Example of using Row Theorem to prove a statement

Span{(110),(101),(011)}=R3?(110101011)→(1100−11011)→(1100−11002)→(1010−11001)→(10001−1001)→(100010001)Because there is a pivot in every row, the span must be equal to R3Span\{\begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}0\\1\\1\end{pmatrix} \} = \mathbb{R}^3?\\ \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 2 \end{pmatrix} \rightarrow \\ \begin{pmatrix} 1 & 0 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \\ \text{Because there is a pivot in every row, the span must be equal to } \mathbb{R}^3

Homogeneous SLE’s

A SLE is homogeneous‾ if its matrix equation is of the form Ax→=0→ and is not homogeneous otherwise.A solution where x→=0→ is a trivial‾ solution, as it will always solve Ax→=0→. Any other solution is nontrivial‾.\text{A SLE is \underline{homogeneous} if its matrix equation is of the form } A\overrightarrow{x} = \overrightarrow{0} \text{ and is not homogeneous otherwise.} \\ \text{A solution where } \overrightarrow{x} = \overrightarrow{0} \text{ is a \underline{trivial} solution, as it will always solve } A\overrightarrow{x} = \overrightarrow{0} \text{. Any other solution is \underline{nontrivial}.}

Example

Does {3x1+5x2−4x3=0−3x1−2x2+4x3=06x1+x2−8x3=0 have a nontrivial solution?{3x1+5x2−4x3=0−3x1−2x2+4x3=06x1+x2−8x3=0→(35−4−3−2461−8)(x1x2x3)=(000)(35−40−3−24061−80)→(35−4003000−300)→(35−4001000000)→(10−43001000000){x1=43x2=0x3=x3=x3(4301) solves the matrix equation.Thus, there ARE nontrivial solutions to the system of equations, as the final vector found is not the 0 vector\text{Does } \begin{cases} 3x_1 + 5x_2 - 4x_3 = 0 \\ -3x_1 -2x_2 + 4x_3 = 0 \\ 6x_1 + x_2 - 8x_3 = 0 \end{cases} \text{ have a nontrivial solution?} \\ \begin{cases} 3x_1 + 5x_2 - 4x_3 = 0 \\ -3x_1 -2x_2 + 4x_3 = 0 \\ 6x_1 + x_2 - 8x_3 = 0 \end{cases} \rightarrow \begin{pmatrix} 3 & 5 & -4 \\ -3 & -2 & 4 \\ 6 & 1 & -8 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ \left( \begin{array}{ccc|c} 3 & 5 & -4 & 0 \\ -3 & -2 & 4 & 0 \\ 6 & 1 & -8 & 0 \\ \end{array} \right) \rightarrow \left( \begin{array}{ccc|c} 3 & 5 & -4 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & -3 & 0 & 0 \\ \end{array} \right) \rightarrow \left( \begin{array}{ccc|c} 3 & 5 & -4 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \rightarrow \left( \begin{array}{ccc|c} 1 & 0 & -\frac{4}{3} & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \\ \begin{cases} x_1 = \frac{4}{3} \\ x_2 = 0 \\ x_3 = x_3 \\ \end{cases} = x_3 \begin{pmatrix} \frac{4}{3} \\ 0 \\ 1 \end{pmatrix} \text{ solves the matrix equation.} \\ \text{Thus, there ARE nontrivial solutions to the system of equations, as the final vector found is not the 0 vector} Ax→=0→ has nontrivial solutions whenever there are free variables but will not if there are no free variables.A\overrightarrow{x} = \overrightarrow{0} \text{ has nontrivial solutions whenever there are free variables but will not if there are no free variables.}

Relating b vector to zero vector

The general solution to Ax→=b→ is x→=x0→+xp→ where xp→ is a particular solution to Ax→=b→ and x0→ is the general solution to Ax→=0→\text{The general solution to } A\overrightarrow{x} = \overrightarrow{b} \text{ is } \overrightarrow{x} = \overrightarrow{x_0} + \overrightarrow{x_p} \\ \text{ where } \overrightarrow{x_p} \text{ is a particular solution to } A\overrightarrow{x} = \overrightarrow{b} \text{ and } \overrightarrow{x_0} \text{ is the general solution to } A\overrightarrow{x} = \overrightarrow{0}

Example

A=(101012344812166121824)b→=(0−2−8−12)Suppose we know that xp→=(10−10) solves Ax→=b→Solve Ax→=0→ and add x0→ to xp→(A0→)=(1010012340481216061218240)→(10100011200000000000)Because there are no pivots in the last two columns, z and w are free variables.x+z=0→x=−zy+z+2w=0→y=−z−2wx0→=(xyzw)=(−z−z−2wzw)=(−z−zz0)+(0−2w0w)=z(−1−110)+w(0−201)x→=x0→+xp→=z(−1−110)+w(0−201)+(10−10)A = \begin{pmatrix} 1 & 0 & 1 & 0 \\ 1 & 2 & 3 & 4 \\ 4 & 8 & 12 & 16 \\ 6 & 12 & 18 & 24 \end{pmatrix} \\ \overrightarrow{b} = \begin{pmatrix} 0\\-2\\-8\\-12\end{pmatrix} \\ \text{Suppose we know that } \overrightarrow{x_p} = \begin{pmatrix}1\\0\\-1\\0\end{pmatrix} \text{ solves } A\overrightarrow{x} = \overrightarrow{b} \\ \text{Solve } A\overrightarrow{x} = \overrightarrow{0} \text{ and add } \overrightarrow{x_0} \text{ to } \overrightarrow{x_p} \\ \left( \begin{array}{c|c} A & \overrightarrow{0} \end{array} \right) = \left( \begin{array}{cccc|c} 1 & 0 & 1 & 0 & 0\\ 1 & 2 & 3 & 4 & 0\\ 4 & 8 & 12 & 16 & 0\\ 6 & 12 & 18 & 24 & 0 \end{array} \right) \rightarrow \left( \begin{array}{cccc|c} 1 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 2 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{array} \right) \\ \text{Because there are no pivots in the last two columns, z and w are free variables.} \\ x + z = 0 \rightarrow x = -z \\ y + z + 2w = 0 \rightarrow y = -z - 2w \\ \overrightarrow{x_0} = \begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix} = \begin{pmatrix} -z \\ -z - 2w\\ z\\ w \end{pmatrix} = \begin{pmatrix} -z \\ -z \\ z\\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ -2w \\ 0\\ w \end{pmatrix} = z \begin{pmatrix} -1 \\ -1 \\ 1\\ 0 \end{pmatrix} + w \begin{pmatrix} 0 \\ -2 \\ 0\\ 1 \end{pmatrix} \\ \overrightarrow{x} = \overrightarrow{x_0} + \overrightarrow{x_p} = z \begin{pmatrix} -1 \\ -1 \\ 1\\ 0 \end{pmatrix} + w \begin{pmatrix} 0 \\ -2 \\ 0\\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 0 \\ -1\\ 0 \end{pmatrix}