Linear Algebra Examples

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

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

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Move $3 x$ to the left side of the equation because it contains a variable.

$8 - 3 x = 3 y$

$1 = 4 x - y$

Move $3 y$ to the left side of the equation because it contains a variable.

$8 - 3 x - 3 y = 0$

$1 = 4 x - y$

$8 - 3 x - 3 y = 0$

$1 = 4 x - y$

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

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Move $4 x$ to the left side of the equation because it contains a variable.

$8 - 3 x - 3 y = 0$

$1 - 4 x = - y$

Move $- y$ to the left side of the equation because it contains a variable.

$8 - 3 x - 3 y = 0$

$1 - 4 x + y = 0$

$8 - 3 x - 3 y = 0$

$1 - 4 x + y = 0$

Move $8$ to the right side of the equation because it does not contain a variable.

$- 3 x - 3 y = - 8$

$1 - 4 x + y = 0$

Move $1$ to the right side of the equation because it does not contain a variable.

$- 3 x - 3 y = - 8$

$- 4 x + y = - 1$

Write the system of equations in matrix form.

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

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{1} = - \frac{1}{3} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - \frac{1}{3} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

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

$R_{1} = - \frac{1}{3} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{1}{3} R_{1}$ .

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

$R_{1} = - \frac{1}{3} R_{1}$

Simplify $R_{1}$ (row $1$ ).

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

Perform the row operation $R_{2} = 4 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = 4 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 1 & \frac{8}{3} \\ 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = 4 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = 4 \cdot R_{1} + R_{2}$ .

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

$R_{2} = 4 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

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

Perform the row operation $R_{2} = \frac{1}{5} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{1}{5} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 1 & \frac{8}{3} \\ \frac{1}{5} R_{2} & \frac{1}{5} R_{2} & \frac{1}{5} R_{2} \end{array}]$

$R_{2} = \frac{1}{5} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{5} R_{2}$ .

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

$R_{2} = \frac{1}{5} R_{2}$

Simplify $R_{2}$ (row $2$ ).

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

Perform the row operation $R_{1} = - 1 \cdot R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - 1 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} \\ 0 & 1 & \frac{29}{15} \end{array}]$

$R_{1} = - 1 \cdot R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - 1 \cdot R_{2} + R_{1}$ .

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

$R_{1} = - 1 \cdot R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

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

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

$x_{1} = \frac{11}{15}$

$x_{2} = \frac{29}{15}$

This expression is the solution set for the system of equations.

${(\frac{11}{15}, \frac{29}{15})}$

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.

$X = [\begin{array}{c} x_{1} \\ x_{2} \end{array}] = [\begin{array}{c} \frac{11}{15} \\ \frac{29}{15} \end{array}]$

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