Respuesta :
Answer:
Let [tex]t\in {\mathbb R}[/tex],
[tex]\vec{u} = (t-2)\vec{u}_1 + (-2t +7)\vec{u}_2 + t\cdot \vec{u}_3[/tex].
[tex]\vec{w}[/tex] is also a linear combination of [tex]\vec{u}_1[/tex], [tex]\vec{u}_2[/tex], [tex]\vec{u}_3[/tex].
Step-by-step explanation:
1.
Write a linear system for [tex]\vec{u} = x_1\cdot\vec{u}_1 + x_2\cdot \vec{u}_2 + x_3\cdot \vec{u}_3[/tex], with one equation for each component. The augmented matrix for the first linear system will be:
[tex]\displaystyle \left[\begin{array}{ccc|c}1 & 2 & 3 & 12\\1 & 3 & 5 & 19\\2 & 5 & 8 & 31\end{array}\right][/tex].
Transform this matrix to its reduced row-echelon form using Gaussian Elimination. Solve for each variable.
[tex]\begin{aligned} &\left[\begin{array}{ccc|c}1 & 2 & 3 & 12\\1 & 3 & 5 & 19\\2 & 5 & 8 & 31\end{array}\right]\\ &\sim \left[\begin{array}{ccc|c}1 & 2 & 3 & 12\\0 & 1 & 2 & 7\\0 & 1 & 2 & 7\end{array}\right]\\&\sim \left[\begin{array}{ccc|c}1 & 2 & 3 & 12\\0 & 1 & 2 & 7\\0 & 0 & 0 & 0\end{array}\right]\\&\sim\left[\begin{array}{ccc|c}1 & 0 & -1 & -2\\0 & 1 & 2 & 7\\0 & 0 & 0 & 0\end{array}\right]\\&\left\{\begin{array}{l}x_1 = t-2\\x_2=- 2t+7\\x_3=t\end{array}\right.\end{aligned}[/tex].
Therefore,
[tex]\vec{u} = (t-2)\vec{u}_1 + (-2t +7)\vec{u}_2 + t\cdot \vec{u}_3[/tex].
2.
Set up a similar augmented matrix for [tex]\vec{w} = x_1\cdot\vec{u}_1 + x_2\cdot \vec{u}_2 + x_3\cdot \vec{u}_3[/tex]:
[tex]\left[\begin{array}{ccc|c}1 & 2 & 3 & 1\\1 & 3 & 5 & 0\\2 & 5 & 8 & 1\end{array}\right][/tex].
The second part of this question isn't concerned about the exact value of [tex]x_1[/tex], [tex]x_2[/tex], or [tex]x_3[/tex]. Therefore, before proceeding with Gaussian Elimination, start by checking the determinant of the coefficient matrix. If this determinant is nonzero, [tex]\vec{w}[/tex] will always be a unique linear combination of [tex]\vec{u}_1[/tex], [tex]\vec{u}_2[/tex], [tex]\vec{u}_3[/tex] now matter what value it takes.
In this case (also as seen in the first part of this question), the determinant of the coefficient matrix for [tex]\vec{u}_1[/tex], [tex]\vec{u}_2[/tex], and [tex]\vec{u}_3[/tex] is zero. Determining whether the linear combination is possible will require elimination.
[tex]\begin{aligned} &\left[\begin{array}{ccc|c}1 & 2 & 3 & 1\\1 & 3 & 5 & 0\\2 & 5 & 8 & 1\end{array}\right]\\ &\sim \left[\begin{array}{ccc|c}1 & 2 & 3 & 1\\0 & 1 & 2 & -1\\0 & 1 & 2 & -1\end{array}\right]\\&\sim \left[\begin{array}{ccc|c}1 & 2 & 3 & 1\\0 & 1 & 2 & -1\\0 & 0 & 0 & 0\end{array}\right]\\&\sim\left[\begin{array}{ccc|c}1 & 0 & -1 & 3\\0 & 1 & 2 & -1\\0 & 0 & 0 & 0\end{array}\right]\\&\left\{\begin{array}{l}x_1 = t+3\\x_2=- 2t-1\\x_3=t\end{array}\right.\end{aligned}[/tex].
Similar to the first part of this question, this linear system is consistent. [tex]\vec{w} = (t+3)\vec{u}_1 + (-2t -1)\vec{u}_2 + t\cdot \vec{u}_3[/tex]. [tex]\vec{w}[/tex] is indeed a linear combination of [tex]\vec{u}_1[/tex], [tex]\vec{u}_2[/tex], [tex]\vec{u}_3[/tex].
