Linear Algebra Examples

[\begin{matrix} 3 & 7 7 & 6 \end{matrix}]

$[\begin{array}{c} 3 & 7 \\ 7 & 6 \end{array}]$

Step 1

Write as an augmented matrix for

A x = 0

$A x = 0$ .

[\begin{matrix} 3 & 7 & 0 7 & 6 & 0 \end{matrix}]

$[\begin{array}{c} 3 & 7 & 0 \\ 7 & 6 & 0 \end{array}]$

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}{3}

$\frac{1}{3}$ 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}{3}

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

1, 1

$1, 1$ a

1

$1$ .

[\begin{matrix} \frac{3}{3} & \frac{7}{3} & \frac{0}{3} 7 & 6 & 0 \end{matrix}]

$[\begin{array}{c} \frac{3}{3} & \frac{7}{3} & \frac{0}{3} \\ 7 & 6 & 0 \end{array}]$

Step 2.1.2

Simplify

R_{1}

$R_{1}$ .

[\begin{matrix} 1 & \frac{7}{3} & 0 7 & 6 & 0 \end{matrix}]

$[\begin{array}{c} 1 & \frac{7}{3} & 0 \\ 7 & 6 & 0 \end{array}]$

[\begin{matrix} 1 & \frac{7}{3} & 0 7 & 6 & 0 \end{matrix}]

$[\begin{array}{c} 1 & \frac{7}{3} & 0 \\ 7 & 6 & 0 \end{array}]$

Step 2.2

Perform the row operation

R_{2} = R_{2} - 7 R_{1}

$R_{2} = R_{2} - 7 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

Tap for more steps...

Step 2.2.1

Perform the row operation

R_{2} = R_{2} - 7 R_{1}

$R_{2} = R_{2} - 7 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & \frac{7}{3} & 0 7 - 7 \cdot 1 & 6 - 7 (\frac{7}{3}) & 0 - 7 \cdot 0 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & \frac{7}{3} & 0 \\ 7 - 7 \cdot 1 & 6 - 7 (\frac{7}{3}) & 0 - 7 \cdot 0 \end{array}]$

Step 2.2.2

Simplify

R_{2}

$R_{2}$ .

⎡ ⎣ \begin{matrix} 1 & \frac{7}{3} & 0 0 & - \frac{31}{3} & 0 \end{matrix} ⎤ ⎦

$[\begin{array}{c} 1 & \frac{7}{3} & 0 \\ 0 & - \frac{31}{3} & 0 \end{array}]$

⎡ ⎣ \begin{matrix} 1 & \frac{7}{3} & 0 0 & - \frac{31}{3} & 0 \end{matrix} ⎤ ⎦

$[\begin{array}{c} 1 & \frac{7}{3} & 0 \\ 0 & - \frac{31}{3} & 0 \end{array}]$

Step 2.3

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{3}{31}

$- \frac{3}{31}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

Tap for more steps...

Step 2.3.1

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{3}{31}

$- \frac{3}{31}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & \frac{7}{3} & 0 - \frac{3}{31} \cdot 0 & - \frac{3}{31} (- \frac{31}{3}) & - \frac{3}{31} \cdot 0 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & \frac{7}{3} & 0 \\ - \frac{3}{31} \cdot 0 & - \frac{3}{31} (- \frac{31}{3}) & - \frac{3}{31} \cdot 0 \end{array}]$

Step 2.3.2

Simplify

R_{2}

$R_{2}$ .

[\begin{matrix} 1 & \frac{7}{3} & 0 0 & 1 & 0 \end{matrix}]

$[\begin{array}{c} 1 & \frac{7}{3} & 0 \\ 0 & 1 & 0 \end{array}]$

Step 2.4

Perform the row operation

$R_{1} = R_{1} - \frac{7}{3} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

Tap for more steps...

Step 2.4.1

Perform the row operation

$R_{1} = R_{1} - \frac{7}{3} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 - \frac{7}{3} \cdot 0 & \frac{7}{3} - \frac{7}{3} \cdot 1 & 0 - \frac{7}{3} \cdot 0 \\ 0 & 1 & 0 \end{array}]$

Step 2.4.2

Simplify

$R_{1}$ .

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

Step 3

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

$x = 0$

$y = 0$

Step 4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 0 \\ 0 \end{array}]$

Step 5

Write as a solution set.

${[\begin{array}{c} 0 \\ 0 \end{array}]}$

Enter YOUR Problem