On the Guided Practice tab,
a system of linear equations in two variables
is generated for you.
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 coefficients matrix,
so that the first coefficient in the matrix
is the coefficient that is multiplied by
the first variable in the variable matrix...
...and the second coefficient in the coefficients matrix
is the coefficient that is multiplied by
the second 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 Calculate to express the matrix
in reduced row echelon form.
In this form, the type of system can be determined by inspection.
Select the type of linear system.
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 row of the matrix of a consistent, dependent system
can be translated back into an equation,
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 coefficient 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.
If the system is consistent and independent,
drag and drop the constants into the boxes.
These values make up the ordered pair that is the solution.
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...
... use the dropdown menus to enter variables.
Then click Check Equations.
Drag the coefficients, variables, and constants
from the equations to the matrices.
Click Calculate to solve the linear system.
The virtual manipulative will determine
what type of system you have entered.
If the system is consistent and independent,
the virtual manipulative will also give you
the constants of the unique solution.
Click Try Another to try another example.