11.8 Solving equation

1. Find all possible solutions of the following system of linear equations with coefficients in \(\mathbb{Z}_7\):
   ​\(\begin{cases} 2x + y + 3z + w = 3 \\ 2x + 4y + 2w = 0 \\ 2y + 3z + w = 1 \end{cases}\)

   After calculation, we have \((5-3w,1-6w,2-w,w)\)​

2. Describe explicitly all \(2 \times 2\) row-reduced echelon matrices.

   \(\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right), \quad \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right), \quad \left( \begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix} \right), \quad \left( \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right) \quad \Rightarrow \quad \left( \begin{matrix} 1 & \alpha \\ 0 & 0 \end{matrix} \right)\)​

3. Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M_2(\mathbb{C})\). Suppose \(A\) is row-reduced and that we have \(a + b + c + d = 0\). Prove that there are exactly 3 matrices which satisfy these conditions.

   \(\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix},\begin{pmatrix}1 & -1\\ 0 & 0\end{pmatrix},\begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix}\)​

4. Let \(A = \begin{pmatrix} 3 & -6 & 2 & -1 \\ -2 & 4 & 1 & 3 \\ 0 & 0 & 1 & 1 \\ 1 & -2 & 1 & 0 \end{pmatrix}\). For which vectors \(y = (y_1, y_2, y_3, y_4)\) does the system of equations \(Ax = y\) have a solution? For these vectors, give all possible solutions.

   We can get the argument matrix hardly: \(\begin{pmatrix}\begin{array}{cccc|c}1 & -2 & 0 & -1 & y_4-3y_4+y_1\\ 0 & 0 & 1 & -1 & 3y_4-y_1\end{array}\end{pmatrix}\)​

   Then we have\(\begin{aligned} & \begin{cases}x_1=2x_2+x_4+y_4,\\ x_2,\\ x_3=-x_4-y_1+3y_4,\\ x_4\end{cases}\\ & \begin{cases}y_2-3y_3+2y_4=0,\\ y_1+y_3-3y_4=0.\end{cases}\end{aligned}\)

5. For each of the following matrices, use elementary row operations to discover whether it is invertible or not. If invertible, find its inverse.

   (a) \(A = \begin{pmatrix} 3 & -1 & 2 \\ 2 & 1 & 1 \\ 1 & -3 & 0 \end{pmatrix}\) Yes! \(A^{-1}=\begin{pmatrix}-\frac12 & 1 & \frac12\\ -\frac16 & \frac13 & -\frac16\\ \frac76 & -\frac43 & -\frac56\end{pmatrix}\)

   \(\begin{pmatrix}\begin{array}{ccc|ccc}3 & -1 & 2 & 1 & 0 & 0\\ 2 & 1 & 1 & 0 & 1 & 0\\ 1 & -3 & 0 & 0 & 0 & 1\end{array}\end{pmatrix}\rightarrow\begin{pmatrix}\begin{array}{ccc|ccc}0 & 4 & 1 & \frac12 & 0 & -\frac32\\ 0 & 7 & 1 & 0 & 1 & -2\\ 1 & -3 & 0 & 0 & 0 & 1\end{array}\end{pmatrix}\rightarrow\begin{pmatrix}\begin{array}{ccc|ccc}0 & 4 & 1 & \frac12 & 0 & -\frac32\\ 0 & 1 & 0 & -\frac16 & \frac13 & -\frac16\\ 1 & -3 & 0 & 0 & 0 & 1\end{array}\end{pmatrix}\rightarrow\begin{pmatrix}\begin{array}{ccc|ccc}0 & 4 & 1 & \frac12 & 0 & -\frac32\\ 0 & 1 & 0 & -\frac16 & \frac13 & -\frac16\\ 1 & 0 & 0 & -\frac12 & 1 & \frac12\end{array}\end{pmatrix}\rightarrow\begin{pmatrix}\begin{array}{ccc|ccc}1 & 0 & 0 & -\frac12 & 1 & \frac12\\ 0 & 1 & 0 & -\frac16 & \frac13 & -\frac16\\ 0 & 0 & 1 & \frac76 & -\frac43 & -\frac56\end{array}\end{pmatrix}\)​

   (b) \(B = \begin{pmatrix} 1 & 1 & 1 & 2 \\ 1 & -3 & 3 & -8 \\ -2 & 1 & 2 & -2 \\ 1 & 2 & 1 & 4 \end{pmatrix}\)

   No!!

   \(\begin{pmatrix}\begin{array}{cccc|cccc}1 & 1 & 1 & 2 & 1 & 0 & 0 & 0\\ 1 & -3 & 3 & -8 & 0 & 1 & 0 & 0\\ -2 & 1 & 2 & -2 & 0 & 0 & 1 & 0\\ 1 & 2 & 1 & 4 & 0 & 0 & 0 & 1\end{array}\end{pmatrix}\rightarrow\begin{pmatrix}\begin{array}{cccc|cccc}1 & 1 & 1 & 2 & 1 & 0 & 0 & 0\\ 1 & -3 & 3 & -8 & 0 & 1 & 0 & 0\\ 0 & 3 & 4 & 2 & 2 & 0 & 1 & 0\\ 0 & 1 & 0 & 2 & -1 & 0 & 0 & 1\end{array}\end{pmatrix}\rightarrow\begin{pmatrix}\begin{array}{cccc|cccc}1 & 1 & 1 & 2 & 1 & 0 & 0 & 0\\ 1 & 0 & 7 & -6 & 2 & 1 & 1 & 0\\ 0 & 0 & 4 & -4 & 5 & 6 & 1 & -3\\ 0 & 1 & 0 & 2 & -1 & 0 & 0 & 1\end{array}\end{pmatrix}\rightarrow\) \(\begin{pmatrix}\begin{array}{cccc|cccc}1 & 0 & 1 & 0 & 2 & 0 & 0 & -1\\ 1 & 0 & 7 & -6 & 2 & 1 & 0 & 0\\ 0 & 0 & 4 & -4 & 5 & 6 & 1 & -3\\ 0 & 1 & 0 & 2 & -1 & 0 & 0 & 1\end{array}\end{pmatrix}\rightarrow\begin{pmatrix}\begin{array}{cccc|cccc}1 & 0 & 1 & 0 & 2 & 0 & 0 & -1\\ 0 & 0 & 6 & -6 & 0 & 1 & 1 & 1\\ 0 & 0 & 4 & -4 & 5 & 6 & 1 & -3\\ 0 & 1 & 0 & 2 & -1 & 0 & 0 & 1\end{array}\end{pmatrix}\)

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