Linear Algebra Examples

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

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

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Subtract $4$ from both sides of the equation.

$0 = 1 - 4$

$6 = 2$

$4 = 6$

Subtract $4$ from $1$ .

$0 = - 3$

$6 = 2$

$4 = 6$

$0 = - 3$

$6 = 2$

$4 = 6$

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

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Subtract $6$ from both sides of the equation.

$0 = - 3$

$0 = 2 - 6$

$4 = 6$

Subtract $6$ from $2$ .

$0 = - 3$

$0 = - 4$

$4 = 6$

$0 = - 3$

$0 = - 4$

$4 = 6$

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

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Subtract $4$ from both sides of the equation.

$0 = - 3$

$0 = - 4$

$0 = 6 - 4$

Subtract $4$ from $6$ .

$0 = - 3$

$0 = - 4$

$0 = 2$

$0 = - 3$

$0 = - 4$

$0 = 2$

Write the system of equations in matrix form.

$[\begin{array}{c} - 3 \\ - 4 \\ 2 \end{array}]$

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 $1$ .

$[\begin{array}{c} - \frac{1}{3} R_{1} \\ - 4 \\ 2 \end{array}]$

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

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

$[\begin{array}{c} (- \frac{1}{3}) \cdot (- 3) \\ - 4 \\ 2 \end{array}]$

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

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

$[\begin{array}{c} 1 \\ - 4 \\ 2 \end{array}]$

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 $0$ .

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

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

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

$[\begin{array}{c} 1 \\ (4) \cdot (1) - 4 \\ 2 \end{array}]$

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

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

$[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}]$

Perform the row operation $R_{3} = - 2 \cdot R_{1} + R_{3}$ 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 $R_{3} = - 2 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 0 \\ - 2 \cdot R_{1} + R_{3} \end{array}]$

$R_{3} = - 2 \cdot R_{1} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 2 \cdot R_{1} + R_{3}$ .

$[\begin{array}{c} 1 \\ 0 \\ (- 2) \cdot (1) + 2 \end{array}]$

$R_{3} = - 2 \cdot R_{1} + R_{3}$

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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