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.