The value of the determinant $∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} {(a_{1} - b_{1})}^{2} & {(a_{1} - b_{2})}^{2} & {(a_{1} - b_{3})}^{2} & {(a_{1} - b_{4})}^{2} {(a_{2} - b_{1})}^{2} & {(a_{2} - b_{2})}^{2} & {(a_{2} - b_{3})}^{2} & {(a_{2} - b_{4})}^{2} {(a_{3} - b_{1})}^{2} & {(a_{3} - b_{2})}^{2} & {(a_{3} - b_{3})}^{2} & {(a_{3} - b_{4})}^{2} {(a_{4} - b_{1})}^{2} & {(a_{4} - b_{2})}^{2} & {(a_{4} - b_{3})}^{2} & {(a_{4} - b_{4})}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$ is

dependent on

a_{i}, i = 1, 2, 3, 4

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

dependent on

b_{i}, i = 1, 2, 3, 4

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

dependent on

a_{i j}, b_{i}, i = 1, 2, 3, 4

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

0

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

Open in App

Solution

The correct option is D $0$

Hence, option 'D' is correct.

Suggest Corrections

Similar questions

\frac{3}{2} x^{2} (x^{2} - 1) + \frac{1}{4} x^{2} (x^{2} + x) - \frac{3}{4} x (x^{3} - 1)

a^{2} (a^{2} - 2 a + 1) - a (a^{4} - 3 a^{3} + a)

a^{3} b^{2} (a^{2} - 1) - a^{2} b (a b + a^{2} b + 1) + a b^{2} (a^{3} - a^{2} + a)

(4 x^{2} - 5 y^{2}) (x^{2} + 2 y^{2})

(2 x^{2} - 1) (4 x^{3} + 5 x^{2})

a^{2} - b^{2} = (a + b) (a - b)

{(7 \frac{3}{4})}^{2} - {(2 \frac{1}{4})}^{2}

1 + (x y) + (x y)^{2} + (x y)^{3} + . . . .

a = 1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . (- 1 < a < 1)

b = 1 + 3 y + 6 y^{2} + 10 y^{3} + . . . (- 1 < b < 1)

{(1, 2)}, a n d B = {1, 3}, t h e n (A \times B) \cup (B \times A)

= {(1, 3), (2, 3), (3, 1), (3, 2), (1, 1), (1, 2), (2, 1),}

Join BYJU'S Learning Program