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Chapter 6: Problem 6

Let \(D_{k}=\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ 2.3^{k} & 16.9^{k} & 26.27^{k} \\ \left(3^{10}-1\right) & 2\left(9^{10}-1\right) & \left(27^{10}-1\right)\end{array}\right|\) then the value of \(\sum_{k=1}^{10} D_{k}\) is (A) \(2(\alpha+\beta+\eta)\) (B) \(\alpha \beta+\alpha \gamma+\beta \gamma\) (C) \(\alpha \beta \gamma\) (D) 0

Short Answer

Expert verified
The sum \( \sum_{k=1}^{10} D_k \) is 0, which corresponds to option (D).

Step by step solution

01

Understanding the Problem

We are given a determinant \( D_k \) which is defined as follows: \[ D_{k}=\begin{vmatrix}\alpha & \beta & \gamma \2.3^{k} & 16.9^{k} & 26.27^{k} \\left(3^{10}-1\right) & 2\left(9^{10}-1\right) & \left(27^{10}-1\right)\end{vmatrix} \]The question asks for the sum of \( D_k \) from \( k=1 \) to \( k=10 \). We need to find if the determinant becomes zero or simplifies to a particular expression for each \( k \).
02

Simplifying the Third Row

Notice the third row elements are powers of 3 and combinations of 1. The row can be rewritten sequentially: \[ (3^{10}-1), 2(9^{10}-1), (27^{10}-1) \]. These are of the form \(a^n-1\), making them potential candidates for cancellation with other rows that involve powers of 3.
03

Comparing Terms and Using Properties of Determinants

Notice the second row: \(3^k = 3^k\), and so forth for 9 and 27's powers. If we divide each element in a respective column by the element of the third row, this step yields zero or significantly simplifies depending on powers and function properties.
04

Calculate the Simplified Determinant

For each \(k\), the determinant simplifies or becomes zero due to column operations or the nature of terms canceling each other out within row operations. Since similar terms and powers appear, this influences zero results as certain rows are scalar multiples.
05

Summing the Determinants

Given \( D_k = 0 \) due to properties and periodicity when \( k \) changes, every integer solution ensures symmetry that zeros the determinant consistently. Hence, for each \( D_k \) across 1-10, the contributing result stays zero, so the sum is zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Properties of Determinants
Determinants are fascinating aspects of linear algebra as they provide insights into matrix properties and operations. Determinants offer several key properties:
  • Scaling Rows: If a matrix row is multiplied by a scalar, the determinant of the matrix is also multiplied by this scalar.

  • Row Swapping: Swapping two rows of a matrix changes the sign of the determinant.

  • Additive Property: If any row of a matrix is expressed as the sum of two instanced rows in another matrix, the determinant of the original matrix is the sum of the determinants of both instances.

  • Zero Row or Column: If a matrix has a zero row or column, its determinant is zero.
These properties are instrumental in simplifying and solving determinant-related problems, like evaluating whether a determinant is nullified under certain conditions. By aligning terms with similar bases or elements, as shown in the exercise, simplifications can lead to zero-valued determinants.
Matrix
A matrix is a rectangular array of numbers, symbols, or expressions organized in rows and columns. In the given exercise, we deal with a 3x3 matrix which is essential in representing and solving systems of equations in linear algebra. Each entry in the matrix is an element, and the arrangement of these elements defines matrix operations:
  • Matrix Addition: Can be performed between matrices of the same dimension by adding corresponding elements.

  • Matrix Multiplication: This process involves a dot product between rows and columns of matrices, generally yielding another matrix.

  • Determinants: Another crucial operation which yields a single number, giving insights about the matrix properties.
Understanding matrices is vital for manipulating determinant problems as determinants are computed from square matrices. In this problem, analyzing and rearranging matrix rows was key to simplifying the determinant.
Linear Algebra
Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and matrices. It provides foundational knowledge for handling systems of linear equations and transformations. Working with matrices is central to linear algebra.
Key concepts in this field include:
  • Vectors: Entities with magnitude and direction, often used to represent data or transformations.

  • Vector Spaces: Collections of vectors that allow linear combinations.

  • Matrices: Tools that facilitate transformations between vector spaces.

  • Eigenvalues and Eigenvectors: Important in determining matrix behaviors, especially in diagonalization and stability analysis.
The exercise involving determinants falls squarely under this realm, showcasing how determinant properties help ascertain solutions in matrices through linear transformations and operations.
Power Functions
Power functions express a variable raised to a particular exponent. These types of functions appear frequently in determinant problems when dealing with elements in matrices as powers. In the provided exercise, third-row operations involve power functions with different bases:
  • Power Laws: Include rules like \[ a^m \times a^n = a^{m+n} \] and \[ (a^m)^n = a^{m \times n} \].

  • Cancellation: Simplification occurs when terms with identical bases appear, such as \[ a^m - a^n = (a^m \times (1 - a^{n-m})) \].

  • Factorization: Can break down power expressions into simpler components for clearer manipulations, as often used in combining terms.
Understanding these properties helps in resolving complex power function cases within matrices, similar to how they helped simplify the third row of the matrix precisely, potentially leading to zero determinants due to cancellations.

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Most popular questions from this chapter

The value of \(\lambda\) for which the equations \(x+y-3=0\), \((1+\lambda) x+(2+\lambda) y-8=0, x-(1+\lambda) y+(2+\lambda)=0\) are consistent is (A) 1 (B) \(5 / 3\) (C) \(-5 / 3\) (D) None of these

Consider the following system of linear equations: \(x_{1}+2 x_{2}+x_{3}=3\) \(2 x_{1}+3 x_{2}+x_{3}=3\) \(3 x_{1}+5 x_{2}+2 x_{3}=1\) The system has (A) exactly 3 solutions (B) a unique solution (C) no solution (D) infinite number of solutions

If maximum and minimum values of the determinant \(\left|\begin{array}{ccc}1+\sin ^{2} x & \cos ^{2} x & \sin 2 x \\ \sin ^{2} x & 1+\cos ^{2} x & \sin 2 x \\ \sin ^{2} x & \cos ^{2} x & 1+\sin 2 x\end{array}\right|\) are \(\alpha\) and \(\beta\), then (A) \(\alpha+\beta^{99}=4\) (B) \(\alpha^{3}-\beta^{17}=26\) (C) \(\left(\alpha^{2 n}-\beta^{2 \pi}\right)\) is always an even integer for \(n \in N\) (D) a triangle can be constructed having its sides as \(\alpha-\beta, \alpha+\beta\) and \(\alpha+3 \beta\)

If \(\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i\end{array}\right|=x+i y\), then (A) \(x=3, y=1\) (B) \(x=1, y=3\) (C) \(x=0, y=3\) (D) \(x=0, y=0\)

If \(x_{1} \neq 0, x_{2} \neq 0, x_{3} \neq 0\), then the determinant \(\left|\begin{array}{ccc}x_{1}+a_{1} b_{1} & a_{1} b_{2} & a_{1} b_{3} \\\ a_{2} b_{1} & x_{2}+a_{2} b_{2} & a_{2} b_{3} \\ a_{3} b_{1} & a_{3} b_{2} & x_{3}+a_{3} b_{3}\end{array}\right|\) is equal to (A) \(x_{1} x_{2} x_{3}\left(1+\frac{a_{1} b_{1}}{x_{1}}+\frac{a_{2} b_{2}}{x_{2}}+\frac{a_{3} b_{3}}{x_{3}}\right)\) (B) \(-x_{1} x_{2} x_{3}\left(1+\frac{a_{1} b_{1}}{x_{1}}+\frac{a_{2} b_{2}}{x_{2}}+\frac{a_{3} b_{3}}{x_{3}}\right)\) (C) \(x_{1} x_{2} x_{3}\left(1-\frac{a_{1} b_{1}}{x_{1}}-\frac{a_{2} b_{2}}{x_{2}}-\frac{a_{3} b_{3}}{x_{3}}\right)\) (D) None of these
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