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On the Guided Practice tab,

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a system of linear equations in two variables

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is generated for you.

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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 coefficients matrix,

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so that the first coefficient in the matrix

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is the coefficient that is multiplied by

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the first variable in the variable matrix...

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...and the second coefficient in the coefficients matrix

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is the coefficient that is multiplied by

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the second 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

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from that equation.

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Click Calculate to express the matrix

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in reduced row echelon form.

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In this form, the type of system can be determined by inspection.

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Select the type of linear system.

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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 on the lowest row

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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 row of the matrix of a consistent, dependent system

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can be translated back into an equation,

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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 on the lowest row

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of the coefficient 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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If the system is consistent and independent,

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drag and drop the constants into the boxes.

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These values make up the ordered pair that is the solution.

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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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... use the dropdown menus to enter variables.

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Then click Check Equations.

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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 Calculate to solve the linear system.

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The virtual manipulative will determine

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what type of system you have entered.

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If the system is consistent and independent,

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the virtual manipulative will also give you

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the constants of the unique solution.

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Click Try Another to try another example.