Determine all values of c so that the set {t+3,2t+c2+2} is linearly independent.

Question

Wendy Boykin · Accepted Answer

Step 1
The set is linearly independent if and only if the value

A_{1} B

in the equation.

A f_{1} + B f_{2} = 0 t + 0

A (t + 3) + B (2 t + c^{2} + 2) = 0 t + 0

Comparing coefficient of t and const term we get

A + 2 B = 0

(1)

3 A + B C c^{2} + 21 = 0

(2)
from (1)

A = - 2 B

put in (2)
Step 2

- 6 B + B (C^{2} + 2) = 0

B (- 6 + C^{2} + 2) = 0

C^{2} - 4 = 0, C^{2} = 4

C = \pm 2

at

C = 2 and - 2

the given set is linearly independent.

Beverly Smith · Accepted Answer

Step 1
To determine all values of c so that the set

{t + 3, 2 t + c^{2} + 2}

is linearly independent.
Step 2
We represent the above vectors

t + 3 and 2 t + c^{2} + 2

as matrix. Each vector is represented by a column in the matrix.
Therefore,

A = [\begin{matrix} 1 & 2 \\ 3 & c^{2} + 2 \end{matrix}]

If the vectors are linearly independent then the determinant of the matrix formed by those vectors is non zero.
Given the vectors are linearly independent.
Hence

det (A) \neq 0

\Rightarrow 1 (c^{2} + 2) - 6 \neq 0

\Rightarrow c^{2} - 4 \neq 0

\Rightarrow c^{2} \neq 4

\Rightarrow c \neq \pm 2

Therefore, the set

{t + 3, 2 t + c^{2} + 2}

is linearly independent for all values of c except +2 and -2.

nick1337 · Accepted Answer

Step 1 The set {t+3,2t+c2+2} is linearly independent if there exist non-zero scalars a and b such that, a(t+3)+b(2t+c2+2)=0 (at+2bt)+(3a+bc2+2b)=0 (a+2b)t+(3a+bc2+2b)=0 Equating the coefficient of t and the constant term in both sides of the above equation with zero we get, a + 2b = 0  a = - 2b Step 2 Again we have, NSK 3a+bc2+2b= 3(−2b)+bc2+2b=0 −6b+bc2+2b=0 bc2−4b=0 b(c2−4)=0 (c2−4)=0 c2=4 c = 2 ; c = -2 Hence the required values of c are c = 2 and c = - 2.

Determine all values of c so that the set \{t+3,

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