If A=abcxyzpqr,B=q−by−pa−xr−cz and if A is invertible. Then which of the following is true?

|B|=q−by−pa−xr−cz _( multiplying _R2 by-1) =−q −b yp −a xr −c zmultiplying C2 by −1 =qbypaxrcz=−paxqbyrcz=−axpbyqczr=−abcxyzpqr=−|A| Hence |A|=−|B|, obviously when |A|≠0,|B|≠0. Also, ∣ adj B|=|B2=(−|A|)2=|A|2

If A=abcxyzpqr,B=q−by−pa−xr−cz and if A is invertible. Then which of the following is true?

|B|=q−by−pa−xr−cz _( multiplying _R2 by-1) =−q −b yp −a xr −c zmultiplying C2 by −1 =qbypaxrcz=−paxqbyrcz=−axpbyqczr=−abcxyzpqr=−|A| Hence |A|=−|B|, obviously when |A|≠0,|B|≠0. Also, ∣ adj B|=|B2=(−|A|)2=|A|2

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If $A = [\begin{matrix} a & b & c \\ x & y & z \\ p & q & r \end{matrix}], B = [\begin{matrix} q & - b & y \\ - p & a & - x \\ r & - c & z \end{matrix}]$ and if A is invertible. Then which of the following is true?

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$| A | = | B |$

$| A | = - | B |$

$| A | = - 2 | B |$

$| A | = 2 | B |$

answer is B.

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Detailed Solution

$| B | = |\begin{array}{ccc} q & - b & y \\ - p & a & - x \\ r & - c & z \end{array}| \underline{} \underline{(multiplying} R_{2} by-1)$

$= - |\begin{array}{lll} q - b y \\ p - a x \\ r - c z \end{array}| (multiplying C_{2} by - 1)$

$\begin{aligned} = |\begin{array}{lll} q & b & y \\ p & a & x \\ r & c & z \end{array}| \\ = - |\begin{array}{lll} p & a & x \\ q & b & y \\ r & c & z \end{array}| \\ = - |\begin{array}{lll} a & x & p \\ b & y & q \\ c & z & r \end{array}| \\ = - |\begin{array}{lll} a & b & c \\ x & y & z \\ p & q & r \end{array}| \\ = - | A | \end{aligned}$