Linear Algebra Examples

$x - 9 = y$ , $x + y = 6$

Step 1

Move variables to the left and constant terms to the right.

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Step 1.1

Subtract $y$ from both sides of the equation.

$x - 9 - y = 0$

$x + y = 6$

Step 1.2

Add $9$ to both sides of the equation.

$x - y = 9$

$x + y = 6$

$x - y = 9$

$x + y = 6$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & - 1 & 9 \\ 1 & 1 & 6 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - 1 & 9 \\ 1 - 1 & 1 + 1 & 6 - 9 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 9 \\ 0 & 2 & - 3 \end{array}]$

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 $1$ .

$[\begin{array}{c} 1 & - 1 & 9 \\ \frac{0}{2} & \frac{2}{2} & \frac{- 3}{2} \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 9 \\ 0 & 1 & - \frac{3}{2} \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 9 - \frac{3}{2} \\ 0 & 1 & - \frac{3}{2} \end{array}]$

Step 3.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{15}{2} \\ 0 & 1 & - \frac{3}{2} \end{array}]$

Step 4

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

$x = \frac{15}{2}$

$y = - \frac{3}{2}$

Step 5

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

$(\frac{15}{2}, - \frac{3}{2})$