On the Guided Practice tab, a system of linear equations in three variables is generated for you. Select a type of linear system. Remember, a consistent and independent system will have a unique solution, a consistent and dependent system will have infinitely many solutions, and an inconsistent system will have no solutions. Drag the coefficients, variables, and constants from the equations to the matrices. The variables should be placed in the variable matrix, which is the second matrix in the matrix equation. The first matrix is the coefficient matrix. The coefficients from an equation should be placed within the same row of the coefficient matrix, so that the first coefficient in the matrix is the coefficient that is multiplied by the first variable in the variable matrix... … the second coefficient in the coefficient matrix is the coefficient that is multiplied by the second variable in the variable matrix... … and the third coefficient in the coefficient matrix is the coefficient that is multiplied by the third variable in the variable matrix. The third matrix is the constant matrix. The constants should be placed in the constant matrix in the same row as the coefficients from that equation. Click “Solve” to express the matrix in reduced row echelon form. The resulting matrix of a consistent independent linear system, which has one solution, will have a coefficient matrix with ones along its diagonal from the top-left to the bottom-right and zeros in all other places. Such a matrix is in reduced row echelon form, and the solution can be determined by translating the matrix back into equations. The resulting matrix of a consistent dependent linear system, which has infinitely many solutions, will have all zeros on the lowest row of the coefficient matrix and a zero in the lowest row of the constant matrix. The upper rows of the matrix of a consistent, dependent system can be translated back into equations, which will describe the infinitely many solutions. The resulting matrix of an inconsistent linear system, which has no solutions, will have all zeros on the lowest row of the coefficients matrix, and a non-zero number in the lowest row of the constant matrix. The final row of the matrix of an inconsistent system can be translated back into an equation that will be mathematically false, thereby indicating that the matrix has no solutions. Click “Try Another” to try another example. On the Workspace tab, you can enter your own linear equations. Click on a box to use the on-screen keypad to enter coefficients and constants. Then click “Submit.” Drag the coefficients, variables, and constants from the equations to the matrices. Click “Solve” to express the matrix in reduced row echelon form. Select the type of linear system. Click “Try Another” to try another example.