co-factors then value new determinant will be (Δn−1) when n is order to determinant=(−7)2=49Option D is correct.

If $Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} x & x^{2} & 1 x^{2} & 1 & x \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 7$ and $Δ_{1} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{3} - 1 & 0 & x - x^{4} 0 & x - x^{4} & x^{1} - 1 x - x^{4} & x^{3} - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣$ then

The correct option is D $Δ_{1} = 49$
If all elements of a determinant $(Δ)$ are replaced by their corresponding co-factors then value new determinant will be $(Δ^{n - 1})$ when n is order to determinant
$\begin{matrix} = {(- 7)}^{2} = 49 \end{matrix}$
Option D is correct.