Linear Algebra Examples

A = [\begin{matrix} 81 \end{matrix}]

$A = [\begin{array}{c} 8 \\ 1 \end{array}]$ ,

x = [\begin{matrix} 3 x + 3 y 4 x - y \end{matrix}]

$x = [\begin{array}{c} 3 x + 3 y \\ 4 x - y \end{array}]$

Step 1

Write as an augmented matrix for

x \cdot x = [\begin{matrix} 81 \end{matrix}]

$x \cdot x = [\begin{array}{c} 8 \\ 1 \end{array}]$ .

[\begin{matrix} 3 x + 3 y & 8 4 x - y & 1 \end{matrix}]

$[\begin{array}{c} 3 x + 3 y & 8 \\ 4 x - y & 1 \end{array}]$

Step 2

Write as a linear system of equations.

8 = 3 x + 3 y

$8 = 3 x + 3 y$

1 = 4 x - y

$1 = 4 x - y$

Step 3

Solve the system of equations.

Tap for more steps...

Step 3.1

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

Tap for more steps...

Step 3.1.1

Move all terms containing variables to the left side of the equation.

Tap for more steps...

Step 3.1.1.1

Subtract

3 x

$3 x$ from both sides of the equation.

8 - 3 x = 3 y

$8 - 3 x = 3 y$

1 = 4 x - y

$1 = 4 x - y$

Step 3.1.1.2

Subtract

3 y

$3 y$ from both sides of the equation.

8 - 3 x - 3 y = 0

$8 - 3 x - 3 y = 0$

1 = 4 x - y

$1 = 4 x - y$

8 - 3 x - 3 y = 0

$8 - 3 x - 3 y = 0$

1 = 4 x - y

$1 = 4 x - y$

Step 3.1.2

Subtract

8

$8$ from both sides of the equation.

- 3 x - 3 y = - 8

$- 3 x - 3 y = - 8$

1 = 4 x - y

$1 = 4 x - y$

Step 3.1.3

Move all terms containing variables to the left side of the equation.

Tap for more steps...

Step 3.1.3.1

Subtract

4 x

$4 x$ from both sides of the equation.

- 3 x - 3 y = - 8

$- 3 x - 3 y = - 8$

1 - 4 x = - y

$1 - 4 x = - y$

Step 3.1.3.2

Add

y

$y$ to both sides of the equation.

- 3 x - 3 y = - 8

$- 3 x - 3 y = - 8$

1 - 4 x + y = 0

$1 - 4 x + y = 0$

- 3 x - 3 y = - 8

$- 3 x - 3 y = - 8$

1 - 4 x + y = 0

$1 - 4 x + y = 0$

Step 3.1.4

Subtract

1

$1$ from both sides of the equation.

- 3 x - 3 y = - 8

$- 3 x - 3 y = - 8$

- 4 x + y = - 1

$- 4 x + y = - 1$

- 3 x - 3 y = - 8

$- 3 x - 3 y = - 8$

- 4 x + y = - 1

$- 4 x + y = - 1$

Step 3.2

Write the system as a matrix.

[\begin{matrix} - 3 & - 3 & - 8 - 4 & 1 & - 1 \end{matrix}]

$[\begin{array}{c} - 3 & - 3 & - 8 \\ - 4 & 1 & - 1 \end{array}]$

Step 3.3

Find the reduced row echelon form.

Tap for more steps...

Step 3.3.1

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 3.3.1.1

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{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 8 - 4 & 1 & - 1 \end{matrix}]

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 8 \\ - 4 & 1 & - 1 \end{array}]$

Step 3.3.1.2

Simplify

R_{1}

$R_{1}$ .

[\begin{matrix} 1 & 1 & \frac{8}{3} - 4 & 1 & - 1 \end{matrix}]

$[\begin{array}{c} 1 & 1 & \frac{8}{3} \\ - 4 & 1 & - 1 \end{array}]$

[\begin{matrix} 1 & 1 & \frac{8}{3} - 4 & 1 & - 1 \end{matrix}]

$[\begin{array}{c} 1 & 1 & \frac{8}{3} \\ - 4 & 1 & - 1 \end{array}]$

Step 3.3.2

Perform the row operation

R_{2} = R_{2} + 4 R_{1}

$R_{2} = R_{2} + 4 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

Tap for more steps...

Step 3.3.2.1

Perform the row operation

R_{2} = R_{2} + 4 R_{1}

$R_{2} = R_{2} + 4 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & \frac{8}{3} - 4 + 4 \cdot 1 & 1 + 4 \cdot 1 & - 1 + 4 (\frac{8}{3}) \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & \frac{8}{3} \\ - 4 + 4 \cdot 1 & 1 + 4 \cdot 1 & - 1 + 4 (\frac{8}{3}) \end{array}]$

Step 3.3.2.2

Simplify

R_{2}

$R_{2}$ .

⎡ ⎣ \begin{matrix} 1 & 1 & \frac{8}{3} 0 & 5 & \frac{29}{3} \end{matrix} ⎤ ⎦

$[\begin{array}{c} 1 & 1 & \frac{8}{3} \\ 0 & 5 & \frac{29}{3} \end{array}]$

⎡ ⎣ \begin{matrix} 1 & 1 & \frac{8}{3} 0 & 5 & \frac{29}{3} \end{matrix} ⎤ ⎦

$[\begin{array}{c} 1 & 1 & \frac{8}{3} \\ 0 & 5 & \frac{29}{3} \end{array}]$

Step 3.3.3

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{5}

$\frac{1}{5}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

Tap for more steps...

Step 3.3.3.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{5}

$\frac{1}{5}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & \frac{8}{3} \frac{0}{5} & \frac{5}{5} & \frac{\frac{29}{3}}{5} \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 1 & 1 & \frac{8}{3} \\ \frac{0}{5} & \frac{5}{5} & \frac{\frac{29}{3}}{5} \end{array}]$

Step 3.3.3.2

Simplify

R_{2}

$R_{2}$ .

⎡ ⎣ \begin{matrix} 1 & 1 & \frac{8}{3} 0 & 1 & \frac{29}{15} \end{matrix} ⎤ ⎦

$[\begin{array}{c} 1 & 1 & \frac{8}{3} \\ 0 & 1 & \frac{29}{15} \end{array}]$

⎡ ⎣ \begin{matrix} 1 & 1 & \frac{8}{3} 0 & 1 & \frac{29}{15} \end{matrix} ⎤ ⎦

$[\begin{array}{c} 1 & 1 & \frac{8}{3} \\ 0 & 1 & \frac{29}{15} \end{array}]$

Step 3.3.4

Perform the row operation

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

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

1, 2

$1, 2$ a

0

$0$ .

Tap for more steps...

Step 3.3.4.1

Perform the row operation

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

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

1, 2

$1, 2$ a

0

$0$ .

⎡ ⎣ \begin{matrix} 1 - 0 & 1 - 1 & \frac{8}{3} - \frac{29}{15} 0 & 1 & \frac{29}{15} \end{matrix} ⎤ ⎦

$[\begin{array}{c} 1 - 0 & 1 - 1 & \frac{8}{3} - \frac{29}{15} \\ 0 & 1 & \frac{29}{15} \end{array}]$

Step 3.3.4.2

Simplify

R_{1}

$R_{1}$ .

⎡ ⎣ \begin{matrix} 1 & 0 & \frac{11}{15} 0 & 1 & \frac{29}{15} \end{matrix} ⎤ ⎦

$[\begin{array}{c} 1 & 0 & \frac{11}{15} \\ 0 & 1 & \frac{29}{15} \end{array}]$

⎡ ⎣ \begin{matrix} 1 & 0 & \frac{11}{15} 0 & 1 & \frac{29}{15} \end{matrix} ⎤ ⎦

$[\begin{array}{c} 1 & 0 & \frac{11}{15} \\ 0 & 1 & \frac{29}{15} \end{array}]$

⎡ ⎣ \begin{matrix} 1 & 0 & \frac{11}{15} 0 & 1 & \frac{29}{15} \end{matrix} ⎤ ⎦

$[\begin{array}{c} 1 & 0 & \frac{11}{15} \\ 0 & 1 & \frac{29}{15} \end{array}]$

Step 3.4

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

x = \frac{11}{15}

$x = \frac{11}{15}$

y = \frac{29}{15}

$y = \frac{29}{15}$

Step 3.5

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

[\begin{matrix} x y \end{matrix}] = [\begin{matrix} \frac{11}{15} \frac{29}{15} \end{matrix}]

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{11}{15} \\ \frac{29}{15} \end{array}]$

Step 3.6

Write as a solution set.

{[\begin{matrix} \frac{11}{15} \frac{29}{15} \end{matrix}]}

${[\begin{array}{c} \frac{11}{15} \\ \frac{29}{15} \end{array}]}$

{[\begin{matrix} \frac{11}{15} \frac{29}{15} \end{matrix}]}

${[\begin{array}{c} \frac{11}{15} \\ \frac{29}{15} \end{array}]}$

Enter YOUR Problem