Problem 2
Question
Evaluate the determinants $$ \left|\begin{array}{ccc} 1 & 1 & 1 \\ x & a & b \\ x^{2} & a^{2} & b^{2} \end{array}\right|,\left|\begin{array}{ccc} x & a & b \\ x^{2} & a^{2} & b^{2} \\ x^{3} & a^{3} & b^{3} \end{array}\right| $$ and factorize both answers.
Step-by-Step Solution
Verified Answer
First determinant evaluates to \((x-a)(x-b)(a-b)\). Second determinant evaluates to \((x-a)(x-b)(a-b)^2\).
1Step 1: Evaluate the first determinant
\(\begin{vmatrix} 1 & 1 & 1 \\ x & a & b \\ x^2 & a^2 & b^2 \end{vmatrix}\)
This is a Vandermonde determinant. Subtracting column 1 from columns 2 and 3, then expanding, we get:
\(= (a-x)(b^2-x^2) - (b-x)(a^2-x^2)\)
\(= (a-x)(b-x)(b+x) - (b-x)(a-x)(a+x)\)
\(= (a-x)(b-x)[(b+x)-(a+x)]\)
\(= (a-x)(b-x)(b-a) = (x-a)(x-b)(a-b)\)
This is a Vandermonde determinant. Subtracting column 1 from columns 2 and 3, then expanding, we get:
\(= (a-x)(b^2-x^2) - (b-x)(a^2-x^2)\)
\(= (a-x)(b-x)(b+x) - (b-x)(a-x)(a+x)\)
\(= (a-x)(b-x)[(b+x)-(a+x)]\)
\(= (a-x)(b-x)(b-a) = (x-a)(x-b)(a-b)\)
2Step 2: Evaluate the second determinant
\(\begin{vmatrix} x & a & b \\ x^2 & a^2 & b^2 \\ x^3 & a^3 & b^3 \end{vmatrix}\)
Factor \(x\) from column 1, \(a\) from column 2, \(b\) from column 3:
\(= xab \begin{vmatrix} 1 & 1 & 1 \\ x & a & b \\ x^2 & a^2 & b^2 \end{vmatrix} = xab(x-a)(x-b)(a-b)\)
Factor \(x\) from column 1, \(a\) from column 2, \(b\) from column 3:
\(= xab \begin{vmatrix} 1 & 1 & 1 \\ x & a & b \\ x^2 & a^2 & b^2 \end{vmatrix} = xab(x-a)(x-b)(a-b)\)
3Step 3: State the factored results
First determinant: \((x-a)(x-b)(a-b)\)
Second determinant: \(xab(x-a)(x-b)(a-b)\)
Second determinant: \(xab(x-a)(x-b)(a-b)\)
Key Concepts
Matrix Factorization3x3 MatricesAlgebraic Expressions
Matrix Factorization
Matrix factorization is more than just finding determinants. It's about expressing a matrix as a product of simpler matrices. This concept is crucial in simplifying complex algebraic problems and making computations more efficient. In the context of evaluating determinants, factorization helps us express complicated determinant expressions in their simplest forms, revealing unique properties or symmetries in matrices.
Factors of matrices can also simplify solving linear equations, computing matrix inverses, and understanding geometric transformations expressed by matrices. By breaking them down into products of matrix factors, each factor retains individual properties aiding computation and understanding.
Matrices can be factorized through:
- LU Decomposition: Factorizing a matrix into lower and upper triangular matrices.
- Cholesky Decomposition: Specific to positive definite matrices, breaking them into a product involving a lower triangular matrix and its transpose.
- QR Decomposition: Useful for orthogonal or orthonormal square matrices, breaking them into orthogonal and upper triangular matrices.
3x3 Matrices
3x3 matrices are square matrices consisting of three rows and three columns. They are common in linear algebra and serve as blocks for representing transformations, systems of equations, and much more. Evaluating determinants of 3x3 matrices provides insight into properties like invertibility.To evaluate the determinant of a 3x3 matrix, one common approach is to use the rule of Sarrus or the cofactor expansion method. This involves mathematical operations on the entries of the matrix, leading to a single value that signifies specific matrix properties.A 3x3 matrix is generally represented as:\[\begin{bmatrix}a_{11} & a_{12} & a_{13} \a_{21} & a_{22} & a_{23} \a_{31} & a_{32} & a_{33}\end{bmatrix}\]Where the determinant \( \text{det}(A) \) is computed via:\[a_{11} \left(a_{22}a_{33} - a_{23}a_{32}\right) - a_{12} \left(a_{21}a_{33} - a_{23}a_{31}\right) + a_{13} \left(a_{21}a_{32} - a_{22}a_{31}\right)\]This process reveals if the matrix can be inverted, with a non-zero determinant confirming invertibility. In our exercise, determinants of 3x3 matrices are evaluated and factorized into simpler algebraic expressions.
Algebraic Expressions
Algebraic expressions are combinations of constants, variables, and operations. They form the foundation of solving determinant problems. When evaluating determinants of matrices, algebraic expressions become essential in executing and simplifying calculations, revealing intrinsic properties of the matrices.
For matrix factorization and determinant evaluation, manipulating algebraic expressions requires techniques of factorization, grouping, and expansion. Often, determinants are reduced to polynomial expressions, which can be simplified by identifying common factors across terms.
In our exercise, after computing the determinants of the given matrices, the results are complex algebraic expressions. Factorizing these expressions means breaking them down into simpler multiplicative forms, typically linear polynomials that represent a "cleaner" algebraic expression. It uncovers operational efficiencies and verifies potential computational errors.
Understanding algebraic expressions allows for:
- Efficient manipulation of equations
- Recognition of patterns within complexities
- Reduction of complex calculations into basic arithmetic
- Enhanced capability to verify or simplify results
Other exercises in this chapter
Problem 1
Let \(\theta\) be the acute angle formed by two diagonals of a cube. Show that \(\cos \theta=1 / 3\), and hence find \(\theta\)
View solution Problem 2
Show that the distance between the point \((-3,0,1)\) and the line given by \((1,0,2)+t(1,1,2)\), where \(t \in \mathbb{R}\), is \(7 / \sqrt{3}\)
View solution Problem 2
Prove the following vector identities: $$ \begin{aligned} &\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+\mathbf{b} \times(\mathbf{c} \times \mathbf{a})+\math
View solution Problem 2
Use the vector product to find the area of the triangle with vertices \((1,2,0)\), \((2,5,2)\) and \((4,-1,2)\).
View solution