if x\(\neq\)y\(\neq\)z and \(\begin{vmatrix} x&x^2&1+x^3 \\ y&y^2&1+y^3 \\ z&z^2&1+z^3 \end{vmatrix}=0\), then the value of xyz is_________
- 2
- 4
- -3
- -1
The Correct Option is D
Solution and Explanation
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Concepts Used:
Determinants
Definition of Determinant
A determinant can be defined in many ways for a square matrix.
The first and most simple way is to formulate the determinant by taking into account the top-row elements and the corresponding minors. Take the first element of the top row and multiply it by its minor, then subtract the product of the second element and its minor. Continue to alternately add and subtract the product of each element of the top row with its respective min or until all the elements of the top row have been considered.
For example let us consider a 1×1 matrix A.
A=[a1…….an]
Read More: Properties of Determinants
Second Method to find the determinant:
The second way to define a determinant is to express in terms of the columns of the matrix by expressing an n x n matrix in terms of the column vectors.
Consider the column vectors of matrix A as A = [ a1, a2, a3, …an] where any element aj is a vector of size x.
Then the determinant of matrix A is defined such that
Det [ a1 + a2 …. baj+cv … ax ] = b det (A) + c det [ a1+ a2 + … v … ax ]
Det [ a1 + a2 …. aj aj+1… ax ] = – det [ a1+ a2 + … aj+1 aj … ax ]
Det (I) = 1
Where the scalars are denoted by b and c, a vector of size x is denoted by v, and the identity matrix of size x is denoted by I.
Read More: Minors and Cofactors
We can infer from these equations that the determinant is a linear function of the columns. Further, we observe that the sign of the determinant can be interchanged by interchanging the position of adjacent columns. The identity matrix of the respective unit scalar is mapped by the alternating multi-linear function of the columns. This function is the determinant of the matrix.