īy this point it is already apparent that the second equation is the same as the first equation,Įxcept for every term being multiplied by 5. Suppose that we wished to solve the system of linear equations defined as The augmented coefficient matrix of the system is Then the system of linear equations is also described by the matrix equation No matter what the solution type ofĪ system of linear equations is, we may always find it by using the same approach to grouping together the coefficients of this system intoĪ particular matrix, which we then manipulate with row operations to find the solution.Ĭonsider a general system of linear equations in the variables □, □, …, □ and the coefficients □ : It is only partway through theĬalculations, when the solution type has been identified, that the method’s approach begins to deviate. Of the solution type, relying on the Gauss–Jordan elimination method for completing the calculations. The method is largely the same and does not require advanced knowledge The method and difficulty for solving a system of equations with an infinite number of solutions are not notablyĭifferent to the situation where there is a unique solution. Generally interested in these types of systems of linear equations. We would say that the system is inconsistent or insoluble. Of linear equations does not have a solution. This is patently ludicrous and is not possible to achieve, meaning that the given system Reading the two equations above, we have to find values of □ and □ that when combined together inĮxactly the same way, somehow give different outputs. Now we have a rather curious pair of equations, which have the same left-hand sides but which have different right-hand sides. We multiply all terms in the second equation by 1 2, giving We can see that the second equation is very similar to the first equation. We consider the system of linear equations It is this type of solution that we will discuss in this explainer.įor completion, we will briefly mention the final possible solution type. We therefore have infinitely many possible solutions. Given that we can pick any value of □ and find a corresponding value of □ from the equation □ = 2 + 3 □, It is also the case that this pair of values solve the original If we picked the example value of □ = 1, then we would find □ = 5Īnd we could check that this pair of values do actually solve the original system. If we were to solve the above equation for □, then we would find □ = 2 + 3 □. □ and □ to take infinitely many values whilst still solving the This shows that, unlike the previous example where we had a unique solution, it is now possible for We lose no information by writing the system of linear equations more concisely as The first and second equations are identical, which means that there is no benefit of writing them both out. Without changing the solution and the system would become In fact, we can multiply each term in the left-hand side by − 1 3 In this instance we can see that the bottom equation is essentially just a copy of the first equation, with every term multiplied by Suppose that we instead have the system of linear equations □ and □ have only one possible value, we would say that the solution is unique. Solving this system of linear equations using any method will give □ = 2 and □ = − 1. Take, for example, the system of linear equations The most familiar case will be a system of linear equations which has a unique solution. When working with a system of linear equations, there are 3 possible categories of solution that are possible to find. In this explainer, we will learn how to find the general solution of a system of linear equations whether it has a unique solution, an infinite number of solutions, or no solution.
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