Linear Algebra Examples

$5 x + 2 y = 8$ , $9 x - 4 y = 2 - 22$

Step 1

Subtract $22$ from $2$ .

$5 x + 2 y = 8, 9 x - 4 y = - 20$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} 5 & 2 & 8 \\ 9 & - 4 & - 20 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{2}{5} & \frac{8}{5} \\ 9 & - 4 & - 20 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{8}{5} \\ 9 & - 4 & - 20 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{8}{5} \\ 9 - 9 \cdot 1 & - 4 - 9 (\frac{2}{5}) & - 20 - 9 (\frac{8}{5}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{8}{5} \\ 0 & - \frac{38}{5} & - \frac{172}{5} \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $- \frac{5}{38}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $- \frac{5}{38}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{8}{5} \\ - \frac{5}{38} \cdot 0 & - \frac{5}{38} (- \frac{38}{5}) & - \frac{5}{38} (- \frac{172}{5}) \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{8}{5} \\ 0 & 1 & \frac{86}{19} \end{array}]$

Step 3.4

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

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Step 3.4.1

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

$[\begin{array}{c} 1 - \frac{2}{5} \cdot 0 & \frac{2}{5} - \frac{2}{5} \cdot 1 & \frac{8}{5} - \frac{2}{5} \cdot \frac{86}{19} \\ 0 & 1 & \frac{86}{19} \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{4}{19} \\ 0 & 1 & \frac{86}{19} \end{array}]$

Step 4

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

$x = - \frac{4}{19}$

$y = \frac{86}{19}$

Step 5

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

$(- \frac{4}{19}, \frac{86}{19})$

Step 6

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - \frac{4}{19} \\ \frac{86}{19} \end{array}]$