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Find the determinants, assuming that \(\left|\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right|=4\) $$\left|\begin{array}{ccc}2 a & b / 3 & -c \\\2 d & e / 3 & -f \\\2 g & h / 3 & -i\end{array}\right|$$

Short Answer

Expert verified
\(-\frac{8}{3}\)

Step by step solution

01

Understand the Property of Determinants

The determinant of a matrix will change based on the following rules. If each element of a single row (or column) of a determinant is multiplied by a constant, the value of the determinant is multiplied by that constant.
02

Identify the Modifications to the Original Matrix

In the given matrix, each element in the first column is multiplied by 2. Elements in the second column are divided by 3, and elements in the third column are multiplied by -1.
03

Compute the Effects on Each Column

From Step 2, the effect of the changes can be calculated as follows:- The first column's change contributes a factor of 2.- The second column's change contributes a factor of \( \frac{1}{3} \).- The third column's change contributes a factor of \( -1 \).
04

Calculate the Overall Determinant Effect

Multiply the constant factors together: \[ 2 \times \frac{1}{3} \times (-1) = -\frac{2}{3} \]
05

Apply the Inferred Changes to the Original Determinant

The determinant of the new matrix is then \[ -\frac{2}{3} \times 4 = -\frac{8}{3} \]

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Key Concepts

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

Matrix Transformations
Matrix transformations involve altering a matrix's structure or elements through specific operations. These operations can include scaling the matrix's elements by a constant, swapping rows or columns, and even multiplying by another matrix. When you change the elements of a matrix, you essentially apply a transformation.

In our exercise, multiple transformations were applied to the original matrix. Each item in the first column was multiplied by 2, effectively scaling it. The second column's elements were divided by 3, reducing their magnitude, and this is a form of linear scaling too. Lastly, the third column had its elements multiplied by -1, which switches their signs.
  • Scaling: Multiplying or dividing a whole row or column by a number.
  • Negating: Changing the sign of elements in a row or column.
  • Swapping: Exchanging positions of rows or columns — not used in this exercise, but useful to note.
Understanding these basic transformations can help in determining how they affect the structure and properties of the matrix, including the determinant.
Properties of Determinants
The determinant is a special number associated with a square matrix. It provides insights into the matrix's properties such as whether the matrix is invertible and its scaling factor for volume transformations.

Several key properties pertain to how determinants react to transformations:
  • If you multiply a row or column by a constant, the determinant is also multiplied by that constant.
  • If you switch two rows, the determinant's sign changes (it is multiplied by -1).
  • The determinant is zero if two rows or columns are identical or if a row or column is all zeros.
In our exercise, each column's transformation in the matrix is a perfect example of these properties:
  • The first column's elements are all scaled by 2, thus multiplying the determinant by 2.
  • The second column's elements are divided by 3, influencing the determinant by a factor of \( \frac{1}{3} \).
  • The third column's elements are multiplied by -1, flipping the sign of the determinant.
These rules highlight how powerful determinant properties are in simplifying calculations and understanding matrix behavior.
Linear Algebra Calculations
Linear algebra deals with calculating aspects of matrices and linear transformations. Such calculations are fundamental to numerous fields, including physics, engineering, computer science, and economics.

When working with determinants in linear algebra, calculations often require understanding how transformations affect the matrix. Here's a breakdown of what we calculate in our particular problem:
  • Step 1: Knowing the original determinant value helps start the calculation.
  • Step 2: Understanding the effects of the scaling factors: each column was altered by multiplying it by a constant. The total effect is the product of all these changes.
  • Step 3: Apply these combined effects to the original determinant value.
Therefore, following these modifications, we arrive at the overall determinant of the new matrix, which is \( \left(-\frac{2}{3} \times 4 = -\frac{8}{3}\right) \).

This demonstrates the sequential thought process of dealing with matrix transformations using linear algebra. It showcases how methodical steps lead to precise results.

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Most popular questions from this chapter

Consider the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\). (a) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 1\end{array}\right]\) (b) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 0\end{array}\right]\) (c) Using eigenvalues and eigenvectors, classify the origin as an attractor, repeller, saddle point, or none of these. (d) Sketch several typical trajectories of the system. $$A=\left[\begin{array}{rr} 0.2 & 0.4 \\ -0.2 & 0.8 \end{array}\right]$$

Species \(X\) preys on species \(Y .\) The sizes of the populations are represented by \(x=x(t)\) and \(y=y(t) .\) The growth rate of each population is governed by the system of differential equations \(\mathbf{x}^{\prime}=A \mathbf{x}+\mathbf{b},\) where \(\mathbf{x}=\left[\begin{array}{l}x \\\ y\end{array}\right]\) and \(\mathbf{b}\) is a constant vector. Determine what happens to the two populations for the given \(A\) and b and initial conditions \(\mathbf{x}(0) .\) (First show that there are constants a and \(b\) such that the substitutions \(x=u+\) a and \(y=v+b\) convert the system into an equivalent one with no constant terms.) $$A=\left[\begin{array}{rr} -1 & 1 \\ -1 & -1 \end{array}\right], \mathbf{b}=\left[\begin{array}{r} 0 \\ 40 \end{array}\right], \mathbf{x}(0)=\left[\begin{array}{l} 10 \\ 30 \end{array}\right]$$

(a) Use mathematical induction to prove that, for \(n \geq 2,\) the companion matrix \(C(p)\) of \(p(x)=x^{n}+\) \(a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\) has characteristic polynomial \((-1)^{n} p(\lambda) .\) [Hint: Expand by cofactors along the last column. You may find it helpful to introduce the polynomial \(q(x)=\left(p(x)-a_{0}\right) / x .\) (b) Show that if \(\lambda\) is an eigenvalue of the companion matrix \(C(p)\) in Equation \((4),\) then an eigenvector corresponding to \(\lambda\) is given by $$\left[\begin{array}{c} \lambda^{n-1} \\ \lambda^{n-2} \\ \vdots \\ \lambda \\ 1 \end{array}\right]$$ If \(p(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\) and \(A\) is a square matrix, we can define a square matrix \(p(A) b y\) $$p(A)=A^{n}+a_{n-1} A^{n-1}+\cdots+a_{1} A+a_{0} I$$ An important theorem in advanced linear algebra says that if \(c_{A}(\lambda)\) is the characteristic polynomial of the matrix \(A\), then \(c_{A}(A)=O\) (in words, every matrix satisfies its characteristic equation). This is the celebrated Cayley-Hamilton Theorem, named after Arthur Cayley \((1821-1895)\) pictured below, and sir William Rowan Hamilton (see page 2 ). Cayley proved this theorem in \(1858 .\) Hamilton discovered it, independently, in his work on quaternions, a generalization of the complex numbers.

The given matrix is of the form \(A=\left[\begin{array}{rr}a & -b \\ b & a\end{array}\right]\). In each case, \(A\) can be factored as the product of a scaling matrix and a rotation matrix. Find the scaling factor r and the angle \(\theta\) of rotation. Sketch the first four points of the trajectory for the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) with \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\ 1\end{array}\right]\) and classify the origin as a spiral attractor, spiral repeller, or orbital center. $$A=\left[\begin{array}{cc} 0 & 0.5 \\ -0.5 & 0 \end{array}\right]$$

Find all (real) values of \(k\) for which \(A\) is diagonalizable. $$A=\left[\begin{array}{lll} 1 & k & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{array}\right]$$

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