Linear Algebra Examples

$- x + y = 5$ , $x - 5 y = - 9$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} - 1 & 1 & 5 \\ 1 & - 5 & - 9 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} - - 1 & - 1 \cdot 1 & - 1 \cdot 5 \\ 1 & - 5 & - 9 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & - 5 \\ 1 & - 5 & - 9 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - 1 & - 5 \\ 1 - 1 & - 5 + 1 & - 9 + 5 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 5 \\ 0 & - 4 & - 4 \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $- \frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 5 \\ - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot - 4 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 5 \\ 0 & 1 & 1 \end{array}]$

Step 2.4

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

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Step 2.4.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & - 5 + 1 \cdot 1 \\ 0 & 1 & 1 \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

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

Step 3

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

$x = - 4$

$y = 1$

Step 4

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

$(- 4, 1)$

Step 5

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 4 \\ 1 \end{array}]$