First slide

Applications of determinant

Question

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?

Moderate

A

$| A | = | B |$
B

$| A | = - | B |$
C

$| A | = - 2 | B |$
D

$| A | = 2 | B |$

Solution

$| B | = |\begin{array}{ccc} q & - b & y \\ - p & a & - x \\ r & - c & z \end{array}| \underline{} \underline{(multiplying} R_{2} by-1)$
$= - |\begin{array}{lll} q - b y \\ p - a x \\ r - c z \end{array}| (multiplying C_{2} by - 1)$
$\begin{aligned} = |\begin{array}{lll} q & b & y \\ p & a & x \\ r & c & z \end{array}| \\ = - |\begin{array}{lll} p & a & x \\ q & b & y \\ r & c & z \end{array}| \\ = - |\begin{array}{lll} a & x & p \\ b & y & q \\ c & z & r \end{array}| \\ = - |\begin{array}{lll} a & b & c \\ x & y & z \\ p & q & r \end{array}| \\ = - | A | \end{aligned}$
$Hence | A | = - | B |, obviously when | A | \neq 0, | B | \neq 0 . Also, ∣ adj {B | = | B|}^{2}$
$\begin{aligned} = (- | A |)^{2} \\ = | A |^{2} \end{aligned}$

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