If then

Engineering

Mathematics

Introduction to Determinants

Question

If $A = ∣ \begin{array}{ccc} a & b & c \\ x & y & z \\ p & q & r \end{array} ∣, B = ∣ \begin{array}{ccc} q & - b & y \\ - p & a & - x \\ r & - c & z \end{array} ∣$ then

|A| = |B|

|A| = – |B|

|A| = 2|B|

A is invertible if and only if B is invertible.

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$∣ B ∣ = ∣ \begin{array}{ccc} q & - b & y \\ - p & a & - x \\ r & - c & z \end{array} ∣ = (- 1) ∣ \begin{array}{ccc} q & - b & y \\ p & - a & x \\ r & - c & z \end{array} ∣$

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

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

= – |A| [∵ |A^T| = |A|]

∴ From here, we get |A| ≠ 0 if and only if |B| ≠ 0.
i.e A is invertible if and only if B is invertible.

Hence, options B and D.

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