Linear Algebra Examples

$x + y = - 4$ , $y = x - 7$

Step 1

Subtract $x$ from both sides of the equation.

$x + y = - 4, y - x = - 7$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & - 4 \\ - 1 & 1 & - 7 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} 1 & 1 & - 4 \\ - 1 + 1 \cdot 1 & 1 + 1 \cdot 1 & - 7 - 4 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & - 4 \\ 0 & 2 & - 11 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & 1 & - 4 \\ \frac{0}{2} & \frac{2}{2} & \frac{- 11}{2} \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & - 4 \\ 0 & 1 & - \frac{11}{2} \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & - 4 + \frac{11}{2} \\ 0 & 1 & - \frac{11}{2} \end{array}]$

Step 3.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{2} \\ 0 & 1 & - \frac{11}{2} \end{array}]$

Step 4

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

$x = \frac{3}{2}$

$y = - \frac{11}{2}$

Step 5

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

$(\frac{3}{2}, - \frac{11}{2})$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{3}{2} \\ - \frac{11}{2} \end{array}]$