The system of equations a x+b y+c z+d=0 -b x+a y-d z+c=0 -c x+d y+a z-b=0 -dx-cy+bz+a=0 for real

The system of equations a x+b y+c z+d=0 -b x+a y-d z+c=0 -c...

Key Concepts

Consistency of Linear Systems

In linear algebra, a system of equations is said to be consistent if there is at least one solution that simultaneously satisfies all of the equations. Analysing consistency involves studying the relationships among the equations, often by looking at the ranks of the coefficient matrix and the augmented matrix. If these ranks are equal the system is consistent; otherwise it is inconsistent, meaning no solution exists.

Overdetermined Systems

An overdetermined system has more equations than unknowns. Such systems may appear to have too many constraints but can be consistent if some of the equations are linearly dependent on others. A careful examination of the relationships among the equations, such as checking for hidden symmetries or dependencies, is essential to determine whether a unique solution exists, an infinite set of solutions exists, or no solution exists at all.

Linear Dependence and Dependency Conditions

Linear dependence among equations means one or more equations can be expressed as a combination of the others. This concept is key when determining the actual number of independent conditions imposed on the unknowns. In overdetermined systems, if some equations are dependent, the effective number of constraints is reduced, potentially allowing consistency even when the number of written equations exceeds the number of unknowns.

Trivial and Nontrivial Solutions

In the context of systems of equations, a trivial solution usually refers to a solution where all the unknowns are zero, which is often the only solution for a homogeneous system. However, when constant terms are nonzero, as they are here, the notion of a trivial solution is replaced by whether any solution exists at all (consistency) and whether that solution is unique or there are infinitely many. 'Nontrivial solution' typically refers to a solution other than the all?zero solution in homogeneous systems, but in nonhomogeneous systems the terminology shifts to simply having a unique or infinitely many solutions.

Transcript

00:01 So in the given question we have a system of equations that are given as ax plus by plus cz plus d is equal to 0, minus bx plus ay minus dz plus c is equal to 0, minus cx plus dy plus az plus c is equal to 0, minus cx plus dy plus az plus az plus minus b is equal to 0 and we have minus d x plus minus d x minus c y plus b z plus a is equal to 0 and we are told to find for non real for non zero for non zero values of the coefficients a, b, c and d, a, b, c and d, we are told to determine whether the given system of equations are consistent or consistent and have a trivial solution, consistent and have infinitely many solution, consistent and have non -trivial solution, or if the system is not consistent at all.

01:36 So these are the four options that are given.

01:38 Given in the question.

01:39 So what we should know is that if a system is consistent, there would be a set of values that would satisfy x, y and z for all the four equations, right? so the condition for consistency is that the determinant that is formed by the coefficient of x, y and z which can be the determinant that is formed by this.

02:08 This system of equations which we can write as a b c d minus b a minus d c minus c d d d a minus c and the last row in the last row we have minus d minus c b so this is the determinant that is formed by the given set of equations.

02:49 So this this determinant would be equal to 0 if the system is consistent, if it is consistent the determinant is equal to 0.

03:06 So that is the condition for consistency, consistent.

03:12 So what we are going to do over here is we can simplify this determinant first and see if the condition holds for this matrix.

03:23 Right.

03:24 So what we have over here is let's multiply the row 2 and row 2, and row 3 and row 4 with minus 1.

03:43 Right or we can say we are taking minus 1 as a common factor from these rows that is row 1 2 and 3 so what we would have then is we would have a b c d we took minus 1 outside from the second row so we have a minus 1 over here inside we have b minus a, d, c.

04:22 Next we have the third row from which we are taking a minus 1 as a common factor.

04:32 Then it would be c minus d minus a b, right? and from the third row we are again taking a minus 1 outside.

04:51 So we have d, c minus b minus a, right? so this is the determinant.

05:05 So now we can write in the next step let's take, let's multiply a in the first row, b in the second row, c in the third row and d in the last row...

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