The value of the determinant | (a-a1)^-2 (a-a1)^-1 a1^-1 (a-a2)^-2 (a-a2)^-1 a

The value of the determinant | (a-a1)^-2 (a-a1)^-1 ...

Key Concepts

Rational Functions in Matrices

When the entries of a matrix are given by rational functions, typically involving expressions like powers or reciprocal terms, special care is required in determinant evaluation. Techniques involve algebraic manipulation to combine terms or extract common factors, and sometimes recognizing underlying structures (such as those resembling Cauchy or Vandermonde matrices) that allow the determinant to be expressed in a factorized form that highlights the roles of the variables in the numerator and denominator.

Row Operations and Factorization in Determinant Evaluation

Row operations and factorization are essential techniques in simplifying the computation of determinants. By performing operations such as adding multiples of one row to another, or factoring common terms from rows or columns, one can reduce a complex determinant into a simpler form that reveals inherent patterns, such as product structures or Vandermonde-like factors, thereby making the evaluation more manageable.

Determinants

Determinants are scalar values computed from square matrices that can provide information about the matrix’s invertibility, eigenvalues, and the volume scaling factor of the linear transformation associated with the matrix. Evaluating determinants often involves methods such as expansion by minors, row and column operations, and factoring out common terms, which facilitate the simplification of complex expressions.

Vandermonde Determinant

A Vandermonde determinant is a specific type of determinant that arises from matrices whose entries are powers of distinct variables. It has a well-known closed-form expression given by the product of the differences between these variables. This type of determinant is central in many areas of algebra and number theory, and its structure is often identified by factors that involve the differences of variables, as seen in many determinant evaluation problems.

Transcript

00:01 Hi, we need to find out the value of the determinant here, the determinant.

00:04 So how we can solve this? the first step is in solving the determinant, what we can do.

00:09 We can factor out here, a negative a1 to the power negative 2 from row 1, then a negative a 2 to the power negative 2 from row 2, then a negative a 3 to the power negative 2 from i'm sorry 2 this is from row 2 and this is factor out from row 3 so we get a determinant this coming out to be it will be 1 1 1 this will be a negative a 1 this will be a 1 to the power negative 1 a negative a 1 whole square you get a negative a 2 a 2 to the power negative 1 a 2 whole square next you will get a negative a 3 a 3 to the power negative 1 a negative a 3 whole square so we got here now if you just work on this now we apply this is one this is one apply now r2 tends to r2 negative r1 and r3 tends to r3 negative r1 so we get the it's coming out to but this will become this will be 1 over a negative a 1 square a negative 82 square a negative a 3 that is given as multiply a negative a i square so that means multiply a negative a 1 square times a negative a 2 square times a negative a 3 square determinant this is 1 this is a negative a 1 so will be a 1 to the power negative 1 a 1 whole square this will be 0 this was coming out to be a 1 negative a 2 this will be 0 is coming up to be a 1 negative a 3 let's find out this term here so if you do let's say we have a 2 to the power negative 1 a negative a 2 whole square negative a 1 to the power negative 1 a negative a 1 whole square if you do this so we get it as a negative a 2 whole square over a 2 negative a 1 whole square over a 1 so this will give we get a 1 a square plus a 2 square negative 2 a a a 2 a a negative we get a 2 a 2 a 1 a 1 or we have a 1 a 2 now we just keep on simplifying this so we get this a square negative a 1 a 2 times a 1 negative a 2 over a 1 8 2 so we get this term we can write this here this term and this term is coming up to be next term would be similarly we'll get it as a square negative a 1 a3 times a 1 negative a 3 over a 1 a 3 this we go so we get determinant is coming out to be expand along c1 we get it as we give 1 over 5 a negative a i square will be a 1 negative a 2 a square will be a 1 negative a 2 a square a 1 negative a 2 over a 1 a 2 will be a 1 negative a 3 a square negative a 3 times a 1 negative a 3 over a 1 a 3 as you have got here we'll expand it along c 1 now from here we can factor out a 1 negative 82 from here row 2 factor out a 1 negative a 3 so when you factor out we get it has a 1 negative a 2 times a 1 negative a 3 over pi a negative a i whole square the moment here we have this is 1 this is 1 here we have a square negative a 1 a 2 over a 1 a 2 now this will be a square negative a 1 a 3 over a 1 a 3 so we got here next part is to solve this so we get a 1 negative a 2 times a 1 negative a 3 over pi a negative a i full square let's simplify this so we have solved this here this cancels out but in the lcm a 1 a2 a3 this cancels out and we'll get it as we'll give a 1 negative a 2 times a 1 negative a 3 times a square a 2 negative a 3 over a 1 a 2 a 3 multiply a negative a a whole square so now this is given as we get negative a square then you with a1 negative a 2 a 2 negative a 3 is negative a 1 right that is given as multiplic n a i negative a j over multiplic n a i a 1 a 2 a 3 for this multiplic n a negative a i whole square so if you see that this should be option b, negative a square, multiplicant a, and negative a .j over multiplicant a .i...

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