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Chapter 6: Problem 4

Compute the inverse matrix. $$ \left[\begin{array}{rr} -1 & 0 \\ 0 & 2 \end{array}\right] $$

Short Answer

Expert verified
The inverse matrix is \[ \begin{pmatrix} -1 & 0 \ 0 & -\frac{1}{2} \end{pmatrix} \].

Step by step solution

01

- Identify the Matrix

Note that you are given the matrix \[ \begin{pmatrix} -1 & 0 \ 0 & 2 \end{pmatrix} \]. This is a 2x2 matrix.
02

- Check for Invertibility

Calculate the determinant of the matrix. If the determinant is non-zero, the matrix is invertible. The determinant of \[ \begin{pmatrix} a & b \ c & d \end{pmatrix} \] is given by \[ ad - bc \]. Substitute the values: \[ (-1 \times 2) - (0 \times 0) = -2 \]. Since the determinant is -2, which is not zero, the matrix is invertible.
03

- Apply the Inverse Formula for a 2x2 Matrix

The inverse of a 2x2 matrix \[ \begin{pmatrix} a & b \ c & d \end{pmatrix} \] is given by \[ \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \].
04

- Substitute the Values

Using the matrix \[ \begin{pmatrix} -1 & 0 \ 0 & 2 \end{pmatrix} \] and its determinant (-2), the inverse matrix is given by \[ \frac{1}{-2} \begin{pmatrix} 2 & 0 \ 0 & -1 \end{pmatrix} \].
05

- Simplify the Expression

Multiply each element of the matrix by the scalar \[ \frac{1}{-2} \]. This results in \[ \begin{pmatrix} \frac{2}{-2} & \frac{0}{-2} \ \frac{0}{-2} & \frac{-1}{-2} \end{pmatrix} = \begin{pmatrix} -1 & 0 \ 0 & -\frac{1}{2} \end{pmatrix} \].

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Key Concepts

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

2x2 Matrix
A 2x2 matrix is a simple form of a matrix consisting of two rows and two columns. The elements of the matrix are arranged in a rectangular array. For example, the matrix from our exercise is written as: $$ \begin{pmatrix} -1 & 0 \ 0 & 2 \end{pmatrix} $$This format is structured as follows:
  • The first number is located in the first row and first column.
  • The second number is located in the first row and second column.
  • The third number is located in the second row and first column.
  • The fourth number is located in the second row and second column.
Understanding this basic structure is the foundation for more advanced matrix operations.
Matrix Determinant
The determinant is a special number that can be calculated from a square matrix. It's crucial for determining whether a matrix is invertible. For a 2x2 matrix, the determinant can be calculated using the formula:\[ \text{Det}(A) = ad - bc \]where the matrix A is:\[ A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \]In our example, the determinant is calculated by:\[ (-1 \times 2) - (0 \times 0) = -2 \]The result is -2. This non-zero value indicates the matrix is invertible. If the determinant were zero, the matrix would not have an inverse.
Invertible Matrix
An invertible matrix (also called a non-singular or non-degenerate matrix) is a square matrix that has an inverse. For a matrix to be invertible, its determinant must be non-zero. In the given exercise, the determinant is -2, hence the matrix is invertible. The main property of an invertible matrix A is that it has an inverse matrix A^{-1} such that:\[ A \times A^{-1} = I \]where I is the identity matrix, which acts like the number 1 in matrix multiplication.The identity matrix for a 2x2 matrix looks like this:\[ I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \]
Inverse Matrix Formula
Finding the inverse of a 2x2 matrix involves a specific formula. For a matrix \[ \begin{pmatrix} a & b \ c & d \end{pmatrix} \]the inverse is calculated as follows:\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]Given our matrix \[ \begin{pmatrix} -1 & 0 \ 0 & 2 \end{pmatrix} \]and its determinant of -2, the inverse is:\[ \frac{1}{-2} \begin{pmatrix} 2 & 0 \ 0 & -1 \end{pmatrix} \]After simplifying, we get\[ \begin{pmatrix} -1 & 0 \ 0 & -\frac{1}{2} \end{pmatrix} \]So, the inverse matrix is\[ \begin{pmatrix} -1 & 0 \ 0 & -\frac{1}{2} \end{pmatrix} \]. This concludes the calculation of the inverse matrix using this formula.

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Most popular questions from this chapter

Dinophilus gyrociliatus is a small species that lives in the fouling community of harbor environments. On average, a female has approximately 30 eggs during her first 6 wk of life. If she survives her first \(6 \mathrm{wk}\), she has on average 15 eggs her second 6 wk of life. Furthermore, approximately \(80 \%\) of the females survive their first \(6 \mathrm{wk}\) and none survive beyond the second \(6 \mathrm{wk} .{ }^{+}\) Assume half the eggs are female and for simplicity, assume that all the eggs are hatched at once at the beginning of each 6-wk period. Ignore the male population and make the two groups females under 6 wk old and females over 6 wk old. a) Draw and label the Leslie diagram. b) Find the Leslie matrix. c) Twenty hatchlings are introduced into an area. Estimate the population of the two groups after 6 wk. d) Estimate the population of the two groups after 12 wk.

Solve using Gaussian elimination. \begin{array}{rr} x-2 y+3 w= & 5 \\ 2 x+3 y-z-2 w= & -7 \\ y-3 z+4 w= & 21 \\ x-2 y+5 z-w= & -16 \end{array}

Assume that \(a>0\) and \(b>0\) and consider the characteristic equation \(r^{2}-a r-b=0\) with roots \(r_{1}\) and \(r_{2}\). Assume that \(a+b>1\). a) Show that \(r_{1}>1\). b) Show that if \(x_{n}=c_{1} r_{1}^{n}+c_{2} r_{2}^{n}\) is a solution of \(x_{n+1}=a x_{n}+b x_{n-1}\), then \(\lim _{n \rightarrow \infty} x_{n}=\infty\)

Find all the eigenvalues and the corresponding eigenvectors for the following matrices. $$ \left[\begin{array}{rrr} 1 & 0 & -1 \\ -2 & -2 & -6 \\ 2 & 0 & 4 \end{array}\right] $$

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 \(\mathrm{B}^{3} \mathrm{v}\).
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