Determine the values of a for which the system has no solutions, exactly one solution, or infini

Determine the values of a for which the system has no...

Key Concepts

Consistency Conditions and Rank

The concept of consistency involves whether a system of equations has at least one solution. By comparing the rank of the coefficient matrix with the rank of the augmented matrix (which includes the constant terms), one can determine if the system is consistent. If the ranks are equal and less than the number of variables, the system has infinitely many solutions; if they are unequal, the system is inconsistent, leading to no solution.

Parametric Dependency in Linear Systems

In many problems, parameters appear within the coefficients or constants, affecting the structure of the system. By identifying the values of the parameter at which the determinant of the coefficient matrix vanishes, one can delineate between cases of unique solutions and degenerate cases. In degenerate cases, it is essential to examine additional consistency conditions to decide whether the system has no solutions or infinitely many solutions.

Coefficient Matrix and Determinant

When a system of linear equations is represented in matrix form, the coefficient matrix contains the coefficients of the variables. The determinant of this matrix plays a key role in determining the nature of the solutions: a nonzero determinant indicates that the system is invertible and thus has a unique solution, whereas a zero determinant signals that the matrix is singular and further analysis is needed to determine if the system is consistent or inconsistent.

System of Linear Equations

This concept involves a set of equations that are linear in the unknowns. In studies of such systems, one is interested in finding values for the variables that satisfy all equations simultaneously. The analysis typically involves methods such as substitution, elimination, or using matrices, and leads to situations with exactly one solution, no solution (inconsistency), or infinitely many solutions (dependency).

Transcript

00:01 All right, we are given a system of equations, and we want to find the values of a that make it have no solution, one solution, infinitely many.

00:11 So, again, a good starting place for these is to aim for reduce estuarial inform.

00:18 In this case, like, ultimately, we're really looking at that last row, and so i didn't go all the way to reduce esthert inform.

00:27 And that's okay.

00:28 So you can go as far as you want with that.

00:31 But again, your focus is really going to be that last row of your matrix.

00:36 So first step, change your, you know, take your coefficients and make them into a matrix.

00:43 And then i did some row operations.

00:47 I did negative one times that first row and added it to row three.

00:51 So i got to this point after doing that.

00:57 Again, the different types of solutions really just coming from the last row.

01:04 So that's my focus.

01:08 We'll start with infinitely many solutions.

01:13 In order for that to be the case, we want all of the last row to be zeros.

01:20 The reason for that is that in our final answer, we'd have a free variable, meaning that we have the option for a one, call this z.

01:31 Z can be anything, and so x and y values can change based off of their relationship to z.

01:37 So all that to say, we want negative a squared plus 2 to be equal to zero, and a minus 2 to equal 0.

01:49 So if we solve this over here, we're going to get the a is equal to positive or negative square root of 2.

02:03 And then over here, a would be equal to 2.

02:06 So notice that there's no overlap.

02:08 There's no way that it's equal, the negative a squared plus 2 is equal to 0 and a minus 2 is equal to 0.

02:17 So our answer for this first part, for infinitely many solutions, is that this is actually not possible...

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