Linear Algebra Examples

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

Step 1

Write as an augmented matrix for $x \cdot x = [\begin{array}{c} 8 \\ 1 \end{array}]$ .

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

Step 2

Write as a linear system of equations.

$8 = 3 x + 3 y$

$1 = 4 x - y$

Step 3

Solve the system of equations.

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Step 3.1

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

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Step 3.1.1

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

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Step 3.1.1.1

Subtract $3 x$ from both sides of the equation.

$8 - 3 x = 3 y$

$1 = 4 x - y$

Step 3.1.1.2

Subtract $3 y$ from both sides of the equation.

$8 - 3 x - 3 y = 0$

$1 = 4 x - y$

$8 - 3 x - 3 y = 0$

$1 = 4 x - y$

Step 3.1.2

Subtract $8$ from both sides of the equation.

$- 3 x - 3 y = - 8$

$1 = 4 x - y$

Step 3.1.3

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

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Step 3.1.3.1

Subtract $4 x$ from both sides of the equation.

$- 3 x - 3 y = - 8$

$1 - 4 x = - y$

Step 3.1.3.2

Add $y$ to both sides of the equation.

$- 3 x - 3 y = - 8$

$1 - 4 x + y = 0$

$- 3 x - 3 y = - 8$

$1 - 4 x + y = 0$

Step 3.1.4

Subtract $1$ from both sides of the equation.

$- 3 x - 3 y = - 8$

$- 4 x + y = - 1$

$- 3 x - 3 y = - 8$

$- 4 x + y = - 1$

Step 3.2

Write the system as a matrix.

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

Step 3.3

Find the reduced row echelon form.

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Step 3.3.1

Multiply each element of $R_{1}$ by $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

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Step 3.3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

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

Step 3.3.1.2

Simplify $R_{1}$ .

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

Step 3.3.2

Perform the row operation $R_{2} = R_{2} + 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.3.2.1

Perform the row operation $R_{2} = R_{2} + 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Step 3.3.2.2

Simplify $R_{2}$ .

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

Step 3.3.3

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

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

Step 3.3.3.2

Simplify $R_{2}$ .

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

Step 3.3.4

Perform the row operation $R_{1} = R_{1} - R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.3.4.1

Perform the row operation $R_{1} = R_{1} - R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Step 3.3.4.2

Simplify $R_{1}$ .

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

Step 3.4

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

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

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

Step 3.5

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

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

Step 3.6

Write as a solution set.

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

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