Linear Algebra Examples

$- 4 = - 2 - 4 - x$ , $y = - 4 - x - 2$

Step 1

Subtract $4$ from $- 2$ .

$- 4 = - 6 - x, y = - 4 - x - 2$

Step 2

Add $4$ to both sides of the equation.

$0 = - 6 - x + 4, y = - 4 - x - 2$

Step 3

Add $x$ to both sides of the equation.

$x = - 6 + 4, y = - 4 - x - 2$

Step 4

Subtract $2$ from $- 4$ .

$x = - 6 + 4, y = - x - 6$

Step 5

Add $x$ to both sides of the equation.

$x = - 6 + 4, y + x = - 6$

Step 6

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 0 & - 6 + 4 \\ 1 & 1 & - 6 \end{array}]$

Step 7

Find the reduced row echelon form.

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Step 7.1

Add $- 6$ and $4$ .

$[\begin{array}{c} 1 & 0 & - 2 \\ 1 & 1 & - 6 \end{array}]$

Step 7.2

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

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Step 7.2.1

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

$[\begin{array}{c} 1 & 0 & - 2 \\ 1 - 1 & 1 - 0 & - 6 + 2 \end{array}]$

Step 7.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & - 2 \\ 0 & 1 & - 4 \end{array}]$

Step 8

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

$x = - 2$

$y = - 4$

Step 9

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

$(- 2, - 4)$

Step 10

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 2 \\ - 4 \end{array}]$