Linear Algebra Examples

$A = [\begin{array}{c} 4 \\ - 8 \\ 2 \end{array}]$ , $x = [\begin{array}{c} 1 \\ 2 \\ 6 \end{array}]$

Move all terms not containing a variable to the right side of the equation.

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Move $4$ to the right side of the equation because it does not contain a variable.

$0 = - 4 + 1$

$- 8 = 2$

$2 = 6$

Add $- 4$ and $1$ .

$0 = - 3$

$- 8 = 2$

$2 = 6$

$0 = - 3$

$- 8 = 2$

$2 = 6$

Move all terms not containing a variable to the right side of the equation.

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

$0 = - 3$

$0 = 8 + 2$

$2 = 6$

Add $8$ and $2$ .

$0 = - 3$

$0 = 10$

$2 = 6$

$0 = - 3$

$0 = 10$

$2 = 6$

Move all terms not containing a variable to the right side of the equation.

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Move $2$ to the right side of the equation because it does not contain a variable.

$0 = - 3$

$0 = 10$

$0 = - 2 + 6$

Add $- 2$ and $6$ .

$0 = - 3$

$0 = 10$

$0 = 4$

$0 = - 3$

$0 = 10$

$0 = 4$

Write the system of equations in matrix form.

$[\begin{array}{c} - 3 \\ 10 \\ 4 \end{array}]$

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 $1$ .

$[\begin{array}{c} - \frac{1}{3} R_{1} \\ 10 \\ 4 \end{array}]$

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

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

$[\begin{array}{c} (- \frac{1}{3}) \cdot (- 3) \\ 10 \\ 4 \end{array}]$

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

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

$[\begin{array}{c} 1 \\ 10 \\ 4 \end{array}]$

Perform the row operation $R_{2} = - 10 \cdot R_{1} + R_{2}$ 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 $R_{2} = - 10 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ - 10 \cdot R_{1} + R_{2} \\ 4 \end{array}]$

$R_{2} = - 10 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 10 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 \\ (- 10) \cdot (1) + 10 \\ 4 \end{array}]$

$R_{2} = - 10 \cdot R_{1} + R_{2}$

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

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

Perform the row operation $R_{3} = - 4 \cdot R_{1} + R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = - 4 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 0 \\ - 4 \cdot R_{1} + R_{3} \end{array}]$

$R_{3} = - 4 \cdot R_{1} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 4 \cdot R_{1} + R_{3}$ .

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

$R_{3} = - 4 \cdot R_{1} + R_{3}$

Simplify $R_{3}$ (row $3$ ).

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

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

$0 = 1$

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

${}$

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.

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

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