Question

Use a linear system to write u = (12, 19, 31) as a linear combination of U1 = (1,1,2), u2 = (2,3,5), and uz = (3,5,8). Is w = (1,0,1) a linear com- bination of u1, U2, U3?

Answer 1

Let
`t\in {\mathbb R}`,

`\vec{u} = (t-2)\vec{u}_1 + (-2t +7)\vec{u}_2 + t\cdot \vec{u}_3`.

`\vec{w}` is also a linear combination of
`\vec{u}_1`,
`\vec{u}_2`,
`\vec{u}_3`.

Explanation:

1.

Write a linear system for
`\vec{u} = x_1\cdot\vec{u}_1 + x_2\cdot \vec{u}_2 + x_3\cdot \vec{u}_3`, with one equation for each component. The augmented matrix for the first linear system will be:

`\displaystyle \left[\begin{array}c1 & 2 & 3 & 12\\1 & 3 & 5 & 19\\2 & 5 & 8 & 31\end{array}\right]`.

Transform this matrix to its reduced row-echelon form using Gaussian Elimination. Solve for each variable.

`\begin{aligned} &\left[\begin{array}c1 & 2 & 3 & 12\\1 & 3 & 5 & 19\\2 & 5 & 8 & 31\end{array}\right]\\ &\sim \left[\begin{array}c1 & 2 & 3 & 12\\0 & 1 & 2 & 7\\0 & 1 & 2 & 7\end{array}\right]\\&\sim \left[\begin{array}ccc1 & 2 & 3 & 12\\0 & 1 & 2 & 7\\0 & 0 & 0 & 0\end{array}\right]\\&\sim\left[\begin{array}ccc1 & 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}`.

Therefore,

`\vec{u} = (t-2)\vec{u}_1 + (-2t +7)\vec{u}_2 + t\cdot \vec{u}_3`.

2.

Set up a similar augmented matrix for
`\vec{w} = x_1\cdot\vec{u}_1 + x_2\cdot \vec{u}_2 + x_3\cdot \vec{u}_3`:

`\left[\begin{array}c1 & 2 & 3 & 1\\1 & 3 & 5 & 0\\2 & 5 & 8 & 1\end{array}\right]`.

The second part of this question isn't concerned about the exact value of
`x_1`,
`x_2`, or
`x_3`. Therefore, before proceeding with Gaussian Elimination, start by checking the determinant of the coefficient matrix. If this determinant is nonzero,
`\vec{w}` will always be a unique linear combination of
`\vec{u}_1`,
`\vec{u}_2`,
`\vec{u}_3` 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
`\vec{u}_1`,
`\vec{u}_2`, and
`\vec{u}_3` is zero. Determining whether the linear combination is possible will require elimination.

`\begin{aligned} &\left[\begin{array}c1 & 2 & 3 & 1\\1 & 3 & 5 & 0\\2 & 5 & 8 & 1\end{array}\right]\\ &\sim \left[\begin{array}c1 & 2 & 3 & 1\\0 & 1 & 2 & -1\\0 & 1 & 2 & -1\end{array}\right]\\&\sim \left[\begin{array}ccc1 & 2 & 3 & 1\\0 & 1 & 2 & -1\\0 & 0 & 0 & 0\end{array}\right]\\&\sim\left[\begin{array}ccc1 & 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}`.

Similar to the first part of this question, this linear system is consistent.
`\vec{w} = (t+3)\vec{u}_1 + (-2t -1)\vec{u}_2 + t\cdot \vec{u}_3`.
`\vec{w}` is indeed a linear combination of
`\vec{u}_1`,
`\vec{u}_2`,
`\vec{u}_3`.