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Homework 4.pdf

1. (20 pts) Let \(A = \begin{pmatrix} 1 & 1 & -1 & 2 \\ 2 & 1 & 1 & 1 \\ 3 & 2 & 0 & 3 \\ 1 & -1 & 1 & 2 \end{pmatrix}\). Determine for which values of \(a\), the system \(Ax = \begin{pmatrix} a \\ 1 \\ 0 \\ 1 \end{pmatrix}\) has a solution. For these values of \(a\), find all possible solutions.

   \(\begin{pmatrix}\begin{array}{cccc|c}1 & 1 & -1 & 2 & a\\ 2 & 1 & 1 & 1 & 1\\ 3 & 2 & 0 & 3 & 0\\ 1 & -1 & 1 & 2 & 1\end{array}\end{pmatrix}\rightarrow R_1-R_4,R_2-2R_4,R_3-3R_4:\begin{pmatrix}\begin{array}{cccc|c}0 & 2 & -2 & 0 & a-1\\ 0 & 3 & -1 & -3 & -1\\ 0 & 5 & -3 & -3 & -3\\ 1 & -1 & 1 & 2 & 1\end{array}\end{pmatrix}\\\rightarrow R_1\leftrightarrow R_4,R_2\leftrightarrow R_3:\begin{pmatrix}\begin{array}{cccc|c}1 & -1 & 1 & 2 & 1\\ 0 & 5 & -3 & -3 & -3\\ 0 & 3 & -1 & -3 & -1\\ 0 & 2 & -2 & 0 & a-1\end{array}\end{pmatrix}\rightarrow R_2-R_3:\begin{pmatrix}\begin{array}{cccc|c}1 & -1 & 1 & 2 & 1\\ 0 & 2 & -2 & 0 & -2\\ 0 & 3 & -1 & -3 & -1\\ 0 & 2 & -2 & 0 & a-1\end{array}\end{pmatrix}\)​

   Then we can observe that \(R_2\) is same with \(R_4\), then we can let \(R_4-R_2\) then we get a zero row.

   \(\begin{pmatrix}\begin{array}{cccc|c}1 & -1 & 1 & 2 & 1\\ 0 & 2 & -2 & 0 & -2\\ 0 & 3 & -1 & -3 & -1\\ 0 & 0 & 0 & 0 & a+1\end{array}\end{pmatrix}\)​

   By theorem, if the system has a solution, then \(a+1=0\Rightarrow a=-1\)

   \(\rightarrow\frac12R_2:\begin{pmatrix}\begin{array}{cccc|c}1 & -1 & 1 & 2 & 1\\ 0 & 1 & -1 & 0 & -1\\ 0 & 3 & -1 & -3 & -1\\ 0 & 0 & 0 & 0 & 0\end{array}\end{pmatrix}\rightarrow R_1+R_2:\begin{pmatrix}\begin{array}{cccc|c}1 & 0 & 0 & 2 & 0\\ 0 & 1 & -1 & 0 & -1\\ 0 & 3 & -1 & -3 & -1\\ 0 & 0 & 0 & 0 & 0\end{array}\end{pmatrix}\)​

   \(\rightarrow R_3-3R_2:\begin{pmatrix}\begin{array}{cccc|c}1 & 0 & 0 & 2 & 0\\ 0 & 1 & -1 & 0 & -1\\ 0 & 0 & 2 & -3 & 2\\ 0 & 0 & 0 & 0 & 0\end{array}\end{pmatrix}\rightarrow R_2+R_3,\frac12R_3:\begin{pmatrix}\begin{array}{cccc|c}1 & 0 & 0 & 2 & 0\\ 0 & 1 & 1 & -3 & -1\\ 0 & 0 & 1 & -\frac32 & 1\\ 0 & 0 & 0 & 0 & 0\end{array}\end{pmatrix}\)

   Finally, let \(R_2-R_3:\begin{pmatrix}\begin{array}{cccc|c}1 & 0 & 0 & 2 & 0\\ 0 & 1 & 0 & -\frac32 & 0\\ 0 & 0 & 1 & -\frac32 & 1\\ 0 & 0 & 0 & 0 & 0\end{array}\end{pmatrix}\)​

   \(\begin{cases}x_1+2x_4=0\\ x_2-\frac32x_4=0\\ x_3-\frac32x_4=1\\ x_4\end{cases}\Rightarrow\begin{cases}x_1=-2x_4\\ x_2=\frac32x_4\\ x_3=1+\frac32x_4\\ x_4\end{cases}\)​

   Thus when \(x_4=0,x_1=0,x_2=0,x_3=1\)​

   Thus the solutions are \(\left\lbrace\left(-2a,\frac32a,1+\frac32a,a\right):a\in\mathbb{R}\right\rbrace\)​

2. (20 pts) Describe explicitly all \(3 \times 3\) row-reduced echelon matrices.

   \(\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & a\\ 0 & 1 & b\\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}1 & a & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}1 & a & b\\ 0 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}0 & 1 & a\\ 0 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}\)​

   \(a,b\in \mathbb{F}\)

3. (30 pts) Consider the system of equations \(Ax = 0\) where \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), is a \(2 \times 2\) matrix over a field \(\mathbb{F}\). Prove the following:

   1. If every entry of \(A\) is \(0\), then every \((x_1, x_2)\) is a solution of \(Ax = 0\).

      Let \(Ax=0\Rightarrow\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}\begin{pmatrix}x_1\\ x_2\end{pmatrix}=0\)​

      Then we have \(\begin{cases}0x_1+0x_2=0\\ 0x_1+0x_2=0\end{cases}\Rightarrow x_1,x_2\) can be any numbers in the field \(\mathbb{F}\)​ 2. If \(ad - bc \neq 0\), the system \(Ax = 0\) has only the trivial solution \((x_1, x_2) = (0, 0)\).

      Solution 1: Define \(A^{-1}\) in the following way: \(A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -a \\ -b & c \end{pmatrix}\)

      Notice that \(A \cdot A^{-1} = I\) therefore, if \(AX = 0\) then \(AA^{-1}X = 0\), i.e. \(IX = X = 0\) 3. If \(ad - bc = 0\) and some entry of \(A\) is not \(0\), then there exists a solution \((x_1^0, x_2^0)\) of the system \(Ax = 0\) such that \((x_1, x_2)\) is a solution of \(Ax = 0\) if and only if \((x_1, x_2) = (\lambda x_1^0, \lambda x_2^0)\) for some scalar \(\lambda \in \mathbb{F}\).

      Solution:

      Suppose that \(a\) is the nonzero entry. Since \(ad - bc = 0\) we know that \(ad = bc\). Thus we can perform the following reduction of \(A\):

      \(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\xrightarrow{\frac{1}{a}I}\begin{pmatrix}1 & \frac{b}{a}\\ c & d\end{pmatrix}\xrightarrow{-cI+II}\begin{pmatrix}1 & \frac{b}{a}\\ 0 & 0\end{pmatrix}\)​

      Thus \(AX = 0\) if and only if \(x_1 = -\frac{b}{a}x_2\), letting \(x_1^0 = -\frac{b}{a}\) and \(x_2^0 = 1\) we obtain what we want. An analogous argument can be given if we assume any other of the entries to be different from 0.

4. (30 pts) Use elementary row operations to discover whether the following matrices are invertible or not. If so, give the inverse explicitly.

   1. \(A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 2 & 4 \\ 0 & 0 & 3 \end{pmatrix}\) where the coefficients of \(A\) are in \(\mathbb{Z}_5\).

      Since we are in the field, thus we can calculate the inverse, then take the modular in the end.

      \(\begin{pmatrix}\begin{array}{ccc|ccc}1 & 2 & 0 & 1 & 0 & 0\\ 0 & 2 & 4 & 0 & 1 & 0\\ 0 & 0 & 3 & 0 & 0 & 1\end{array}\end{pmatrix}\rightarrow R_1-2R_2,\frac12R_2,\frac13R_3:\begin{pmatrix}\begin{array}{ccc|ccc}1 & 0 & -4 & 1 & -1 & 0\\ 0 & 1 & 2 & 0 & \frac12 & 0\\ 0 & 0 & 1 & 0 & 0 & \frac13\end{array}\end{pmatrix}\\\rightarrow R_1+4R_3,R_2-2R_3:\begin{pmatrix}\begin{array}{ccc|ccc}1 & 0 & 0 & 1 & -1 & \frac43\\ 0 & 1 & 0 & 0 & \frac12 & -\frac23\\ 0 & 0 & 1 & 0 & 0 & \frac13\end{array}\end{pmatrix}\)

      Thus \(A^{-1}=\begin{pmatrix}1 & -1 & \frac43\\ 0 & \frac12 & -\frac23\\ 0 & 0 & \frac13\end{pmatrix}=\begin{pmatrix}1 & 4 & 4\cdot2\\ 0 & 1\cdot3 & -2\cdot2\\ 0 & 0 & 1\cdot2\end{pmatrix}=\begin{pmatrix}1 & 4 & 8\\ 0 & 3 & -4\\ 0 & 0 & 2\end{pmatrix}=\begin{pmatrix}1 & 4 & 3\\ 0 & 3 & 1\\ 0 & 0 & 2\end{pmatrix}\)

      Since \(2\cdot3\equiv1(\mod5)\), thus the inverse of \(3\) is \(2\). 2. \(B = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 6 \\ 5 & 0 & 2 \end{pmatrix}\) where the coefficients of \(B\) are in \(\mathbb{Z}_7\).

      Since we are in the field, thus we can calculate the inverse, then take the modular in the end.

      \(\begin{pmatrix}\begin{array}{ccc|ccc}1 & 2 & 0 & 1 & 0 & 0\\ 0 & 3 & 6 & 0 & 1 & 0\\ 5 & 0 & 2 & 0 & 0 & 1\end{array}\end{pmatrix}\rightarrow\frac13R_2:\begin{pmatrix}\begin{array}{ccc|ccc}1 & 2 & 0 & 1 & 0 & 0\\ 0 & 1 & 2 & 0 & \frac13 & 0\\ 5 & 0 & 2 & 0 & 0 & 1\end{array}\end{pmatrix}\\\rightarrow R_3-5R_1:\begin{pmatrix}\begin{array}{ccc|ccc}1 & 2 & 0 & 1 & 0 & 0\\ 0 & 1 & 2 & 0 & \frac13 & 0\\ 0 & -10 & 2 & -5 & 0 & 1\end{array}\end{pmatrix}\rightarrow R_2-R_3\begin{pmatrix}\begin{array}{ccc|ccc}1 & 2 & 0 & 1 & 0 & 0\\ 0 & 11 & 0 & 5 & \frac13 & -1\\ 0 & -10 & 2 & -5 & 0 & 1\end{array}\end{pmatrix}\)​

      \(\rightarrow\frac{R_2}{11},\frac{R_3}{2}:\begin{pmatrix}\begin{array}{ccc|ccc}1 & 2 & 0 & 1 & 0 & 0\\ 0 & 1 & 0 & \frac{5}{11} & \frac{1}{33} & -\frac{1}{11}\\ 0 & -5 & 1 & -\frac52 & 0 & \frac12\end{array}\end{pmatrix}\rightarrow R_1-2R_2:\begin{pmatrix}\begin{array}{ccc|ccc}1 & 0 & 0 & \frac{1}{11} & -\frac{2}{33} & \frac{2}{11}\\ 0 & 1 & 0 & \frac{5}{11} & \frac{1}{33} & -\frac{1}{11}\\ 0 & -5 & 1 & -\frac52 & 0 & \frac12\end{array}\end{pmatrix}\\\rightarrow R_3+5R_2:\begin{pmatrix}\begin{array}{ccc|ccc}1 & 0 & 0 & \frac{1}{11} & -\frac{2}{33} & \frac{2}{11}\\ 0 & 1 & 0 & \frac{5}{11} & \frac{1}{33} & -\frac{1}{11}\\ 0 & 0 & 1 & -\frac{5}{22} & \frac{5}{33} & \frac{1}{22}\end{array}\end{pmatrix}\)​

      Thus \(B^{-1}=\begin{pmatrix}\frac{1}{11} & -\frac{2}{33} & \frac{2}{11}\\ \frac{5}{11} & \frac{1}{33} & -\frac{1}{11}\\ -\frac{5}{22} & \frac{5}{33} & \frac{1}{22}\end{pmatrix}=\begin{pmatrix}2 & -\frac43 & 4\\ 10 & \frac23 & -2\\ -5 & \frac{10}{3} & 1\end{pmatrix}=\begin{pmatrix}2 & -20 & 4\\ 10 & 10 & -2\\ -5 & 50 & 1\end{pmatrix}=\begin{pmatrix}2 & 1 & 4\\ 3 & 3 & 5\\ 2 & 1 & 1\end{pmatrix}\)

      Since \(11\cdot2\equiv1(\mod7)\), thus the inverse of \(11\) is \(2\).

      Since \(3\cdot5\equiv1(\mod7)\), thus the inverse of \(3\) is \(5\).

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