(i) Find the values of λfor which the system of equations ( 10-λ -2 -5 -2 2

(i) Find the values of λfor which the system of equations ( ...

Key Concepts

Eigenvalues

Eigenvalues are scalars associated with a matrix that satisfy the equation (A - ?I)x = 0, where A is a given matrix, I is the identity matrix, ? is a scalar, and x is a non-zero vector. The existence of non-trivial solutions (x ? 0) occurs if and only if the determinant of (A - ?I) is zero, making the corresponding scalar an eigenvalue of A.

Eigenvectors

Eigenvectors are the non-zero vectors that satisfy the equation (A - ?I)x = 0 for a particular eigenvalue ?. They represent the directions in which the linear transformation associated with the matrix A acts by merely stretching or compressing, without changing direction.

Characteristic Polynomial

The characteristic polynomial is a polynomial obtained by computing the determinant of (A - ?I). Setting this polynomial equal to zero yields the eigenvalue equation whose roots (the eigenvalues) indicate the values of ? for which non-trivial solutions exist.

Normalization

Normalization involves scaling vectors so that their length (or norm) becomes one. When eigenvectors are normalized, they become unit vectors, facilitating the construction of an orthonormal basis which is often desirable for simplifying many matrix computations.

Orthogonal Matrices

An orthogonal matrix is a square matrix whose columns (and rows) are orthonormal vectors. Such matrices satisfy the property that their transpose is also their inverse. In the context of eigenvector problems, forming a matrix with normalized eigenvectors of a symmetric matrix often results in an orthogonal matrix, which simplifies diagonalization and many other operations in linear algebra.

Transcript

00:01 So in the given question there are three parts that we need to solve right and the first part of the question what we are given is we are told to find the value of lambda a constant lambda from the system of equations that are given as 10 minus lambda minus 2 minus 5 minus 2 minus lambda 3 minus 5 3 5 minus lambda and over here we have times x the matrix x y z is equal to 0 0 .0 so this is the system of equations that have been given and we have to find the value of lambda such that such that the given system has non -trivial solutions as non -trivial solutions.

01:28 So for a system of equations to have a non -trivial solution it is there is a condition that for a system of equation of the form a x equal to 0 for it to have non -trivial solutions the determinant of a should be equal to zero non trivial solution right so let's first take the determinant of a which is this matrix over here right so we can take the determinant of this matrix as we can write it as the determinant of 10 minus lambda minus 2 minus 5 minus 2 minus 2 minus lambda 3 minus 5 3 5 minus 5 minus lambda is equal to when we find the matrix of a find the determinant of this matrix what we get is minus lambda cube plus 17 lambda squared minus 4.

02:59 2 lambda right and this is equal to 0 right since we have to have we have to the system is said to have a non -trivial solutions right so then we can write we can just take lambda as a common factor minus lambda as a common factor and we would have minus lambda times lambda square minus 17 lambda minus minus 2 right equal to 0 and we can further simplify this equation and write it as lambda minus 3 minus lambda times lambda minus 3 times lambda minus 14 is equal to 0 and from this we can write that the values the possible values of lambda are 0 3 and 14 right so for these values of lambda the given system has non -trivial solutions.

04:02 So this is the first part of the question.

04:06 Now next we have the second part of the question and in the second part we are told to find the non -trivial solutions of the of the type x1 is equal to x1 is equal to the matrix a, i, b, i, c -i, c -i, transpose and the solutions of the order x i is equal to a i b i the matrix a i b i the transpose of this matrix where i is what is i so each solutions corresponding to corresponding to lambda equal to lambda i where i is one 2 and 3 right so if lambda equal to lambda 1 we can take the value of lambda as 0 if it is lambda equal to 2 it is 3 if it is lambda equal to lambda 3 it is 14 and that in that way right so what we can do over here is first we can take lambda equal to 0 first right and when lambda equal to 0 what do we have? we have the matrix, the matrix given in the question would turn to 10 minus 2 minus 5, minus 2, 2, 2, 3, minus 5, 3, 5, 5, 5, 5, 5, 5, z is equal to the matrix 0, 0, 0, 0, 0.

06:14 Right and now we can find the solutions of this matrix by taking we can take x divided by this matrix that is the determinant of this matrix that is 2 3 3 5 right next up we have the determinant y divided by the determinant of minus 2 minus 2 minus 5 minus 2 minus 5, 3 5, 3 5.

06:59 And this is equal to z divided by the matrix minus 2 to minus 5 the determinant of this matrix.

07:12 So how did we form this determinants? that's one thing we need to discuss and what we did is it is a method of a final the solutions from this type of a system that is first what we do is when we take x divided by a certain determinant what we will have is we have 10 minus 2 to 3 and the third row is minus 5 3 5 right so when we take x divided by the matrix 2 335 what we are doing is we are teaching this determinant, right? this the matrix, the determinant of this matrix.

08:04 So x divided by this matrix is equal to y divided by this matrix minus 2, 3 minus 5, 5.

08:16 And that is equal to z divided by this matrix which is the determinant of this matrix that is minus 2 to minus 5 to minus 5 .3.

08:28 This is a method of finding the solutions from the given system and this is equal to we can say some constant c.

08:38 So this proportion, this ratios are equal to some constant c.

08:44 And when we take this determinant, determinants what we will have is x by 1 is equal to y by minus 5 is equal to z by 4.

09:00 Right and now since all this are equal to some constant c we can write x is equal to c y equal to minus 5 c and z equal to 4c right so the solution matrix x1 is of the form x y z x y z which is now equal to x is c let's say this constant over here is we can take it as k1 right let's take this constant as k1 so we have c minus 5 c which is minus 5 k1 and 4 k1 instead of c we are just taking it as k1 that is the constant corresponding to the first value of lambda so we will have over here we can take the solution x one as c times the matrix 1 minus 5 4, 1 minus 5 4, right? we replaced instead of c we replaced and placed k1 instead of c.

10:27 We are writing k1 instead of c.

10:30 So this is the required solution, right? but in the question we are told that the solutions the solutions should be written.

10:39 In the form a .i.

10:42 B, i, c, i, the transpose of the matrix a .i.

10:49 B, i, c i, c...

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