Let p λ4+q λ3+r λ2+s λ+t =|λ2+3 λ λ-1 λ-3 λ-1 -2 λ λ-4 λ-3 λ+4 3 λ | where p, q, r, s and t are constants. Then value of t is

Question

Let p λ4+q λ3+r λ2+s λ+t =|λ2+3 λ λ-1 λ-3 λ-1 -2 λ λ-4 λ-3 λ+4 3 λ | where p, q, r, s and t are constants. Then value of t is (A) 0 (B) -1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

Putting λ=0, we obtain t=|0 -1 3 1 0 -4 -3 4 0|=0 as it is a skew symmetric determinant of odd order.

Q. Let $p λ^{4} + q λ^{3} + r λ^{2} + s λ + t$ $= ∣ ∣ λ^{2} + 3 λ λ - 1 λ - 3 λ - 1 - 2 λ λ + 4 λ - 3 λ - 4 3 λ ∣ ∣$
where $p, q, r, s$ and $t$ are constants. Then value of $t$ is

0

-1

2

3

Putting $λ = 0$ , we obtain
$t = ∣ ∣ 01 - 3 - 1 04 3 - 4 0 ∣ ∣ = 0$
as it is a skew symmetric determinant of odd order.

Q. Let pλ4+qλ3+rλ2+sλ+t =∣∣​λ2+3λλ−1λ−3​λ−1−2λλ+4​λ−3λ−43λ​∣∣​ where p,q,r,s and t are constants. Then value of t is

0

-1

2

3

Putting λ=0, we obtain t=∣∣​01−3​−104​3−40​∣∣​=0 as it is a skew symmetric determinant of odd order.

Q. Let $p λ^{4} + q λ^{3} + r λ^{2} + s λ + t$ $= ∣ ∣ λ^{2} + 3 λ λ - 1 λ - 3 λ - 1 - 2 λ λ + 4 λ - 3 λ - 4 3 λ ∣ ∣$
where $p, q, r, s$ and $t$ are constants. Then value of $t$ is

Putting $λ = 0$ , we obtain
$t = ∣ ∣ 01 - 3 - 1 04 3 - 4 0 ∣ ∣ = 0$
as it is a skew symmetric determinant of odd order.