Let \( \mathbf{u}=\left[\begin{array}{l}3 \\ 0\end{array}\right] \) and \( \mathbf{v}=\left[\begin{array}{l}5 \\ 4\end{array}\right] \). Compute the area of the parallelogram determined by \( \mathbf{u}, \mathbf{v}, \mathbf{u}+\mathbf{v} \), and \( \mathbf{0} \), and compute the determinant of \( [\mathbf{u} v \mathbf{v}] \). How do they compare? Replace the first entry of \( \mathbf{v} \) by an arbitrary number x , and repeat the problem. Draw a picture and explain what you find. Select the correct choice below and fill in the answer box(es) to complete your choice. (Simplify your answer.) The area of the parallelogram, \( \square \) , is greater than the A. determinant of \( [\mathbf{u} \mathbf{v}] \), \( \square \) The area of the parallelogram, \( \square \) , is less than the B. determinant of \( [\mathbf{u} \mathbf{v}] \), \( \square \) . The trea of the parallelogram and the determinant of C. \( [\mathbf{u} \mathbf{v}] \) both equal \( \square \)

To find the area of the parallelogram formed by \( \mathbf{u} \) and \( \mathbf{v} \), we can compute the determinant of the matrix formed by these vectors. The vectors are given by: \[ \mathbf{u} = \begin{bmatrix} 3 \\ 0 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 5 \\ 4 \end{bmatrix} \] The area of the parallelogram is given by the absolute value of the determinant of the matrix: \[ \text{Area} = \left| \det \begin{bmatrix} \mathbf{u} & \mathbf{v} \end{bmatrix} \right| = \left| \det \begin{bmatrix} 3 & 5 \\ 0 & 4 \end{bmatrix} \right| = \left| (3)(4) - (0)(5) \right| = \left| 12 \right| = 12 \] Now, we compute the determinant of \( [\mathbf{u} \, \mathbf{v}] \): \[ \det \begin{bmatrix} 3 & 5 \\ 0 & 4 \end{bmatrix} = 12 \] Both the area of the parallelogram and the determinant equal \( 12 \). Thus, we can conclude: The area of the parallelogram and the determinant of \( [\mathbf{u} \mathbf{v}] \) both equal \( 12 \). Now, substituting \( x \) for the first entry of \( \mathbf{v} \), we have: \[ \mathbf{v} = \begin{bmatrix} x \\ 4 \end{bmatrix} \] The new determinant will be: \[ \det \begin{bmatrix} 3 & x \\ 0 & 4 \end{bmatrix} = 12 \] The area of the parallelogram remains 12 as well, as the vectors remain in a similar position regardless of the first entry of \( \mathbf{v} \). From this, you'll observe that while the first element of \( \mathbf{v} \) changes, the area of the parallelogram does not change provided that \( y \) remains non-zero, demonstrating how the second component primarily influences the area. In summary, the area of the parallelogram and the determinant are the same and reveal fascinating stability, even as we tweak one of the components of \( \mathbf{v} \)!