3. Consider the matrix \[ M=\left[\begin{array}{rrrrrrr}1 & 2 & 1 & 3 & 6 & 0 & 5 \\ 0 & -2 & 1 & -3 & 5 & 1 & 4 \\ 1 & 1 & -2 & -3 & -4 & 0 & -1 \\ 2 & 3 & -3 & 1 & -2 & 0 & 1 \\ 3 & 7 & -3 & 7 & -1 & -1 & 2\end{array}\right] \] (a) (1 pt) Use Matlab to find the reduced row echelon form of \( M \). (b) ( 2 pts ) Use your answer to (a) to find a basis for nul( \( M \) ). (c) (2 pts) Suppose that \( M \) is the augmented matrix of a linear system. Use your answer to (a) to describe the set of solutions of this system as a set of vectors in \( \mathbb{R}^{6} \). Comment: You should take a screen shot of your Matlab work on (a), and include it with your submission. Parts (b), (c) and (d) should be done 'by hand'.

To solve the problem step by step, we will follow the instructions for each part. ### Part (a) To find the reduced row echelon form (RREF) of the matrix \( M \), we would typically use MATLAB. However, since I cannot run MATLAB directly, I will guide you on how to do it. 1. Open MATLAB. 2. Define the matrix \( M \): ```matlab M = [1 2 1 3 6 0 5; 0 -2 1 -3 5 1 4; 1 1 -2 -3 -4 0 -1; 2 3 -3 1 -2 0 1; 3 7 -3 7 -1 -1 2]; ``` 3. Use the `rref` function to find the reduced row echelon form: ```matlab R = rref(M); ``` 4. The output \( R \) will be the reduced row echelon form of \( M \). Make sure to take a screenshot of the output and include it in your submission. ### Part (b) To find a basis for the null space (nul) of \( M \), we will use the RREF obtained in part (a). 1. Let’s denote the RREF of \( M \) as \( R \). 2. Identify the pivot columns in \( R \). The columns without pivots correspond to the free variables. 3. Set up the equations based on the RREF to express the pivot variables in terms of the free variables. 4. Solve for the free variables to express the null space. Assuming the RREF of \( M \) is: \[ R = \begin{bmatrix} 1 & 0 & a_1 & 0 & a_2 & 0 & b_1 \\ 0 & 1 & a_3 & 0 & a_4 & 0 & b_2 \\ 0 & 0 & 0 & 1 & a_5 & 0 & b_3 \\ 0 & 0 & 0 & 0 & 0 & 1 & b_4 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \] where \( a_i \) and \( b_i \) are constants derived from the RREF. Let \( x_3, x_5, x_6 \) be free variables. Then we can express the solution in terms of these variables: \[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} = \begin{bmatrix} b_1 + a_1 x_3 + a_2 x_5 + a_3 x_6 \\ b_2 + a_4 x_5 + a_5 x_6 \\ x_3 \\ b_3 + a_6 x_6 \\ x_5 \\ x_6 \end{bmatrix} \] The basis for the null space will consist of vectors corresponding to each free variable. ### Part (c) To describe the set of solutions of the linear system represented by the augmented matrix \( M \), we will use the RREF obtained in part (a). 1. The general solution can be expressed as: \[ \mathbf{x} = \mathbf{p} + x_3 \mathbf{v_3} + x_5 \mathbf{v_5} + x_6 \mathbf{v_6} \] where \( \mathbf{p} \) is a particular solution and \( \mathbf{v_i} \) are the basis vectors for the null space corresponding to the free variables. 2. The solution set can be described as a linear combination of the basis vectors in \( \mathbb{R}^6 \). 3. Each free variable contributes a dimension to the solution space, and the number of free variables indicates the dimension of the null space. ### Summary - For part (a), run the MATLAB code to find the RREF and take a screenshot. - For part (b), identify the free variables from the RREF and express the null space. - For part (c), describe the solution set in terms of the basis vectors and a particular solution. If you provide the RREF from MATLAB, I can help you further with parts (b) and (c) based on that specific output.