Linear Algebra Examples

2 x + y = - 2

$2 x + y = - 2$ ,

x + 2 y = 2

$x + 2 y = 2$

Step 1

Write the system of equations in matrix form.

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

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

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Step 2.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{array}{c} \frac{2}{2} & \frac{1}{2} & \frac{- 2}{2} \\ 1 & 2 & 2 \end{array}]

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

Step 2.1.2

Simplify

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & \frac{1}{2} & - 1 \\ 1 & 2 & 2 \end{array}]

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

[\begin{array}{c} 1 & \frac{1}{2} & - 1 \\ 1 & 2 & 2 \end{array}]

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

Step 2.2

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Step 2.2.1

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

[\begin{array}{c} 1 & \frac{1}{2} & - 1 \\ 1 - 1 & 2 - \frac{1}{2} & 2 + 1 \end{array}]

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

Step 2.2.2

Simplify

R_{2}

$R_{2}$ .

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

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

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

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

Step 2.3

Multiply each element of

R_{2}

$R_{2}$ by

\frac{2}{3}

$\frac{2}{3}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Step 2.3.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{2}{3}

$\frac{2}{3}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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

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

Step 2.3.2

Simplify

R_{2}

$R_{2}$ .

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

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

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

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

Step 2.4

Perform the row operation

R_{1} = R_{1} - \frac{1}{2} R_{2}

$R_{1} = R_{1} - \frac{1}{2} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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Step 2.4.1

Perform the row operation

R_{1} = R_{1} - \frac{1}{2} R_{2}

$R_{1} = R_{1} - \frac{1}{2} R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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

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

Step 2.4.2

Simplify

R_{1}

$R_{1}$ .

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

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

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

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

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

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

Step 3

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

x = - 2

$x = - 2$

y = 2

$y = 2$

Step 4

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

(- 2, 2)

$(- 2, 2)$

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} - 2 \\ 2 \end{array}]

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