Determine if the Vector is in the Column Space A=[[17,18],[2,4]] , x=[[8],[3]]

$A = [\begin{array}{c} 17 & 18 \\ 2 & 4 \end{array}]$ , $x = [\begin{array}{c} 8 \\ 3 \end{array}]$

$C_{1} \cdot [\begin{array}{c} 17 \\ 2 \end{array}] + C_{2} \cdot [\begin{array}{c} 18 \\ 4 \end{array}] = [\begin{array}{c} 8 \\ 3 \end{array}]$

$2 C_{1} + 4 C_{2} = 3 17 C_{1} + 18 C_{2} = 8$

Write the system of equations in matrix form.

$[\begin{array}{c} 17 & 18 & 8 \\ 2 & 4 & 3 \end{array}]$

Find the reduced row echelon form of the matrix.

Perform the row operation $R_{1} = \frac{1}{17} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{17} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} \frac{1}{17} R_{1} & \frac{1}{17} R_{1} & \frac{1}{17} R_{1} \\ 2 & 4 & 3 \end{array}]$

$R_{1} = \frac{1}{17} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{17} R_{1}$ .

$[\begin{array}{c} (\frac{1}{17}) \cdot (17) & (\frac{1}{17}) \cdot (18) & (\frac{1}{17}) \cdot (8) \\ 2 & 4 & 3 \end{array}]$

$R_{1} = \frac{1}{17} R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & \frac{18}{17} & \frac{8}{17} \\ 2 & 4 & 3 \end{array}]$

Perform the row operation $R_{2} = - 2 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 2 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & \frac{18}{17} & \frac{8}{17} \\ - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 2 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 2 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & \frac{18}{17} & \frac{8}{17} \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (\frac{18}{17}) + 4 & (- 2) \cdot (\frac{8}{17}) + 3 \end{array}]$

$R_{2} = - 2 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & \frac{18}{17} & \frac{8}{17} \\ 0 & \frac{32}{17} & \frac{35}{17} \end{array}]$

Perform the row operation $R_{2} = \frac{17}{32} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{17}{32} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & \frac{18}{17} & \frac{8}{17} \\ \frac{17}{32} R_{2} & \frac{17}{32} R_{2} & \frac{17}{32} R_{2} \end{array}]$

$R_{2} = \frac{17}{32} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{17}{32} R_{2}$ .

$[\begin{array}{c} 1 & \frac{18}{17} & \frac{8}{17} \\ (\frac{17}{32}) \cdot (0) & (\frac{17}{32}) \cdot (\frac{32}{17}) & (\frac{17}{32}) \cdot (\frac{35}{17}) \end{array}]$

$R_{2} = \frac{17}{32} R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & \frac{18}{17} & \frac{8}{17} \\ 0 & 1 & \frac{35}{32} \end{array}]$

Perform the row operation $R_{1} = - \frac{18}{17} R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - \frac{18}{17} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{18}{17} R_{2} + R_{1} & - \frac{18}{17} R_{2} + R_{1} & - \frac{18}{17} R_{2} + R_{1} \\ 0 & 1 & \frac{35}{32} \end{array}]$

$R_{1} = - \frac{18}{17} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{18}{17} R_{2} + R_{1}$ .

$[\begin{array}{c} (- \frac{18}{17}) \cdot (0) + 1 & (- \frac{18}{17}) \cdot (1) + \frac{18}{17} & (- \frac{18}{17}) \cdot (\frac{35}{32}) + \frac{8}{17} \\ 0 & 1 & \frac{35}{32} \end{array}]$

$R_{1} = - \frac{18}{17} R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & - \frac{11}{16} \\ 0 & 1 & \frac{35}{32} \end{array}]$

Use the result matrix to declare the final solutions to the system of equations.

$C_{1} = - \frac{11}{16}$

$C_{2} = \frac{35}{32}$

The solution is the set of ordered pairs that makes the system true.

$(- \frac{11}{16}, \frac{35}{32})$

The vector is in the column space because there is a transformation of the vector that exists. This was determined by solving the system and showing there is a valid result.

In the Column Space

Determine if the Vector is in the Column Space A=[[17,18],[2,4]] , x=[[8],[3]]

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