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On the Guided Practice tab,

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a system of linear equations in three variables

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is generated for you.

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Select a type of linear system.

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Remember, a consistent and independent system

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will have a unique solution,

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a consistent and dependent system

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will have infinitely many solutions,

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and an inconsistent system will have no solutions.

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Drag the coefficients, variables, and constants

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from the equations to the matrices.

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The variables should be placed in the variable matrix,

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which is the second matrix in the matrix equation.

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The first matrix is the coefficient matrix.

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The coefficients from an equation should be placed

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within the same row of the coefficient matrix,

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so that the first coefficient in the matrix

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is the coefficient that is multiplied

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by the first variable in the variable matrix...

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…the second coefficient in the coefficient matrix

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is the coefficient that is multiplied

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by the second variable in the variable matrix...

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…and the third coefficient in the coefficient matrix

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is the coefficient that is multiplied by

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the third variable in the variable matrix.

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The third matrix is the constant matrix.

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The constants should be placed in the constant matrix

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in the same row as the coefficients from that equation.

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Click “Solve” to express the matrix

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in reduced row echelon form.

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The resulting matrix of

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a consistent independent linear system,

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which has one solution,

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will have a coefficient matrix with ones along its diagonal

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from the top-left to the bottom-right

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and zeros in all other places.

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Such a matrix is in reduced row echelon form,

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and the solution can be determined

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by translating the matrix back into equations.

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The resulting matrix of

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a consistent dependent linear system,

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which has infinitely many solutions,

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will have all zeros

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on the lowest row of the coefficient matrix

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and a zero in the lowest row of the constant matrix.

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The upper rows of the matrix of a consistent, dependent system

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can be translated back into equations,

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which will describe the infinitely many solutions.

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The resulting matrix of an inconsistent linear system,

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which has no solutions,

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will have all zeros

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on the lowest row of the coefficients matrix,

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and a non-zero number

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in the lowest row of the constant matrix.

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The final row of the matrix of an inconsistent system

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can be translated back into an equation

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that will be mathematically false,

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thereby indicating that the matrix has no solutions.

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Click “Try Another” to try another example.

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On the Workspace tab, you can enter your own linear equations.

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Click on a box to use the on-screen keypad

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to enter coefficients and constants.

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Then click “Submit.”

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Drag the coefficients, variables, and constants

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from the equations to the matrices.

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Click “Solve” to express the matrix

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in reduced row echelon form.

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Select the type of linear system.

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Click “Try Another” to try another example.