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Chapter 6: Problem 3

Compute the inverse matrix. $$ \left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] $$

Short Answer

Expert verified
The inverse matrix is \(\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}\).

Step by step solution

01

- Write Down the Given Matrix

The matrix provided is: \[\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}\]
02

- Identify the Determinant of the Matrix

Calculate the determinant of the matrix. For a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is given by \(AD - BC \). Substituting the values from the matrix, we get: Det = (0 \cdot 0) - (1 \cdot 1) = -1
03

- Create the Adjugate Matrix

The adjugate matrix for a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix} \) is obtained by swapping the elements on the main diagonal and changing the signs of the off-diagonal elements. For the given matrix, the adjugate matrix is:\[\begin{pmatrix} 0 & -1 \ -1 & 0 \end{pmatrix}\]
04

- Calculate the Inverse Matrix

The inverse of a matrix is given by dividing the adjugate matrix by the determinant. Therefore, the inverse matrix is:\(\frac{1} {-1} \cdot\begin{pmatrix} 0 & -1 \ -1 & 0 \end{pmatrix}\) =\[\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}\]

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Key Concepts

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

Determinant
A determinant is a special number that can be calculated from a square matrix. It tells us various properties of the matrix, including whether it has an inverse. For a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is calculated using the formula: Det = ad - bc.

In our given matrix \(\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}\), we substitute the values into the formula:

Det = (0 \cdot 0) - (1 \cdot 1) = -1

The determinant is -1. This non-zero value confirms that our matrix has an inverse.
Adjugate Matrix
The adjugate of a matrix is the transpose of its cofactor matrix. However, for a 2x2 matrix, finding the adjugate is simpler. You just swap the elements of the main diagonal and change the signs of the off-diagonal elements.

For matrix \(\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}\), the adjugate is calculated as follows:

Original matrix: \(\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}\)

Adjugate matrix: \(\begin{pmatrix} 0 & -1 \ -1 & 0 \end{pmatrix}\)
This adjugate matrix will be used in the final step to find the inverse.
Matrix Inversion
Matrix inversion is the process of finding a matrix that, when multiplied by the original matrix, yields the identity matrix. For a 2x2 matrix, the inverse is computed by dividing the adjugate matrix by the determinant.

The formula for the inverse of a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\) is:

Inverse = \(\frac{1}{det(A)} \cdot adj(A)\)

In our case, the determinant is -1 and the adjugate matrix is \(\begin{pmatrix} 0 & -1 \ -1 & 0 \end{pmatrix}\). Therefore,

Inverse matrix = \(\frac{1}{-1} \cdot \begin{pmatrix} 0 & -1 \ -1 & 0 \end{pmatrix}\)

Perform the multiplication:

Inverse matrix = \(\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}\)
2x2 Matrix
A 2x2 matrix is a matrix with two rows and two columns. It is one of the simplest forms of a square matrix and is often used to introduce basic concepts of linear algebra.

In our exercise, we are given a 2x2 matrix:

\(\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}\)

This matrix has the following properties:
  • Two rows and two columns
  • The elements are arranged in a grid format.
  • It can represent transformations in a 2-dimensional space.


Understanding 2x2 matrices helps build a foundation for grasping more complex matrix concepts and operations.

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Most popular questions from this chapter

Assume that \(\left(x_{0}, y_{0}, z_{0}\right)\) and \(\left(x_{1}, y_{1}, z_{1}\right)\) are both solutions to \(a x+b y+c z=d\). a) Show that \(\left(t x_{0}+(1-t) x_{1}, t y_{0}+(1-t) y_{1},\right.\), \(\left.t z_{0}+(1-t) z_{1}\right)\) is also a solution to \(a x+\) \(b y+c z=d\) for any value of \(t\). b) If a system of linear equations in three variables has two solutions, explain why it has an infinite number of solutions. c) Explain why part (b) generalizes to any number of variables.

The matrix \(A\) has eigenvalues \(r_{1}\) and \(r_{2}\) with corresponding eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\), respectively. Compute \(\mathbf{A}^{n} \mathbf{w}\). $$ \begin{array}{l} r_{1}=-2, r_{2}=0, v_{1}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right], v_{2}=\left[\begin{array}{l} 0 \\ 1 \end{array}\right], n=10 \\ w=\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \end{array} $$

Let \(A=\left[\begin{array}{ll}4 & -1 \\ 7 & -9\end{array}\right], B=\left[\begin{array}{rr}3 & 9 \\ 2 & -2\end{array}\right], C=\left[\begin{array}{rr}8 & 3 \\ 0 & -3\end{array}\right]\) \(D=\left[\begin{array}{rrr}5 & -6 & 1 \\ 10 & 3 & -1\end{array}\right], E=\left[\begin{array}{rr}-7 & 4 \\ -3 & 2 \\ 2 & -1\end{array}\right]\) \(F=\left[\begin{array}{rrr}-4 & 2 & 3 \\ 0 & -1 & 2 \\ -7 & -2 & 5\end{array}\right]\), and \(v=\left[\begin{array}{r}2 \\\ -3\end{array}\right]\). Compute \(\mathbf{A}^{3} \mathbf{v}\).

Let \(A=\left[\begin{array}{ll}4 & -1 \\ 7 & -9\end{array}\right], B=\left[\begin{array}{rr}3 & 9 \\ 2 & -2\end{array}\right], C=\left[\begin{array}{rr}8 & 3 \\ 0 & -3\end{array}\right]\) \(D=\left[\begin{array}{rrr}5 & -6 & 1 \\ 10 & 3 & -1\end{array}\right], E=\left[\begin{array}{rr}-7 & 4 \\ -3 & 2 \\ 2 & -1\end{array}\right]\) \(F=\left[\begin{array}{rrr}-4 & 2 & 3 \\ 0 & -1 & 2 \\ -7 & -2 & 5\end{array}\right]\), and \(v=\left[\begin{array}{r}2 \\\ -3\end{array}\right]\). Compute AD.

Multiply the matrix and the vector to determine if the vector is an eigenvector. If so, what is the eigenvalue? $$ \left[\begin{array}{rrr} 5 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \end{array}\right],\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] $$
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