To determine the value of [tex]\( t \)[/tex] such that the vector [tex]\((4, 6, t)\)[/tex] is a linear combination of the vectors [tex]\((1, 3, 1)\)[/tex] and [tex]\((2, 8, -1)\)[/tex], we can take the following steps:
1. Express the Problem in Terms of a Linear Combination:
We need to find scalars [tex]\( a \)[/tex] and [tex]\( b \)[/tex] such that:
[tex]\[
a \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} + b \begin{pmatrix} 2 \\ 8 \\ -1 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \\ t \end{pmatrix}
\][/tex]
2. Set Up the System of Equations:
This leads to three equations, one for each coordinate:
[tex]\[
a \cdot 1 + b \cdot 2 = 4 \quad \quad (1)
\][/tex]
[tex]\[
a \cdot 3 + b \cdot 8 = 6 \quad \quad (2)
\][/tex]
[tex]\[
a \cdot 1 + b \cdot (-1) = t \quad \quad (3)
\][/tex]
3. Solve for [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
Using equations (1) and (2), we can solve for [tex]\( a \)[/tex] and [tex]\( b \)[/tex].
From equation (1), we have:
[tex]\[
a + 2b = 4 \quad \quad (4)
\][/tex]
From equation (2), we have:
[tex]\[
3a + 8b = 6 \quad \quad (5)
\][/tex]
4. Eliminate One Variable:
To solve the system, we can eliminate [tex]\( a \)[/tex] by multiplying equation (4) by 3 and subtracting it from equation (5):
[tex]\[
3(a + 2b) = 3 \cdot 4
\][/tex]
So,
[tex]\[
3a + 6b = 12
\][/tex]
Subtract equation (5):
[tex]\[
(3a + 8b) - (3a + 6b) = 6 - 12
\][/tex]
[tex]\[
2b = -6
\][/tex]
[tex]\[
b = -3
\][/tex]
5. Substitute [tex]\( b \)[/tex] back into Equation (4):
Using [tex]\( b = -3 \)[/tex]:
[tex]\[
a + 2(-3) = 4
\][/tex]
[tex]\[
a - 6 = 4
\][/tex]
[tex]\[
a = 10
\][/tex]
6. Determine [tex]\( t \)[/tex]:
Now that we have [tex]\( a \)[/tex] and [tex]\( b \)[/tex], substitute these back into equation (3):
[tex]\[
a \cdot 1 + b \cdot (-1) = t
\][/tex]
[tex]\[
10 \cdot 1 + (-3) \cdot (-1) = t
\][/tex]
[tex]\[
10 + 3 = t
\][/tex]
[tex]\[
t = 13
\][/tex]
Thus, the value of [tex]\( t \)[/tex] is [tex]\( 13 \)[/tex]. Therefore, the correct answer is:
E. 13
