Let $\mathbf{v} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $\mathbf{w} = \begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix}.$  The columns of a matrix are $\mathbf{u},$ $\mathbf{v},$ and $\mathbf{w},$ where $\mathbf{u}$ is a unit vector.  Find the largest possible determinant of the matrix.
To find the determinant of the matrix, I need to use the formula involving the dot product and the cross product of the vectors. The determinant is equal to $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}).$ The cross product of $\mathbf{v}$ and $\mathbf{w}$ is a vector that is perpendicular to both of them, and its magnitude is the area of the parallelogram spanned by them. To find the cross product, I can use the determinant of a 3x3 matrix with the standard basis vectors in the first row, and $\mathbf{v}$ and $\mathbf{w}$ in the second and third rows. The cross product is $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix} = (1 \cdot 3 - 0 \cdot (-1))\mathbf{i} - (2 \cdot 3 - 1 \cdot (-1))\mathbf{j} + (2 \cdot 0 - 1 \cdot 1)\mathbf{k} = 3\mathbf{i} - 7\mathbf{j} - \mathbf{k}.$ The dot product of $\mathbf{u}$ and the cross product is the same as the magnitude of $\mathbf{u}$ times the magnitude of the cross product times the cosine of the angle between them. The magnitude of $\mathbf{u}$ is 1, since it is a unit vector. The magnitude of the cross product is $\sqrt{3^2 + 7^2 + 1^2} = \sqrt{59}.$ The cosine of the angle between them is maximized when the angle is zero, or when they are parallel. Therefore, the largest possible value of the dot product is the magnitude of the cross product, or $\sqrt{59}.$ 