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 \( [\mathbf{u} \mathbf{v} \) ]. 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. 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 \( \left[\begin{array}{ll}\mathbf{u} & \mathbf{v}\end{array}\right] \), \( \square \) The area of the parallelogram and the determinant of C. \( [\mathbf{u} \mathbf{v}] \) both equal 12 . Replace the first entry of \( \mathbf{v} \) by an arbitrary number x to make \( \mathbf{w}=\left[\begin{array}{l}\mathrm{x} \\ 4\end{array}\right] \). 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 less than the A. determinant of \( \left[\begin{array}{ll}\mathbf{u} & \mathbf{w}\end{array} \mathbf{H}\right] \), \( \square \) \( \square \). The area of the parallelogram and the determinant of B. \( [\mathbf{u} \mathbf{w}] \) both equal \( \square \) . C. determinant of \( \left[\begin{array}{ll}\mathbf{u} & \mathbf{w}] \text {, }\end{array}\right. \) \( \square \)

First, let’s find the area of the parallelogram determined by vectors \( \mathbf{u} \) and \( \mathbf{v} \). The area can be computed using the magnitude of the cross product of the two vectors when visualized in 3D, or more simply for 2D vectors using the determinant formula: \[ \text{Area} = | \mathbf{u} \times \mathbf{v} | = |\det([\mathbf{u} \mathbf{v}])| \] Calculating the determinant: \[ \det\left(\begin{bmatrix} 3 & 5 \\ 0 & 4 \end{bmatrix}\right) = (3)(4) - (0)(5) = 12 \] Thus, the area of the parallelogram is 12. Comparing the area with the determinant, they are equal, confirming: C. The area of the parallelogram and the determinant of \( [\mathbf{u} \mathbf{v}] \) both equal 12. Now, let's replace the first entry of \( \mathbf{v} \) with an arbitrary number \( x \) to make it \( \mathbf{w}=\left[\begin{array}{l}x \\ 4\end{array}\right] \). We need to calculate the new area: \[ \det\left(\begin{bmatrix} 3 & x \\ 0 & 4 \end{bmatrix}\right) = (3)(4) - (0)(x) = 12 \] Now, regardless of what \( x \) is, the determinant remains 12. This means: B. The area of the parallelogram and the determinant of \( [\mathbf{u} \mathbf{w}] \) both equal 12. When we draw the parallelogram formed by \( \mathbf{u} \) and \( \mathbf{w} \), we can show how changing \( x \) moves \( \mathbf{w} \) along the line defined by a fixed y-coordinate of 4, but it does not affect the area since the height (determined by the length of \( \mathbf{u} \) and the perpendicular dropped from the tail of \( \mathbf{u} \) to the line of \( \mathbf{w} \)) remains unchanged within limits of linear transformations. In conclusion, adjusting \( \mathbf{v} \) or \( \mathbf{w} \) while maintaining linear independence does not change the area of the parallelogram defined by these vectors, as confirmed by the consistent determinant value.