Let's solve the problem in **two steps** as you requested: --- ### **Step 1: Analyze the Row Equivalence and Linear Dependence Condition** - The matrix \( A = [u_1\ u_2\ u_3\ u_4\ u_5] \) must be **row equivalent** to \[ R = \begin{bmatrix} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 2 & 3 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} \] - The vectors \( u_2 - u_5, u_3, u_1 + u_4 \) **must be linearly dependent**. **Explanation:** Row equivalence means there is a sequence of elementary row operations that can transform \( A \) into \( R \). The linear dependence requirement means there exist scalars (not all zero) such that \(\alpha(u_2-u_5) + \beta u_3 + \gamma(u_1+u_4) = 0\). --- ### **Step 2: Try to Construct Such a Matrix or Explain Why It's Impossible** #### **Examine the Row-Reduced Form** - The matrix \( R \) has 3 pivots (one in each of the first three columns), so its rank is 3. - The columns \( u_1, u_2, u_3 \) are linearly independent (since they correspond to the pivot columns). - \( u_4 \) and \( u_5 \) can be written as linear combinations of the first three columns. Let's write \( R \) as column vectors: \[ u_1 = \begin{bmatrix}1\\0\\0\\0\end{bmatrix},\ u_2 = \begin{bmatrix}0\\1\\0\\0\end{bmatrix},\ u_3 = \begin{bmatrix}0\\0\\1\\0\end{bmatrix},\ u_4 = \begin{bmatrix}1\\2\\1\\0\end{bmatrix},\ u_5 = \begin{bmatrix}0\\3\\2\\0\end{bmatrix} \] Now, let's check if \( u_2-u_5, u_3, u_1+u_4 \) are linearly dependent. Let’s compute: \[ u_2 - u_5 = \begin{bmatrix}0\\1\\0\\0\end{bmatrix} - \begin{bmatrix}0\\3\\2\\0\end{bmatrix} = \begin{bmatrix}0\\-2\\-2\\0\end{bmatrix} \] \[ u_1 + u_4 = \begin{bmatrix}1\\0\\0\\0\end{bmatrix} + \begin{bmatrix}1\\2\\1\\0\end{bmatrix} = \begin{bmatrix}2\\2\\1\\0\end{bmatrix} \] \( u_3 = \begin{bmatrix}0\\0\\1\\0\end{bmatrix} \) Suppose \( a(u_2-u_5) + b u_3 + c(u_1+u_4) = 0 \): \[ a\begin{bmatrix}0\\-2\\-2\\0\end{bmatrix} + b\begin{bmatrix}0\\0\\1\\0\end{bmatrix} + c\begin{bmatrix}2\\2\\1\\0\end{bmatrix} = \begin{bmatrix}0\\0\\0\\0\end{bmatrix} \] This gives: - Row 1: \( 0a + 0b + 2c = 0 \implies c = 0 \) - Row 2: \( -2a + 0b + 2c = 0 \implies -2a = 0 \implies a = 0 \) - Row 3: \( -2a + 1b + 1c = 0 \implies b = 0 \) Thus, the only solution is \( a = b = c = 0 \). So, **the vectors are linearly independent**. --- ## **Final Answer** **It is impossible** to find such a matrix \( A \) that is row equivalent to the given matrix **and** for which \( u_2-u_5, u_3, u_1+u_4 \) are linearly dependent. **Explanation:** The row-reduced form has 3 pivot columns, making the first three columns linearly independent. The given vectors \( u_2-u_5, u_3, u_1+u_4 \) are thus also independent as shown, so the linear dependence condition cannot be satisfied.