Linear Algebra Examples

A = ⎡ ⎢ ⎣ \begin{matrix} - 2 & 2 - 2 & - 1 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

x = ⎡ ⎢ ⎣ \begin{matrix} - 12 0 - 4 \end{matrix} ⎤ ⎥ ⎦

$x = [\begin{array}{c} - 12 \\ 0 \\ - 4 \end{array}]$

Step 1

Write as an augmented matrix for

A x = ⎡ ⎢ ⎣ \begin{matrix} - 12 0 - 4 \end{matrix} ⎤ ⎥ ⎦

$A x = [\begin{array}{c} - 12 \\ 0 \\ - 4 \end{array}]$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} - 2 & 2 & - 12 - 2 & - 1 & 0 0 & 1 & - 4 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} - 2 & 2 & - 12 \\ - 2 & - 1 & 0 \\ 0 & 1 & - 4 \end{array}]$

Step 2

Find the reduced row echelon form.

Tap for more steps...

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

$1$ .

Tap for more steps...

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{matrix} - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 2 & - \frac{1}{2} \cdot - 12 - 2 & - 1 & 0 0 & 1 & - 4 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 2 & - \frac{1}{2} \cdot - 12 \\ - 2 & - 1 & 0 \\ 0 & 1 & - 4 \end{array}]$

Step 2.1.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 6 \\ - 2 & - 1 & 0 \\ 0 & 1 & - 4 \end{array}]$

Step 2.2

Perform the row operation

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

$2, 1$ a

$0$ .

Tap for more steps...

Step 2.2.1

Perform the row operation

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

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & - 1 & 6 \\ - 2 + 2 \cdot 1 & - 1 + 2 \cdot - 1 & 0 + 2 \cdot 6 \\ 0 & 1 & - 4 \end{array}]$

Step 2.2.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 6 \\ 0 & - 3 & 12 \\ 0 & 1 & - 4 \end{array}]$

Step 2.3

Multiply each element of

$R_{2}$ by

$- \frac{1}{3}$ to make the entry at

$2, 2$ a

$1$ .

Tap for more steps...

Step 2.3.1

Multiply each element of

$R_{2}$ by

$- \frac{1}{3}$ to make the entry at

$2, 2$ a

$1$ .

$[\begin{array}{c} 1 & - 1 & 6 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 12 \\ 0 & 1 & - 4 \end{array}]$

Step 2.3.2

Simplify

$R_{2}$ .

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

Step 2.4

Perform the row operation

$R_{3} = R_{3} - R_{2}$ to make the entry at

$3, 2$ a

$0$ .

Tap for more steps...

Step 2.4.1

Perform the row operation

$R_{3} = R_{3} - R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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

Step 2.4.2

Simplify

$R_{3}$ .

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

Step 2.5

Perform the row operation

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

$1, 2$ a

$0$ .

Tap for more steps...

Step 2.5.1

Perform the row operation

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

$1, 2$ a

$0$ .

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

Step 2.5.2

Simplify

$R_{1}$ .

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

Step 3

Write the matrix as a system of linear equations.

$x = 2$

$y = - 4$

$0 = 0$

Step 4

Write the solutions as a set of vectors.

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

Enter YOUR Problem