Algebra Examples

$A = [\begin{array}{c} - 1 & 2 \\ 5 & 9.1 \end{array}]$ , $x = [\begin{array}{c} 8 \\ 2 \end{array}]$

Step 1

$C_{1} \cdot [\begin{array}{c} - 1 \\ 5 \end{array}] + C_{2} \cdot [\begin{array}{c} 2 \\ 9.1 \end{array}] = [\begin{array}{c} 8 \\ 2 \end{array}]$

Step 2

$5 C_{1} + 9.1 C_{2} = 2 - C_{1} + 2 C_{2} = 8$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} - 1 & 2 & 8 \\ 5 & 9.1 & 2 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

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

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Step 4.1.1

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

$[\begin{array}{c} - - 1 & - 1 \cdot 2 & - 1 \cdot 8 \\ 5 & 9.1 & 2 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 2 & - 8 \\ 5 & 9.1 & 2 \end{array}]$

Step 4.2

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.2.1

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 2 & - 8 \\ 5 - 5 \cdot 1 & 9.1 - 5 \cdot - 2 & 2 - 5 \cdot - 8 \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & - 8 \\ 0 & 19.1 & 42 \end{array}]$

Step 4.3

Multiply each element of $R_{2}$ by $\frac{1}{19.1}$ to make the entry at $2, 2$ a $1$ .

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Step 4.3.1

Multiply each element of $R_{2}$ by $\frac{1}{19.1}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 2 & - 8 \\ \frac{0}{19.1} & \frac{19.1}{19.1} & \frac{42}{19.1} \end{array}]$

Step 4.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & - 8 \\ 0 & 1 & 2.19895287 \end{array}]$

Step 4.4

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

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Step 4.4.1

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

$[\begin{array}{c} 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & - 8 + 2 \cdot 2.19895287 \\ 0 & 1 & 2.19895287 \end{array}]$

Step 4.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 3.60209424 \\ 0 & 1 & 2.19895287 \end{array}]$

Step 5

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

$C_{1} = - 3.60209424$

$C_{2} = 2.19895287$

Step 6

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

$(- 3.60209424, 2.19895287)$

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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