Fall 2026: Math 590 Quiz 2 Solutions

Problem 1

Let \(V\) be a vector space over \(F\) and \(v_1, \ldots, v_r\in V\). Define what it means for the given vectors to be linearly dependent, and then give an equivalent condition. (4 points)

Solution. \(v_1, \ldots, v_r\) are linearly dependent if and only if some \(v_i\) can be written as a linear combination of the remaining vectors \(v_1, \ldots, v_{i-1}, v_{i+1}, \ldots, v_r\). Equivalently, the vectors \(v_1, \ldots, v_r\) are said to be linearly dependent if there exist scalars \(\alpha_1, \ldots, \alpha_r\in F\), not all zero, such that

\[\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_rv_r = 0.\]

Problem 2

Let \(v_1 = \begin{pmatrix} 1\\1\\1\end{pmatrix}\), \(v_2 = \begin{pmatrix} 0\\1\\1\end{pmatrix}\), \(v_3 = \begin{pmatrix} 1\\1\\0\end{pmatrix}\). Use Gaussian elimination to determine if these vectors are linearly independent or linearly dependent. (3 points)

Solution. Form the matrix \(A\) with columns \(v_1, v_2, v_3\) and row reduce:

\[\begin{aligned} A = \begin{pmatrix} 1 & 0 & 1\\1 & 1 & 1\\1 & 1 & 0\end{pmatrix} &\xrightarrow{\substack{-R_1+R_2\\-R_1+R_3}} \begin{pmatrix} 1 & 0 & 1\\0 & 1 & 0\\0 & 1 & -1\end{pmatrix} \xrightarrow{-R_2+R_3} \begin{pmatrix} 1 & 0 & 1\\0 & 1 & 0\\0 & 0 & -1\end{pmatrix}\\[6pt] &\xrightarrow{\substack{-1\cdot R_3\\-R_3+R_1}} \begin{pmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}. \end{aligned}\]

Thus the system of equations with coefficient matrix \(A\) has only the zero solution, so the vectors are linearly independent.

Problem 3

For \(w = \begin{pmatrix} 4\\6\\3\end{pmatrix}\), use Gaussian elimination to write \(w\) as a linear combination of the vectors \(v_1, v_2, v_3\) given in the previous problem. (3 points)

Solution. We seek \(\alpha_1, \alpha_2, \alpha_3\) such that \(\alpha_1v_1+\alpha_2v_2+\alpha_3v_3 = w\). Row reduce the augmented matrix \([v_1\mid v_2\mid v_3\mid w]\):

\[\begin{aligned} \left[\begin{array}{ccc|c} 1 & 0 & 1 & 4\\1 & 1 & 1 & 6\\1 & 1 & 0 & 3\end{array}\right] &\xrightarrow{\substack{-R_1+R_2\\-R_1+R_3}} \left[\begin{array}{ccc|c} 1 & 0 & 1 & 4\\0 & 1 & 0 & 2\\0 & 1 & -1 & -1\end{array}\right] \xrightarrow{-R_2+R_3} \left[\begin{array}{ccc|c} 1 & 0 & 1 & 4\\0 & 1 & 0 & 2\\0 & 0 & -1 & -3\end{array}\right]\\[6pt] &\xrightarrow{\substack{-1\cdot R_3\\-R_3+R_1}} \left[\begin{array}{ccc|c} 1 & 0 & 0 & 1\\0 & 1 & 0 & 2\\0 & 0 & 1 & 3\end{array}\right]. \end{aligned}\]

Thus \(\alpha_1 = 1\), \(\alpha_2 = 2\), \(\alpha_3 = 3\), so

\[w = 1\cdot v_1+2\cdot v_2+3\cdot v_3,\]

which can be checked directly: \(1\begin{pmatrix} 1\\1\\1\end{pmatrix}+2\begin{pmatrix} 0\\1\\1\end{pmatrix}+3\begin{pmatrix} 1\\1\\0\end{pmatrix} = \begin{pmatrix} 1+0+3\\1+2+3\\1+2+0\end{pmatrix} = \begin{pmatrix} 4\\6\\3\end{pmatrix} = w\).