Examples

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

Step 1

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

Step 2

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

Step 3

Write the system of equations in matrix form.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of $R_{1}$ by $\frac{1}{17}$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $\frac{1}{17}$ to make the entry at $1, 1$ a $1$ .

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

Step 4.1.2

Simplify $R_{1}$ .

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

Step 4.2

Perform the row operation $R_{2} = R_{2} - 2 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.2.1

Perform the row operation $R_{2} = R_{2} - 2 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Step 4.2.2

Simplify $R_{2}$ .

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

Step 4.3

Multiply each element of $R_{2}$ by $\frac{17}{32}$ to make the entry at $2, 2$ a $1$ .

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Step 4.3.1

Multiply each element of $R_{2}$ by $\frac{17}{32}$ to make the entry at $2, 2$ a $1$ .

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

Step 4.3.2

Simplify $R_{2}$ .

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

Step 4.4

Perform the row operation $R_{1} = R_{1} - \frac{18}{17} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 4.4.1

Perform the row operation $R_{1} = R_{1} - \frac{18}{17} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Step 4.4.2

Simplify $R_{1}$ .

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

Step 5

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

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

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

Step 6

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

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

Step 7

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

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