Linear Algebra Examples

$A = [\begin{array}{c} 8 \\ 1 \\ - 2 \end{array}]$ , $x = [\begin{array}{c} 4 \\ - 1 \\ 1.5 \end{array}]$

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.

$0 = - 8 + 4$

$1 = - 1$

$- 2 = 1.5$

Add $- 8$ and $4$ .

$0 = - 4$

$1 = - 1$

$- 2 = 1.5$

$0 = - 4$

$1 = - 1$

$- 2 = 1.5$

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

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

$0 = - 4$

$0 = - 1 - 1$

$- 2 = 1.5$

Subtract $1$ from $- 1$ .

$0 = - 4$

$0 = - 2$

$- 2 = 1.5$

$0 = - 4$

$0 = - 2$

$- 2 = 1.5$

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.

$0 = - 4$

$0 = - 2$

$0 = 2 + 1.5$

Add $2$ and $1.5$ .

$0 = - 4$

$0 = - 2$

$0 = 3.5$

$0 = - 4$

$0 = - 2$

$0 = 3.5$

Write the system of equations in matrix form.

$[\begin{array}{c} - 4 \\ - 2 \\ 3.5 \end{array}]$

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{1} = - \frac{1}{4} R_{1}$ 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 $R_{1} = - \frac{1}{4} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} - \frac{1}{4} R_{1} \\ - 2 \\ 3.5 \end{array}]$

$R_{1} = - \frac{1}{4} R_{1}$

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

$[\begin{array}{c} (- \frac{1}{4}) \cdot (- 4) \\ - 2 \\ 3.5 \end{array}]$

$R_{1} = - \frac{1}{4} R_{1}$

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

$[\begin{array}{c} 1 \\ - 2 \\ 3.5 \end{array}]$

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

$[\begin{array}{c} 1 \\ 2 \cdot R_{1} + R_{2} \\ 3.5 \end{array}]$

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

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

$[\begin{array}{c} 1 \\ (2) \cdot (1) - 2 \\ 3.5 \end{array}]$

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

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

$[\begin{array}{c} 1 \\ 0 \\ 3.5 \end{array}]$

Perform the row operation $R_{3} = - 3.5 \cdot R_{1} + R_{3}$ 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 $R_{3} = - 3.5 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 0 \\ - 3.5 \cdot R_{1} + R_{3} \end{array}]$

$R_{3} = - 3.5 \cdot R_{1} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 3.5 \cdot R_{1} + R_{3}$ .

$[\begin{array}{c} 1 \\ 0 \\ (- 3.5) \cdot (1) + 3.5 \end{array}]$

$R_{3} = - 3.5 \cdot R_{1} + R_{3}$

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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