How many solutions does a linear system have

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Interactive System of Linear Equations. Case I: 1 Solution. How can we find solutions to systems of equations? To find the solution to systems of linear equations, you can any of the methods below: Solve by Graphing Solve by Elmination Solve by Substitution Interactive System of Linear Equations.

Video on Solutions of Systems of Equations. Now subtract times row 3 from row 1, and then add times row 3 to row 2 to get. The corresponding equations are , , and , which give the unique solution. In Example 1. A matrix is said to be in row-echelon form and will be called a row-echelon matrix if it satisfies the following three conditions:.

A row-echelon matrix is said to be in reduced row-echelon form and will be called a reduced row-echelon matrix if, in addition, it satisfies the following condition:. Each leading is the only nonzero entry in its column.

Entries above and to the right of the leading s are arbitrary, but all entries below and to the left of them are zero.

Hence, a matrix in row-echelon form is in reduced form if, in addition, the entries directly above each leading are all zero. Note that a matrix in row-echelon form can, with a few more row operations, be carried to reduced form use row operations to create zeros above each leading one in succession, beginning from the right.

In fact we can give a step-by-step procedure for actually finding a row-echelon matrix. Observe that while there are many sequences of row operations that will bring a matrix to row-echelon form, the one we use is systematic and is easy to program on a computer.

Note that the algorithm deals with matrices in general, possibly with columns of zeros. Step 2. Otherwise, find the first column from the left containing a nonzero entry call it , and move the row containing that entry to the top position. Step 3. Now multiply the new top row by to create a leading. Step 4. By subtracting multiples of that row from rows below it, make each entry below the leading zero. This completes the first row, and all further row operations are carried out on the remaining rows.

The process stops when either no rows remain at step 5 or the remaining rows consist entirely of zeros. Observe that the gaussian algorithm is recursive: When the first leading has been obtained, the procedure is repeated on the remaining rows of the matrix. This makes the algorithm easy to use on a computer. Note that the solution to Example 1. The reason for this is that it avoids fractions. However, the general pattern is clear: Create the leading s from left to right, using each of them in turn to create zeros below it.

Here is one example. Now subtract times row 1 from row 2, and subtract times row 1 from row 3. In other words, the two have the same solutions.

But this last system clearly has no solution the last equation requires that , and satisfy , and no such numbers exist. Hence the original system has no solution. To solve a linear system, the augmented matrix is carried to reduced row-echelon form, and the variables corresponding to the leading ones are called leading variables. Because the matrix is in reduced form, each leading variable occurs in exactly one equation, so that equation can be solved to give a formula for the leading variable in terms of the nonleading variables.

Every choice of these parameters leads to a solution to the system, and every solution arises in this way. This procedure works in general, and has come to be called. There is a variant of this procedure, wherein the augmented matrix is carried only to row-echelon form.

The nonleading variables are assigned as parameters as before. Then the last equation corresponding to the row-echelon form is used to solve for the last leading variable in terms of the parameters. This last leading variable is then substituted into all the preceding equations. Then, the second last equation yields the second last leading variable, which is also substituted back.

The process continues to give the general solution. This procedure is called back-substitution. This procedure can be shown to be numerically more efficient and so is important when solving very large systems. It can be proven that the reduced row-echelon form of a matrix is uniquely determined by. That is, no matter which series of row operations is used to carry to a reduced row-echelon matrix, the result will always be the same matrix.

By contrast, this is not true for row-echelon matrices: Different series of row operations can carry the same matrix to different row-echelon matrices. Indeed, the matrix can be carried by one row operation to the row-echelon matrix , and then by another row operation to the reduced row-echelon matrix. However, it is true that the number of leading 1s must be the same in each of these row-echelon matrices this will be proved later.

Hence, the number depends only on and not on the way in which is carried to row-echelon form. The reduction of to row-echelon form is. Because this row-echelon matrix has two leading s, rank. Suppose that rank , where is a matrix with rows and columns.

Then because the leading s lie in different rows, and because the leading s lie in different columns. Moreover, the rank has a useful application to equations. Recall that a system of linear equations is called consistent if it has at least one solution. Suppose a system of equations in variables is consistent , and that the rank of the augmented matrix is.

The fact that the rank of the augmented matrix is means there are exactly leading variables, and hence exactly nonleading variables. These nonleading variables are all assigned as parameters in the gaussian algorithm, so the set of solutions involves exactly parameters.

Hence if , there is at least one parameter, and so infinitely many solutions. If , there are no parameters and so a unique solution. Therefore, they have no points in common, and therefore no common solutions. An easy way to check for parallel lines is to check if your two equations have the same slope, but different y intercepts. How many solutions can a system of linear equations have?

Darshan Senthil. Oct 22, With linear systems of equations, there are three possible outcomes in terms of number of solutions: One solution.