Which of the following equations shows commutativity in multiplication:

$a \times (b \times c) = (a \times b) \times c$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$a \times b = b \times a$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$a \times (b + c) = (a \times b + a \times c$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$2 \times 3 = 5 + 1$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
$a \times b = b \times a$

As per commutative law for multiplication, even if we change the order of the variables, their product will remain the same.

Suggest Corrections

Similar questions

∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + c b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3 (b - c) (c - a) (a - b) (a + b + c) (b c + c a + a b)

|\begin{matrix} a + b + c & - c & - b \\ - c & a + b + c & - a \\ - b & - a & a + b + c \end{matrix}| = 2 (a + b) (b + c) (c + a)

{tan}^{- 1} \sqrt{\frac{a (a + b + c)}{b c}} + {tan}^{- 1} \sqrt{\frac{b (a + b + c)}{c a}} + {tan}^{- 1} \sqrt{\frac{c (a + b + c)}{a b}}

a, b, c > 0

|\begin{matrix} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix}| = 0

a + b + c = 0 or, a = b = c

a, b

c

Δ = ∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b c + a & a + b & b + c a + b & b + c & c + a \end{matrix} ∣ ∣ ∣ ∣ = 0

a + b + c = 0

a = b = c

Join BYJU'S Learning Program