Problem 113
Question
If \(D=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|\) for \(x \neq 0, y \neq 0\) then \(\mathrm{D}\) is (A) divisible by neither \(x\) not \(y\) (B) divisible by both \(\bar{x}\) and \(y\) (C) divisible by \(x\) but not \(y\) (D) divisible by \(y\) but \(\operatorname{not} x\)
Step-by-Step Solution
Verified Answer
Option (C) is correct: divisible by \(x\) but not \(y\).
1Step 1: Identify the Determinant
The given determinant is \( D = \begin{vmatrix} 1 & 1 & 1 \ 1 & 1+x & 1 \ 1 & 1 & 1+y \end{vmatrix} \). Our task is to evaluate this determinant to determine its divisibility by \( x \) and \( y \).
2Step 2: Expand the Determinant
To expand the determinant, we use the standard method of cofactor expansion along the first row. The determinant can be written as: \[ D = 1(1(1+y) - 1(1)) - 1(1(1) - 1(1+y)) + 1(1(1) - 1(1+x)) \].
3Step 3: Simplify the Expression
Simplifying each minor, we get: \( 1(1+y - 1) = y \), \( -1(1 - (1+y)) = y \), and \( 1((1) - (1+x)) = -x \). Therefore, \( D = y - y - x \).
4Step 4: Final Expression
Combining the terms, the final expression for the determinant \( D \) is \( -x \).
5Step 5: Determine Divisibility
Since \( D = -x \), it is clear that the determinant is divisible by \( x \) but not by \( y \). This means that option (C) "divisible by \(x\) but not \(y\)" is the correct choice.
Key Concepts
Matrix AlgebraCofactor ExpansionMathematical Divisibility
Matrix Algebra
Matrix algebra is a fundamental part of linear algebra which is used to perform operations on matrices. In the realm of matrix algebra, you can add, subtract, multiply matrices, and even find their determinants and inverses. Each operation follows specific rules and properties that ensure mathematical consistency.
When dealing with matrices, the determinant is a special value that can be calculated from a square matrix. It plays a vital role in various matrix operations. A determinant gives users information about the matrix, such as whether it is invertible and how it behaves under certain transformations.
For example, if the determinant of a matrix is zero, the matrix is singular and does not have an inverse. In problems like the one in our original exercise, understanding properties of determinants is key to solving them, particularly when calculating whether a matrix's determinant shows divisibility by certain variables.
When dealing with matrices, the determinant is a special value that can be calculated from a square matrix. It plays a vital role in various matrix operations. A determinant gives users information about the matrix, such as whether it is invertible and how it behaves under certain transformations.
For example, if the determinant of a matrix is zero, the matrix is singular and does not have an inverse. In problems like the one in our original exercise, understanding properties of determinants is key to solving them, particularly when calculating whether a matrix's determinant shows divisibility by certain variables.
Cofactor Expansion
Cofactor expansion, also known as Laplace expansion, is a powerful technique for calculating the determinant of a matrix. This method allows you to break down a complex determinant calculation into simpler parts that are easier to manage.
In cofactor expansion, you choose a specific row or column of the matrix to expand along. The determinant is then expressed as a sum of products; each product consists of an element from the chosen row or column and its corresponding minor (the determinant of a smaller matrix obtained by removing the row and column of that element) multiplied by a cofactor. The cofactor includes a sign that alternates based on the position within the matrix.
For our exercise, cofactor expansion was applied along the first row. This process helped simplify the expression by reducing the larger 3x3 determinant into manageable calculations, leading us to the crucial divisibility result for the exercise at hand.
In cofactor expansion, you choose a specific row or column of the matrix to expand along. The determinant is then expressed as a sum of products; each product consists of an element from the chosen row or column and its corresponding minor (the determinant of a smaller matrix obtained by removing the row and column of that element) multiplied by a cofactor. The cofactor includes a sign that alternates based on the position within the matrix.
For our exercise, cofactor expansion was applied along the first row. This process helped simplify the expression by reducing the larger 3x3 determinant into manageable calculations, leading us to the crucial divisibility result for the exercise at hand.
Mathematical Divisibility
Mathematical divisibility is a concept that describes one number being divisible by another without leaving a remainder. It is a cornerstone in number theory and is essential in solving a variety of algebraic problems.
In the context of matrix determinants, determining if a determinant is divisible by a given variable involves simplifying the determinant expression and checking the terms. If the final expression of the determinant includes the variable as a factor, then it is divisible by that variable. If not, it isn't divisible.
In our exercise, after simplifying the determinant expression, we identified it as \(-x\). This result means the expression is divisible by \(x\)—as \(-x\) implies \(x\) is a factor—but not by \(y\), as \(y\) is absent from the simplified expression. Understanding these kinds of divisibility concepts allows students to tackle similar algebraic challenges and further strengthens their foundation in algebra.
In the context of matrix determinants, determining if a determinant is divisible by a given variable involves simplifying the determinant expression and checking the terms. If the final expression of the determinant includes the variable as a factor, then it is divisible by that variable. If not, it isn't divisible.
In our exercise, after simplifying the determinant expression, we identified it as \(-x\). This result means the expression is divisible by \(x\)—as \(-x\) implies \(x\) is a factor—but not by \(y\), as \(y\) is absent from the simplified expression. Understanding these kinds of divisibility concepts allows students to tackle similar algebraic challenges and further strengthens their foundation in algebra.
Other exercises in this chapter
Problem 111
If \(1, \omega, \omega^{2}\) are the cube roots of unity, then \(\left|\begin{array}{lll}1 & \omega^{n} & \omega^{2 \pi} \\ \omega^{n} & \omega^{2 n} & 1 \\ \om
View solution Problem 112
If \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\) are in G.P., then the value of the determinant \(\quad\) \(\left|\begin{array}{lll}\log a_{n} & \log a_{n+1} &
View solution Problem 114
Let \(a, b, c\) be any real numbers. Suppose that there are real numbers \(x, y, z\) not all zero such that \(x=c y+b z\), \(y=a z+c x\) and \(z=b x+a y .\) The
View solution Problem 115
Let \(\mathrm{A}\) be a square matrix all of whose entries are integers. Then which one of the following is true? (A) If \(\operatorname{det} \mathrm{A}=\pm 1\)
View solution