Question

How do you know if a matrix has an infinite solution?

Answer 1

In simple words, when a system is consistent, and the number of variables is more than the number of nonzero rows in the RREF (Reduced Row-Echelon Form) of the matrix, the matrix equation will have infinitely many solutions. There will be infinite solutions if and only if there is at least one solution of the linear equation $AX=0$.

Complete step-by-step solution:
A matrix equation or the system of equations of the form AX = B may have one solution, no solution and infinitely many solutions based on the behavior of free variables in the RREF (reduced row-echelon form) form of a matrix.
In simple words, an infinite solution can be defined as the number of variables is more than the number of non-zero rows in the reduced row echelon form.
Consider an example of reduced row-echelon form for more understanding of infinite number of solution:

1 & 0 & 3 \\\ 0 & 1 & -2 \\\ 0 & 0 & 0 \\\ \end{matrix} \right|\begin{matrix} 4 \\\ 5 \\\ 0 \\\ \end{matrix} \right]$$ It has a solution set $$(4-3z,\text{ }5+2z,\text{ }z)$$ where z can be any real number. In this case $$z$$ is called the parameter. The parameter is usually a letter not representing the original variables for example $$(4-3t,\text{ }5+2t,\text{ t})$$ where $$\text{t}$$ can be any real number. The row of 0's only means that one of the original equations was redundant. The solution set would be exactly the same if it were removed. The following examples show how to get the infinite solution set starting from the reduced row-echelon form of the augmented matrix for the system of equations. $$~\left[ \left. \begin{matrix} 1 & 0 & 0 & -3 \\\ 0 & 1 & 0 & 2 \\\ 0 & 0 & 1 & 1 \\\ 0 & 0 & 0 & 0 \\\ \end{matrix} \right|\begin{matrix} 4 \\\ 1 \\\ 2 \\\ 0 \\\ \end{matrix} \right]$$ The system is consistent since there are no inconsistent rows It has 4 variables and only 3 nonzero rows so there will be one parameter. Write each row as an equation in the original variables and solve for the variables in terms of the parameter. Row 1: $$x-3w=4$$ Row2: $$y+2w=1$$ Row3: $$z+w=2$$ Because of the echelon form, the most convenient parameter is w. Solving each row equation in terms of w we have: $$x=4+3w$$ $$y=1-2w$$ $$z=2-w$$ $$w$$ Can be any real number Last of all put this into an ordered 4-tuple, $$~\left( 4+3w,\text{ }1-2w,\text{ }2-w,\text{ }w \right)$$ where,$$w$$ can be any real number. So, in this way we have calculate infinite solutions of a matrix **Note:** To know about the infinite solution of a matrix first we have to check nonzero rows in the matrix. That means if the number of variables is more than nonzero rows then that matrix has an infinite solution. Don’t make silly mistakes while calculating the infinite solution set and don’t confuse yourself between parameters which have any real number and original variable.