begin vmatrixa-b - b-c - c-ax-y - y-z - z-xp-q - q-r - r-pend vmatrix

The correct option is B 0a b amp; b c amp; c a x y amp; y z amp; z x p q amp; q r amp; r p = 0 amp; b c amp; c a 0 amp; y z am ...

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Sarrus Rule

begin vmatrix...

Question

$∣ ∣ ∣ ∣ \begin{matrix} a - b & b - c & c - a x - y & y - z & z - x p - q & q - r & r - p \end{matrix} ∣ ∣ ∣ ∣$

a(x+y+z)+b(p+q+r)+c

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abc+xyz+pqr

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None of these

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Solution

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∣ ∣ ∣ ∣ \begin{matrix} a & b & c x & y & z p & q & r \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} y & b & q x & a & p z & c & r \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} x & y & z p & q & r a & b & c \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} a & p & x b & q & y c & r & z \end{matrix} ∣ ∣ ∣ ∣ = 16

∣ ∣ ∣ ∣ \begin{matrix} p + x & a + x & a + p q + y & b + y & b + q r + z & c + z & c + r \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b q + r & r + p & p + q y + z & z + x & x + y \end{matrix} ∣ ∣ ∣ ∣ = 2 ∣ ∣ ∣ ∣ \begin{matrix} a & b & c p & q & r x & y & z \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b q + r & r + p & p + q y + z & z + x & x + y \end{matrix} ∣ ∣ ∣ ∣ = 2 ∣ ∣ ∣ ∣ \begin{matrix} a & b & c p & q & r x & y & z \end{matrix} ∣ ∣ ∣ ∣ .

∣ ∣ ∣ ∣ \begin{matrix} b + c & p + r & y + z c + a & r + p & z + x a + b & p + q & x + y \end{matrix} ∣ ∣ ∣ ∣ = 2 ∣ ∣ ∣ ∣ \begin{matrix} a & p & x b & q & y c & r & z \end{matrix} ∣ ∣ ∣ ∣

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