Examples

$A = [\begin{array}{c} 1 & - 1 & - 8 \\ 1 & 2 & 6 \end{array}]$ ,

$x = [\begin{array}{c} 12 \\ - 3 \end{array}]$

Step 1

$C_{1} \cdot [\begin{array}{c} 1 \\ 1 \end{array}] + C_{2} \cdot [\begin{array}{c} - 1 \\ 2 \end{array}] + C_{3} \cdot [\begin{array}{c} - 8 \\ 6 \end{array}] = [\begin{array}{c} 12 \\ - 3 \end{array}]$

Step 2

$C_{1} + 2 C_{2} + 6 C_{3} = - 3 C_{1} - C_{2} - 8 C_{3} = 12$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & - 1 & - 8 & 12 \\ 1 & 2 & 6 & - 3 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Perform the row operation

$R_{2} = R_{2} - R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 4.1.1

Perform the row operation

$R_{2} = R_{2} - R_{1}$ to make the entry at

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & - 1 & - 8 & 12 \\ 1 - 1 & 2 + 1 & 6 + 8 & - 3 - 12 \end{array}]$

Step 4.1.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 8 & 12 \\ 0 & 3 & 14 & - 15 \end{array}]$

Step 4.2

Multiply each element of

$R_{2}$ by

$\frac{1}{3}$ to make the entry at

$2, 2$ a

$1$ .

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Step 4.2.1

Multiply each element of

$R_{2}$ by

$\frac{1}{3}$ to make the entry at

$2, 2$ a

$1$ .

$[\begin{array}{c} 1 & - 1 & - 8 & 12 \\ \frac{0}{3} & \frac{3}{3} & \frac{14}{3} & \frac{- 15}{3} \end{array}]$

Step 4.2.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 8 & 12 \\ 0 & 1 & \frac{14}{3} & - 5 \end{array}]$

Step 4.3

Perform the row operation

$R_{1} = R_{1} + R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Step 4.3.1

Perform the row operation

$R_{1} = R_{1} + R_{2}$ to make the entry at

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & - 8 + \frac{14}{3} & 12 - 5 \\ 0 & 1 & \frac{14}{3} & - 5 \end{array}]$

Step 4.3.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{10}{3} & 7 \\ 0 & 1 & \frac{14}{3} & - 5 \end{array}]$

Step 5

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

$C_{1} - \frac{10 C_{3}}{3} = 7$

$C_{2} + \frac{14 C_{3}}{3} = - 5$

Step 6

Add

$\frac{10 C_{3}}{3}$ to both sides of the equation.

$C_{1} = 7 + \frac{10 C_{3}}{3}$

$C_{2} + \frac{14 C_{3}}{3} = - 5$

Step 7

Subtract

$\frac{14 C_{3}}{3}$ from both sides of the equation.

$C_{2} = - 5 - \frac{14 C_{3}}{3}$

$C_{1} = 7 + \frac{10 C_{3}}{3}$

Step 8

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

$(7 + \frac{10 C_{3}}{3}, - 5 - \frac{14 C_{3}}{3}, C_{3})$

Step 9

There is not a transformation of the vector that exists because there was no unique solution to the system of equations. Since there is no linear transformation, the vector is not in the column space.

Not in the Column Space

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